Let S be a set of 5 elements and $\mathrm{P}(\mathrm{S})$ denote the power set of S . Let E be an event of choosing an ordered pair (A, B) from the set $\mathrm{P}(\mathrm{S}) \times \mathrm{P}(\mathrm{S})$ such that $\mathrm{A} \cap \mathrm{B}=\emptyset$. If the probability of the event $E$ is $\frac{3^p}{2^q}$, where $p, q \in N$, then $p+q$ is equal to
Explanation:
$ \begin{aligned} & \text { Let } S=\left\{a_1, a_2, a_3, a_4, a_5\right\} \\ & n(S)=5 \\ & P(S)=\text { power set of } S \\ & n P(S)=2^{n(S)}=2^5=32 \\ & n(T)=n[P(S) \times P(S)]=32 \times 32 \quad \text { (Total outcomes) } \end{aligned} $
$E=$ event that choosing ordered pair ( $A, B$ ) from $P(S) \times P(S)$ such that $A \cap B=\phi$
for each element $x \in S, n(S)=5$
there are 4 possibilities such that $A \cap B=\phi$
1.$x \in A$ and $x \notin B$
2.$x \notin A$ and $x \in B$
3.$x \notin A$ and $x \notin B$
4.$x \in A$ and $x \in B$
for $A \cap B=\phi$, exclude the fourth.
so, for each of 5 elements of $S$ there are 3 choices.
The number of favourable outcome $n(E)=3 \times 3 \times 3 \times 3 \times 3$
$ P(E)=\frac{\text { favourable outcome }}{\text { Total outcome }}=\frac{3^5}{32 \times 32}=\frac{3^5}{2^{10}} $
It is given that $P(E)=\frac{3^p}{2^q}$
compare $p=5, q=10 \Longrightarrow p+q=15$ Ans
From the first 100 natural numbers, two numbers first $a$ and then $b$ are selected randomly without replacement. If the probability that $\mathrm{a}-\mathrm{b} \geqslant 10$ is $\frac{\mathrm{m}}{\mathrm{n}}, \operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$, then $\mathrm{m}+\mathrm{n}$ is equal to
$\_\_\_\_$ .
Explanation:
Let $b=1$
$\Rightarrow a>11 \Rightarrow$ a can take 90 values from $\{11,12, \ldots 100\}$
Let $b=2 \Rightarrow a>12 \Rightarrow 8>$ values and so on till $b=99 \Rightarrow$ a can only take $a=100$, only 1 value ⇒ favourable cases $=1+2+3+\ldots .+90$
$ =\frac{\frac{90 \times 91}{2}}{100 \times 99}=\frac{91}{220} \Rightarrow m+n=311 $
A card from a pack of 52 cards is lost. From the remaining 51 cards, n cards are drawn and are found to be spades. If the probability of the lost card to be a spade is $\frac{11}{50}$, then n is equal to ________ .
Explanation:
$P\left(\frac{\text { Lost }_{\text {(spade })}}{\mathrm{n} \text { cards are spade }}\right)$
$\begin{aligned} & =\frac{P\left(\frac{n_s}{L_s}\right) P\left(L_s\right)}{P\left(\frac{n_s}{L_s}\right) P\left(L_s\right)+P\left(\frac{n_s}{\bar{L}_s}\right) P\left(\bar{L}_s\right)} \\ & =\frac{\frac{{ }^{12} C_n}{{ }^{51} C_n} \times \frac{1}{4}}{\frac{{ }^{12} C_n}{{ }^{51} C_n} \times \frac{1}{4}+\frac{3}{4} \times \frac{{ }^{13} C_n}{{ }^{51} C_n}}=\frac{1}{1+3 \cdot \frac{{ }^{13} C_n}{{ }^{12} C_n}}=\frac{13-n}{52-n} \\ & \Rightarrow \frac{13-n}{52-n}=\frac{11}{50} \\ & \Rightarrow n=2 \end{aligned}$
Three distinct numbers are selected randomly from the set $\{1,2,3, \ldots, 40\}$. If the probability, that the selected numbers are in an increasing G.P., is $\frac{m}{n}, \operatorname{gcd}(m, n)=1$, then $m+n$ is equal to __________ .
Explanation:
Total number of ways of selecting $3$ distinct numbers from $\{1,2,\dots,40\}$ is
$\binom{40}{3}=\frac{40\cdot 39\cdot 38}{6}=9880.$
Now, three numbers in increasing G.P. can be written as
$a,\; ar,\; ar^2 \quad (r>1).$
Let $r=\dfrac{p}{q}$ in lowest form, where $p>q$ and $\gcd(p,q)=1$.
Then
$a,\; a\frac{p}{q},\; a\frac{p^2}{q^2}$
must all be integers. So $a$ must be a multiple of $q^2$. Write
$a=kq^2 \quad (k\in\mathbb{N}).$
Then the triple becomes
$kq^2,\; kpq,\; kp^2,$
which is strictly increasing since $p>q$.
Also all must be $\le 40$. Since
$q^2 < pq < p^2,$
it is enough to ensure
$kp^2 \le 40 \;\Rightarrow\; k \le \left\lfloor \frac{40}{p^2}\right\rfloor.$
So for each coprime pair $(p,q)$ with $p>q$, the number of valid triples equals $\left\lfloor \dfrac{40}{p^2}\right\rfloor$.
Now $p^2\le 40 \Rightarrow p\le 6$. Count for each $p$:
$p=2$: $q=1$ (1 value), $\left\lfloor \frac{40}{4}\right\rfloor=10$ $\Rightarrow 10$
$p=3$: $q=1,2$ (2 values), $\left\lfloor \frac{40}{9}\right\rfloor=4$ $\Rightarrow 2\cdot 4=8$
$p=4$: $q=1,3$ (2 values), $\left\lfloor \frac{40}{16}\right\rfloor=2$ $\Rightarrow 2\cdot 2=4$
$p=5$: $q=1,2,3,4$ (4 values), $\left\lfloor \frac{40}{25}\right\rfloor=1$ $\Rightarrow 4\cdot 1=4$
$p=6$: $q=1,5$ (2 values), $\left\lfloor \frac{40}{36}\right\rfloor=1$ $\Rightarrow 2\cdot 1=2$
Hence number of favourable triples:
$10+8+4+4+2=28.$
Therefore required probability is
$\frac{28}{9880}=\frac{7}{2470} \quad (\gcd(7,2470)=1).$
So $m=7,\; n=2470 \Rightarrow m+n=2477.$
Let $\mathrm{a}, \mathrm{b}$ and $\mathrm{c}$ denote the outcome of three independent rolls of a fair tetrahedral die, whose four faces are marked $1,2,3,4$. If the probability that $a x^2+b x+c=0$ has all real roots is $\frac{m}{n}, \operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$, then $\mathrm{m}+\mathrm{n}$ is equal to _________.
Explanation:
A quadratic equation $ax^2 + bx + c = 0$ has real roots if and only if its discriminant is non-negative. The discriminant $\Delta$ of the quadratic equation is given by:
$\Delta = b^2 - 4ac$
For the quadratic equation to have all real roots, the discriminant must be non-negative:
$\Delta \geq 0$
That means:
$b^2 - 4ac \geq 0$
Given that $a, b, c$ are the outcomes of rolling a fair tetrahedral die, they can each be one of the numbers 1, 2, 3, or 4. Our task is to determine the probability that this condition holds.
We need to analyze the cases where $b^2 \geq 4ac$.
Let’s consider all possible values for $a$, $b$, and $c$, and count how many of them satisfy the condition. Since there are 4 choices for each of the variables, there are a total of $4 \times 4 \times 4 = 64$ possible combinations.
Now, we count the valid combinations where $b^2 \geq 4ac$:
- For $a = 1$: $b^2 \geq 4c$
- $b = 1: 1 \geq 4c \rightarrow \text{(Not possible since } c \ \text{must be } \geq 1 \text{ and not zero)}$
- $b = 2: 4 \geq 4c \rightarrow c \leq 1 \rightarrow c = 1$ (1 case)
- $b = 3: 9 \geq 4c \rightarrow c \leq 2 \rightarrow c = 1 \text{ or } 2$ (2 cases)
- $b = 4: 16 \geq 4c \rightarrow c \leq 4 \rightarrow c = 1, 2, 3, 4$ (4 cases)
- For $a = 2$: $b^2 \geq 8c$
- $b = 1: 1 \geq 8c \rightarrow \text{(Not possible)}$
- $b = 2: 4 \geq 8c \rightarrow \text{(Not possible)}$
- $b = 3: 9 \geq 8c \rightarrow c \leq 1$ (1 case)
- $b = 4: 16 \geq 8c \rightarrow c \leq 2$ (2 cases)
- For $a = 3$: $b^2 \geq 12c$
- $b = 1: 1 \geq 12c \rightarrow \text{(Not possible)}$
- $b = 2: 4 \geq 12c \rightarrow \text{(Not possible)}$
- $b = 3: 9 \geq 12c \rightarrow \text{(Not possible)}$
- $b = 4: 16 \geq 12c \rightarrow c \leq 1$ (1 case)
- For $a = 4$: $b^2 \geq 16c$
- $b = 1: 1 \geq 16c \rightarrow \text{(Not possible)}$
- $b = 2: 4 \geq 16c \rightarrow \text{(Not possible)}$
- $b = 3: 9 \geq 16c \rightarrow \text{(Not possible)}$
- $b = 4: 16 \geq 16c \rightarrow c \leq 1$ (1 case)
Total for $a = 1 = 1 + 2 + 4 = 7$
Total for $a = 2 = 1 + 2 = 3$
Total for $a = 3 = 1$
Total for $a = 4 = 1$
Adding up all the valid cases:
$7 + 3 + 1 + 1 = 12$
The total number of valid combinations is 12 out of 64. Thus, the probability is:
$\frac{12}{64} = \frac{3}{16}$
The value of $\mathrm{m} = 3$ and $\mathrm{n} = 16$. The sum $\mathrm{m} + \mathrm{n} = 3 + 16 = 19$.
Hence, the answer is 19.
Three balls are drawn at random from a bag containing 5 blue and 4 yellow balls. Let the random variables $X$ and $Y$ respectively denote the number of blue and yellow balls. If $\bar{X}$ and $\bar{Y}$ are the means of $X$ and $Y$ respectively, then $7 \bar{X}+4 \bar{Y}$ is equal to ___________.
Explanation:
| $X$ | 3 | 2 | 1 | 0 |
|---|---|---|---|---|
| $Y$ | 0 | 1 | 2 | 3 |
$\begin{aligned} & \bar{X}=\sum X p(X) \\ & \bar{Y}=\sum Y p(Y) \\ & P(X=3)=P(Y=0)=\frac{{ }^5 C_3 \cdot C_0}{{ }^9 C_3}=\frac{{ }^5 C_2}{{ }^9 C_3}=\frac{5}{42} \\ & P(X=2)=P(Y=1)=\frac{{ }^5 C_2 \cdot C_1}{{ }^9 C_3}=\frac{10}{21} \\ & P(X=1)=P(Y=2)=\frac{{ }^5 C_1 \cdot C_2}{{ }^9 C_3}=\frac{5}{14} \\ & P(X=0)=P(Y=3)=\frac{{ }^5 C_0 \cdot C_3}{{ }^9 C_3}=\frac{4}{84}=\frac{1}{21} \\ & \bar{X}=3 \times \frac{5}{42}+2 \times \frac{10}{21}+\frac{5}{14}+0 \times \frac{1}{21}=\frac{15+40+15}{42}=\frac{70}{42} \\ & \bar{Y}=0 \times \frac{5}{42}+1 \times \frac{10}{21}+2 \times \frac{5}{14}+3 \times \frac{1}{21}=\frac{20+30+6}{42}=\frac{56}{42} \\ & 7 \bar{X}+4 \bar{Y}=17 \end{aligned}$
From a lot of 12 items containing 3 defectives, a sample of 5 items is drawn at random. Let the random variable $X$ denote the number of defective items in the sample. Let items in the sample be drawn one by one without replacement. If variance of $X$ is $\frac{m}{n}$, where $\operatorname{gcd}(m, n)=1$, then $n-m$ is equal to _________.
Explanation:
Given a lot of 12 items, 3 are defective.
Good items, $12-3=9$
Let $X$ denote the number of defective items.
So, value of $X=0,1,2,3$
A sample of $S$ items is drawn.
$P(X=0)=G G G G G$
(here $G$ is good item and $d$ is defective)
$\begin{aligned} & \frac{9}{12} \cdot \frac{8}{11} \cdot \frac{7}{10} \cdot \frac{6}{9} \cdot \frac{5}{8}=\frac{21}{132}=\frac{7}{44} \\ & P(X=1)=5\left[\frac{9 \cdot 8 \cdot 7 \cdot 6 \cdot 3}{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8}\right]=\frac{21}{44} \\ & P(X=2)=5\left[\frac{9 \cdot 8 \cdot 7 \cdot 3 \cdot 2}{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8}\right]=\frac{14}{44} \\ & P(X=3)=5\left[\frac{3 \cdot 2 \cdot 1 \cdot 9 \cdot 8}{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8}\right]=\frac{2}{44} \\ & P(X=4)=0 \\ & P(X=5)=0 \end{aligned}$
| $X$ | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| $P(X)$ | $ \frac{7}{44} $ |
$ \frac{21}{44} $ |
$ \frac{14}{44} $ |
$ \frac{2}{44} $ |
0 | 0 |
| $XP(X)$ | 0 | $ \frac{21}{44} $ |
$ \frac{28}{44} $ |
$ \frac{6}{44} $ |
0 | 0 |
| $X^2P(X)$ | 0 | $ \frac{21}{44} $ |
$ \frac{56}{44} $ |
$ \frac{18}{44} $ |
0 | 0 |
$\begin{aligned} & \sigma_x^2=\sum X^2 P(x)-\left(\sum x P(x)\right)^2 \\ & =\frac{95}{44}-\left(\frac{55}{44}\right)^2 \\ & =\frac{4180-3025}{1936}=\frac{1155}{1936}=\frac{105}{176}=\frac{m}{n} \\ & =n-m=71 \end{aligned}$
From a lot of 10 items, which include 3 defective items, a sample of 5 items is drawn at random. Let the random variable $X$ denote the number of defective items in the sample. If the variance of $X$ is $\sigma^2$, then $96 \sigma^2$ is equal to __________.
Explanation:
| $x$ | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| $P(x)$ | $ \frac{{ }^7 C_5}{{ }^{10} C_5}=\frac{1}{12} $ |
$ \frac{C_4 \cdot{ }^3 C_1}{{ }^{10} C_5}=\frac{5}{12} $ |
$ \frac{{ }^7 C_3 \cdot{ }^3 C_2}{{ }^{10} C_5}=\frac{5}{12} $ |
$ \frac{{ }^7 C_2 \cdot{ }^3 C_3}{{ }^{10} C_5}=\frac{1}{12} $ |
| $xP(x)$ | 0 | $\frac{5}{12}$ | $\frac{10}{12}$ | $\frac{3}{12}$ |
$\begin{aligned} & \mu=\sum x P(x)=0+\frac{5}{12}+\frac{10}{12}+\frac{3}{12}=\frac{3}{2} \\ & \sigma^2=\sum(x-\mu) P(x)=\sum\left(x-\frac{3}{2}\right)^2 P(x) \\ & =\frac{9}{4} \times \frac{1}{12}+\frac{1}{4} \times \frac{5}{12}+\frac{1}{4} \times \frac{5}{12}+\frac{9}{4} \times \frac{1}{12}=\frac{7}{12} \end{aligned}$
$\Rightarrow \sigma^2 \cdot 96=8 \times 7=56$
In a tournament, a team plays 10 matches with probabilities of winning and losing each match as $\frac{1}{3}$ and $\frac{2}{3}$ respectively. Let $x$ be the number of matches that the team wins, and $y$ be the number of matches that team loses. If the probability $\mathrm{P}(|x-y| \leq 2)$ is $p$, then $3^9 p$ equals _________.
Explanation:
$\begin{aligned} & x+y=10 \\ & A=x-y \\ & P(|A|<2) \text { is } P \\ & \Rightarrow|A|=2,1,0 \Rightarrow A=0,1,-1,2,-2 \\ & \Rightarrow x=\frac{10+A}{2} \Rightarrow A \in \text { even as } x \in \text { integer } \\ & \Rightarrow A=0,-2,2 \\ & \Rightarrow P(|A| \leq 2)=P(A=0)+P(A=-2)+P(A=2) \end{aligned}$
(1) $A=0 \Rightarrow x=5=y$
$P(A=0)={ }^{10} C_5\left(\frac{1}{3}\right)^5\left(\frac{2}{3}\right)^5$
(2) $A=-2$
$\Rightarrow x=4$ and $y=6$
$P(A=-2)={ }^{10} C_4 \cdot\left(\frac{1}{3}\right)^4\left(\frac{2}{3}\right)^6$ and
$\begin{aligned} & \text { Similarly, } P(A=2)={ }^{10} C_6\left(\frac{1}{3}\right)^6\left(\frac{2}{3}\right)^4 \\ & \Rightarrow P(|A| \leq 2) 3^9=3\left({ }^{10} C_5 \cdot 2^5+{ }^{10} C_4 \cdot 2^6+{ }^{10} C_6 \cdot 2^4\right) \\ & =8288 \end{aligned}$
A group of 40 students appeared in an examination of 3 subjects - Mathematics, Physics and Chemistry. It was found that all students passed in atleast one of the subjects, 20 students passed in Mathematics, 25 students passed in Physics, 16 students passed in Chemistry, atmost 11 students passed in both Mathematics and Physics, atmost 15 students passed in both Physics and Chemistry, atmost 15 students passed in both Mathematics and Chemistry. The maximum number of students passed in all the three subjects is _________.
Explanation:
$ \begin{aligned} & n(C \cup P \cup M) \leq n(U)=40 . \\\\ & n(C)+n(P)+n(M)-n(C \cap M)-n(P \cap M)-n(C \cap \\\\ & P)+n(C \cap P \cap M) \leq 40 \\\\ & 20+25+16-11-15-15+x \leq 40 \\\\ & x \leq 20 \end{aligned} $
But $11-x \geq 0$ and $15-x \geq 0$
$ \Rightarrow x \geq 11 $
$a=P(X=3), b=P(X \geqslant 3)$ and $c=P(X \geqslant 6 \mid X>3)$. Then $\frac{b+c}{a}$ is equal to __________.
Explanation:
To solve this problem, we need to compute the probabilities $a$, $b$, and $c$, and then plug those values into the expression $\frac{b+c}{a}$.
Let's begin by defining each of the variables:
- $a = P(X=3)$: This is the probability that the first six appears on the third toss.
- $b = P(X \geqslant 3)$: This is the probability that the first six appears on the third toss or later.
- $c = P(X \geqslant 6 \mid X>3)$: This is the probability that the first six appears on the sixth toss or later, given that it has not appeared in the first three tosses.
Since we're dealing with a fair die, each side has an equal probability of $\frac{1}{6}$ of landing face up. Let's find the probabilities step by step:
Calculating $a$:
The probability of rolling anything other than a six is $\frac{5}{6}$. So for the first six to show up exactly on the third roll, the sequence of rolls must be NN6, where N is anything but a six (i.e., the results of the first two rolls). Thus,
$a = P(X=3) = \left(\frac{5}{6}\right) \cdot \left(\frac{5}{6}\right) \cdot \left(\frac{1}{6}\right)$
Calculating $b$:
For the first six to appear on the third roll or later, we can think of two cases: when the first six appears on the third roll (which we've already calculated, $a$), and when it appears after the third roll. To combine these probabilities, we can use the fact that $P(X \geqslant 3) = 1 - P(X < 3)$, where $P(X < 3)$ is the probability that the first six appears on either the first or the second roll. So we calculate the latter first:
$ P(X < 3) = P(X=1) + P(X=2) $
$ P(X < 3) = \left(\frac{1}{6}\right) + \left(\frac{5}{6}\right) \cdot \left(\frac{1}{6}\right) $
Thus,
$ b = P(X \geqslant 3) = 1 - P(X < 3) = 1 - \left[ \left(\frac{1}{6}\right) + \left(\frac{5}{6}\right) \cdot \left(\frac{1}{6}\right) \right] $
Calculating $c$:
This is the probability that the first six appears on or after the sixth roll, given that it hasn't appeared in the first three rolls. Since $X>3$, the first three outcomes must not be a six, which occurs with probability $\left(\frac{5}{6}\right)^3$. The subsequent outcomes until (and including) the fifth roll also must not be a six. So,
$c = P(X \geqslant 6 \mid X>3) = \left(\frac{5}{6}\right)^2$
Notice here, we did not include the probability of rolling a six, because we are looking for the probability that we have not yet rolled a six after the fifth roll.
Now we can calculate $a$, $b$, and $c$:
$a = \left(\frac{5}{6}\right)^2 \cdot \frac{1}{6}$
$b = 1 - \left[ \left(\frac{1}{6}\right) + \left(\frac{5}{6}\right)\cdot\left(\frac{1}{6}\right) \right]$
$c = \left(\frac{5}{6}\right)^2$
Now we'll substitute to find $\frac{b+c}{a}$:
$\frac{b+c}{a} = \frac{1 - \left[ \left(\frac{1}{6}\right) + \left(\frac{5}{6}\right)\cdot\left(\frac{1}{6}\right) \right] + \left(\frac{5}{6}\right)^2}{\left(\frac{5}{6}\right)^2 \cdot \frac{1}{6}}$
Simplifying the numerator:
$1 - \left[ \left(\frac{1}{6}\right) + \left(\frac{5}{6}\right)\cdot\left(\frac{1}{6}\right) \right] + \left(\frac{5}{6}\right)^2$
$= 1 - \left[\frac{1}{6} + \frac{5}{36}\right] + \frac{25}{36}$
$= 1 - \left[\frac{6}{36} + \frac{5}{36}\right] + \frac{25}{36}$
$= 1 - \frac{11}{36} + \frac{25}{36}$
$= \frac{36}{36} - \frac{11}{36} + \frac{25}{36}$
$= \frac{50}{36}$
Now, substitute this back into the expression and solve:
$\frac{b+c}{a} = \frac{\frac{50}{36}}{\left(\frac{5}{6}\right)^2 \cdot \frac{1}{6}}$
$\frac{b+c}{a} = \frac{50}{36} \cdot \frac{6}{\left(\frac{5}{6}\right)^2}$
$\frac{b+c}{a} = \frac{50 \cdot 6}{25}$
$\frac{b+c}{a} = \frac{300}{25}$
$\frac{b+c}{a} = 12$
Therefore, $\frac{b+c}{a} = 12$.
A fair $n(n > 1)$ faces die is rolled repeatedly until a number less than $n$ appears. If the mean of the number of tosses required is $\frac{n}{9}$, then $n$ is equal to ____________.
Explanation:
$ \text { Mean }=\sum\limits_{i=1}^{\infty} p_i x_i=1 \cdot \frac{n-1}{n}+\frac{2}{n} \cdot\left(\frac{n-1}{n}\right)+\frac{3}{n^2}\left(\frac{n-1}{n}\right)+\ldots $
$ \frac{n}{9}=\left(1-\frac{1}{n}\right) S $ ......(1)
where
$ \begin{aligned} & S=1+\frac{2}{n}+\frac{3}{n^2}+\frac{4}{n^3}+\ldots \\\\ & \frac{1}{n} S=\frac{1}{n}+\frac{2}{n^2}+\frac{3}{n^3}+\ldots \\\\ & ----------------\\\\ & \left(1-\frac{1}{n}\right) S=1+\frac{1}{n}+\frac{1}{n^2}+\frac{1}{n^3}+\ldots \end{aligned} $
$ \begin{aligned} & \Rightarrow \left(1-\frac{1}{n}\right) S=\frac{1}{1-\frac{1}{n}} \\\\ & \Rightarrow \frac{n}{9}=\left(1-\frac{1}{n}\right) \times \frac{1}{\left(1-\frac{1}{n}\right)^2}=\frac{n}{n-1} \end{aligned} $
$ \Rightarrow $ n = 10
Let the probability of getting head for a biased coin be $\frac{1}{4}$. It is tossed repeatedly until a head appears. Let $\mathrm{N}$ be the number of tosses required. If the probability that the equation $64 \mathrm{x}^{2}+5 \mathrm{Nx}+1=0$ has no real root is $\frac{\mathrm{p}}{\mathrm{q}}$, where $\mathrm{p}$ and $\mathrm{q}$ are coprime, then $q-p$ is equal to ________.
Explanation:
This gives us :
$(5N)^2 - 4\times64\times1 < 0$
$\Rightarrow 25N^2 < 256$
$\Rightarrow N^2 < \frac{256}{25}$
$\Rightarrow N < \sqrt{\frac{256}{25}} = \frac{16}{5}$
Since $N$ must be an integer (as it represents the number of tosses), the possible values of $N$ are 1, 2, or 3.
The probability of getting the first head on the $n$-th toss (given the probability of getting a head is $1/4$) is given by the geometric distribution formula, $(1 - p)^{n-1}\times p$.
So, the probability for our specific values of $N$ is:
$P(N=1) = (1 - 1/4)^{1-1}\times(1/4) = 1/4$
$P(N=2) = (1 - 1/4)^{2-1}\times(1/4) = 3/4 \times 1/4 = 3/16$
$P(N=3) = (1 - 1/4)^{3-1}\times(1/4) = (3/4)^2 \times 1/4 = 9/64$
Therefore, the total probability (p/q) is :
$p/q = P(N=1) + P(N=2) + P(N=3)$
$= 1/4 + 3/16 + 9/64$
$= 16/64 + 12/64 + 9/64$
$= 37/64$
So, $p = 37$, $q = 64$ and $q-p = 64 - 37 = 27$.
Therefore, $q-p$ is equal to $27$.
Explanation:
$ \begin{aligned} & \mathrm{P}(\mathrm{A})=\frac{\operatorname{ar}(\mathrm{OACDEG})}{(\mathrm{OBDF})} \\\\ & =\frac{\operatorname{ar}(\mathrm{OBDF})-\operatorname{ar}(\mathrm{ABC})-\operatorname{ar}(\mathrm{EFG})}{\operatorname{ar}(\mathrm{OBDF})} \\\\ & \Rightarrow \frac{11}{36}=\frac{(60)^2-\frac{1}{2}(60-\mathrm{a})^2-\frac{1}{2}(60-\mathrm{a})^2}{3600} \\\\ & \Rightarrow 1100=3600-(60-\mathrm{a})^2 \\\\ & \Rightarrow (60-\mathrm{a})^2=2500 \Rightarrow 60-\mathrm{a}=50 \\\\ & \Rightarrow \mathrm{a}=10 \end{aligned} $
Explanation:
$p = {6 \over {36}} = {1 \over 6}$
$q = {{{}^6{C_1} \times {}^5{C_1} \times {{4!} \over {3!}}} \over {{6^4}}} = {{120} \over {1296}} = {5 \over {54}}$
${p \over q} = {{{1 \over 6}} \over {{5 \over {54}}}} = {{54} \over {6 \times 5}} = {9 \over 5} = {m \over n}$
$m + n = 14$
25% of the population are smokers. A smoker has 27 times more chances to develop lung cancer than a non smoker. A person is diagnosed with lung cancer and the probability that this person is a smoker is $\frac{k}{10}%$. Then the value of k is __________.
Explanation:
Probability of a person being non-smoker $=\frac{3}{4}$
$P\left(\frac{\text { Person is smoker }}{\text { Person diagonsed with cancer }}\right)=\frac{\frac{1}{4} \cdot 27 P}{\frac{1}{4} \cdot 27 P+\frac{3 P}{4}}$
$=\frac{9}{10}=\frac{k}{10}$
$\Rightarrow k=9$
Three urns A, B and C contain 4 red, 6 black; 5 red, 5 black; and $\lambda$ red, 4 black balls respectively. One of the urns is selected at random and a ball is drawn. If the ball drawn is red and the probability that it is drawn from urn C is 0.4 then the square of the length of the side of the largest equilateral triangle, inscribed in the parabola $y^2=\lambda x$ with one vertex at the vertex of the parabola, is :
Explanation:
$E_{2}:$ Ball is drawn from urn $B(5 R+5 B)$
$E_{3}$ : Ball is drawn from urn $C(\lambda R+4 B)$
$A \rightarrow$ Ball drawn is red.
Required probability $=P\left(\frac{E_{3}}{A}\right)$
$=\frac{\frac{1}{3} \times \frac{\lambda}{\lambda+4}}{\frac{1}{3} \times \frac{4}{10}+\frac{1}{3} \times \frac{5}{10}+\frac{1}{3} \times \frac{\lambda}{\lambda+4}}=\frac{2}{5}$
$\Rightarrow\frac{10 \lambda}{19 \lambda+36}=\frac{2}{5}$
$\Rightarrow \lambda=6$
So, parabola $y^2=6 x$
Let side length of the triangle be $l$.
$ \begin{aligned} & \tan 30^{\circ}=\frac{3 t} {\frac{3}{2} t^2} \\\\ & \Rightarrow \frac{1}{\sqrt{3}}=\frac{2}{t} \\\\ & \therefore t=2 \sqrt{3} \\\\ & \text { So, }\left(\frac{3}{2} t^2, 3 t\right) \\\\ & =(18,6 \sqrt{3}) \end{aligned} $
$ \text { Now, } l^2=18^2+(6 \sqrt{3})^2=324+108=432 $
The sum and product of the mean and variance of a binomial distribution are 82.5 and 1350 respectively. Then the number of trials in the binomial distribution is ____________.
Explanation:
Given $n p+n p q=82.5$
and $n p(n p q)=1350$
$ \therefore $ Quadratic equation is
$ x^{2}-82.5 x+1350=0$
$\Rightarrow x^{2}-22.5 x-60 x+1350=0$
$\Rightarrow x-(x-22.5)-60(x-22.5)=0$
Mean $=60$ and Variance $=22.5$
$ n p=60, n p q=22.5 $
$ \Rightarrow q=\frac{9}{24}=\frac{3}{8}, p=\frac{5}{8} $
$ \therefore \quad n \frac{5}{8}=60 \quad \Rightarrow n=96 $
A bag contains 4 white and 6 black balls. Three balls are drawn at random from the bag. Let $\mathrm{X}$ be the number of white balls, among the drawn balls. If $\sigma^{2}$ is the variance of $\mathrm{X}$, then $100 \sigma^{2}$ is equal to ________.
Explanation:
$X = $ Number of white ball drawn
$P(X = 0) = {{{}^6{C_3}} \over {{}^{10}{C_3}}} = {1 \over 6}$
$P(X = 1) = {{{}^6{C_2} \times {}^4{C_1}} \over {{}^{10}{C_3}}} = {1 \over 2},$
$P(X = 2) = {{{}^6{C_1} \times {}^4{C_2}} \over {{}^{10}{C_3}}} = {3 \over {10}}$
and $P(X = 3) = {{{}^6{C_0} \times {}^4{C_3}} \over {{}^{10}{C_3}}} = {1 \over {30}}$
Variance $ = {\sigma ^2} = \sum {{P_i}X_i^2 - {{\left( {\sum {{P_i}{X_i}} } \right)}^2}} $
${\sigma ^2} = {1 \over 2} + {{12} \over {10}} + {3 \over {10}} - {\left( {{1 \over 2} + {6 \over {10}} + {1 \over {10}}} \right)^2}$
$ = {{56} \over {100}}$
$100{\sigma ^2} = 56.$
The probability distribution of X is :
| X | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| P(X) | ${{1 - d} \over 4}$ | ${{1 + 2d} \over 4}$ | ${{1 - 4d} \over 4}$ | ${{1 + 3d} \over 4}$ |
For the minimum possible value of d, sixty times the mean of X is equal to _______________.
Explanation:
| X | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| $P(x)$ | ${{1 - d} \over 4}$ | ${{1 + 2d} \over 4}$ | ${{1 - 4d} \over 4}$ | ${{1 + 3d} \over 4}$ |
We know, $0 \le P(x) \le 1$
$\therefore$ $0 \le {{1 - d} \over 4} \le 1$
$ \Rightarrow 0 \le 1 - d \le 4$
$ \Rightarrow - 1 \le - d \le 3$
$ \Rightarrow 1 \ge d \ge - 3$
Also,
$0 \le {{1 + 2d} \over 4} \le 1$
$ \Rightarrow 0 \le 1 + 2d \le 4$
$ \Rightarrow - 1 \le 2d \le 3$
$ \Rightarrow - {1 \over 2} \le d \le {3 \over 2}$
Also,
$0 \le {{1 - 4d} \over 4} \le 1$
$ \Rightarrow 0 \le 1 - 4d \le 4$
$ \Rightarrow - 1 \le - 4d \le 3$
$ \Rightarrow 1 \ge 4d \ge - 3$
$ \Rightarrow {1 \over 4} \ge d \ge - {3 \over 4}$
And,
$0 \le {{1 + 3d} \over 4} \le 1$
$ \Rightarrow 0 \le 1 + 3d \le 4$
$ \Rightarrow - 1 \le 3d \le 3$
$ \Rightarrow - {1 \over 3} \le d \le 1$
Common range is $ = - {1 \over 3}$ to ${1 \over 4}$
$\therefore$ $d\, \in \left[ { - {1 \over 3},{1 \over 4}} \right]$
$\therefore$ Minimum value of $d = - {1 \over 3}$
We know, mean
$E(x) = \sum {x\,.\,P(x)} $
$ = 0 \times {{1 - d} \over 4} + 1 \times {{1 + 2d} \over 4} + 2 \times {{1 - 4d} \over 4} + 3 \times {{1 + 3d} \over 4}$
$ = {{1 + 2d + 2 - 8d + 3 + 9d} \over 4}$
$ = {{6 + 3d} \over 4}$
For $d = - {1 \over 3}$, $E(x) = {{6 + 3 \times - {1 \over 3}} \over 4} = {5 \over 4}$
$\therefore$ $60E(x) = 60 \times {5 \over 4} = 75$
Let S = {E1, E2, ........., E8} be a sample space of a random experiment such that $P({E_n}) = {n \over {36}}$ for every n = 1, 2, ........, 8. Then the number of elements in the set $\left\{ {A \subseteq S:P(A) \ge {4 \over 5}} \right\}$ is ___________.
Explanation:
Here $P({E_n}) = {n \over {36}}$ for n = 1, 2, 3, ......, 8
Here $P(A) = {{Any\,possible\,sum\,of\,(1,2,3,\,...,\,8)( = a\,say)} \over {36}}$
$\because$ ${a \over {36}} \ge {4 \over 5}$
$\therefore$ $a \ge 29$
If one of the number from {1, 2, ......, 8} is left then total $a \ge 29$ by 3 ways.
Similarly by leaving terms more 2 or 3 we get 16 more combinations.
$\therefore$ Total number of different set A possible is 16 + 3 = 19
If the probability that a randomly chosen 6-digit number formed by using digits 1 and 8 only is a multiple of 21 is p, then 96 p is equal to _______________.
Explanation:
Total number of numbers from given
Condition = n(s) = 26.
Every required number is of the form
A = 7 . (10a1 + 10a2 + 10a3 + .......) + 111111
Here 111111 is always divisible by 21.
$\therefore$ If A is divisible by 21 then
10a1 + 10a2 + 10a3 + ....... must be divisible by 3.
For this we have 6C0 + 6C3 + 6C6 cases are there
$\therefore$ n(E) = 6C0 + 6C3 + 6C6 = 22
$\therefore$ Required probability = ${{22} \over {{2^6}}} = p$
$\therefore$ ${{11} \over {32}} = p$
$\therefore$ $96p = 33$
In an examination, there are 10 true-false type questions. Out of 10, a student can guess the answer of 4 questions correctly with probability ${3 \over 4}$ and the remaining 6 questions correctly with probability ${1 \over 4}$. If the probability that the student guesses the answers of exactly 8 questions correctly out of 10 is ${{{{27}k}} \over {{4^{10}}}}$, then k is equal to ___________.
Explanation:
Student guesses only two wrong. So there are three possibilities.
(i) Student guesses both wrong from 1st section
(ii) Student guesses both wrong from 2nd section
(iii) Student guesses two wrong one from each section
Required probabilities
$ = {}^4{C_2}{\left( {{3 \over 4}} \right)^2}{\left( {{1 \over 4}} \right)^2}{\left( {{1 \over 6}} \right)^6} + {}^6{C_2}{\left( {{3 \over 4}} \right)^2}{\left( {{1 \over 4}} \right)^4}{\left( {{3 \over 4}} \right)^4} + {}^4{C_1}\,.\,{}^6{C_1}\left( {{3 \over 4}} \right)\left( {{1 \over 4}} \right){\left( {{3 \over 4}} \right)^3}{\left( {{1 \over 4}} \right)^5}$
$ = {1 \over {{4^{10}}}}\left[ {6 \times 9 + 15 \times {9^4} + 24 \times {9^2}} \right]$
$ = {{27} \over {{4^{10}}}}\left[ {2 + 27 \times 15 + 72} \right]$
$ = {{27 \times 479} \over {{4^{10}}}}$
| x | $ - $2 | $ - $1 | 3 | 4 | 6 |
|---|---|---|---|---|---|
| P(X = x) | ${1 \over 5}$ | a | ${1 \over 3}$ | ${1 \over 5}$ | b |
If the mean of X is 2.3 and variance of X is $\sigma$2, then 100 $\sigma$2 is equal to :
Explanation:
| x | $ - $2 | $ - $1 | 3 | 4 | 6 |
|---|---|---|---|---|---|
| P(X = x) | ${1 \over 5}$ | a | ${1 \over 3}$ | ${1 \over 5}$ | b |
$\overline X $ = 2.3
$-$a + 6b = ${9 \over {10}}$ ..... (1)
$\sum {{P_i} = {1 \over 5} + a + {1 \over 3} + {1 \over 5} + b = 1} $
$a + b = {4 \over {15}}$ .... (2)
From equation (1) and (2)
$a = {1 \over {10}},b = {1 \over 6}$
${\sigma ^2} = \sum {{p_i}x_i^2 - {{(\overline X )}^2}} $
${1 \over 5}(4) + a(1) + {1 \over 3}(9) + {1 \over 5}(16) + b(36) - {(2.3)^2}$
$ = {4 \over 5} + a + 3 + {{16} \over 5} + 36b - {(2.3)^2}$
$ = 4 + a + 3 + 36b - {(2.3)^2}$
$ = 7 + a + 36b - {(2.3)^2}$
$ = 7 + {1 \over {10}} + 6 - {(2.3)^2}$
$ = 13 + {1 \over {10}} - {\left( {{{23} \over {10}}} \right)^2}$
$ = {{131} \over {10}} - {\left( {{{23} \over {10}}} \right)^2}$
$ = {{1310 - {{(23)}^2}} \over {100}}$
$ = {{1310 - 529} \over {100}}$
${\sigma ^2} = {{781} \over {100}}$
$100{\sigma ^2} = 781$
Explanation:
I2 = second unit is functioning
P(I1) = 0.9, P(I2) = 0.8
P($\overline {{I_1}} $) = 0.1, P($\overline {{I_2}} $) = 0.2
$P = {{0.8 \times 0.1} \over {0.1 \times 0.2 + 0.9 \times 0.2 + 0.1 \times 0.8}} = {8 \over {28}}$
$98P = {8 \over {28}} \times 98 = 28$
| X | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| P(X) | K | 2K | 2K | 3K | K |
Let p = P(1 < X < 4 | X < 3). If 5p = $\lambda$K, then $\lambda$ equal to ___________.
Explanation:
$ \Rightarrow k = {1 \over 9}$
Now, $p = P\left( {{{kx < 4} \over {X < 3}}} \right) = {{P(X = 2)} \over {P(X < 3)}} = {{{{2k} \over {9k}}} \over {{k \over {9k}} + {{2k} \over {9k}}}} = {2 \over 3}$
$ \Rightarrow p = {2 \over 3}$
Now, $5p = \lambda k$
$ \Rightarrow (5)\left( {{2 \over 3}} \right) = \lambda (1/9)$
$ \Rightarrow \lambda = 30$
Explanation:
1 $-$ P(All tail) $\ge$ 0.9
$1 - {\left( {{1 \over 2}} \right)^n}$ $\ge$ 0.9
$ \Rightarrow {\left( {{1 \over 2}} \right)^n} \le {1 \over {10}}$
$\Rightarrow$ nmin = 4
($\alpha$ $-$ 2$\beta$)p = $\alpha$$\beta$ and ($\beta$ $-$ 3$\gamma$)p = 2$\beta$$\gamma$. All the given probabilities are assumed to lie in the interval (0, 1).
Then, $\frac{Probability\ of\ occurrence\ of\ E_{1}}{Probability\ of\ occurrence\ of\ E_{3}} $ is equal to _____________.
Explanation:
$\alpha $ = P$\left( {{E_1} \cap {{\overline E }_2} \cap {{\overline E }_3}} \right)$ = $P\left( {{E_1}} \right).P\left( {{{\overline E }_2}} \right).P\left( {{{\overline E }_3}} \right)$
$ \Rightarrow $ $\alpha $ = x(1 $-$ y) (1 $-$ z) ......(i)
Similarly
β = (1 – x).y(1 – z) ...(ii)
$\gamma $ = (1 – x)(1 – y).z ...(iii)
p = (1 – x)(1 – y)(1 – z) ...(iv)
From (i) and (iv)
${x \over {1 - x}} = {\alpha \over p}$
$ \Rightarrow $ x = ${\alpha \over {\alpha + p}}$
From (iii) and (iv)
${z \over {1 - z}} = {\gamma \over p}$
$ \Rightarrow $ z = ${\gamma \over {\gamma + p}}$
$ \therefore $ ${{P\left( {{E_1}} \right)} \over {P\left( {{E_3}} \right)}} = {x \over z} = {{{\alpha \over {\alpha + p}}} \over {{\gamma \over {\gamma + p}}}}$ $ = {{{{\gamma + p} \over \gamma }} \over {{{\alpha + p} \over \alpha }}} = {{1 + {p \over \gamma }} \over {1 + {p \over \alpha }}}$ ..(v)
Also given,
($\alpha$ $-$ 2$\beta$)p = $\alpha$$\beta$ $ \Rightarrow $ $\alpha $p = ($\alpha $ + 2p)$\beta $ ....(vi)
$\beta$ $-$ 3$\gamma$)p = 2$\beta$$\gamma$ $ \Rightarrow $ 3$\gamma $p = (p - 2$\gamma $)$\beta $ .....(vii)
From (vi) and (vii),
${\alpha \over {3\gamma }} = {{\alpha + 2p} \over {p - 2\gamma }}$
$ \Rightarrow $ p$\alpha $ - 6p$\gamma $ = 5$\gamma $$\alpha $
$ \Rightarrow $ ${p \over \gamma } - {{6p} \over \alpha } = 5$
$ \Rightarrow $ ${p \over \gamma } + 1 = 6\left( {{p \over \alpha } + 1} \right)$ ....(viii)
Now from (v) and (viii),
${{P\left( {{E_1}} \right)} \over {P\left( {{E_3}} \right)}}$ = 6
Then ${{P\left( {{B_1}} \right)} \over {P\left( {{B_3}} \right)}}$ is equal to ________.
Explanation:
$\alpha $ = P(B1 $ \cap $ $\overline {{B_2}} \cap \overline {{B_3}} $) = $P\left( {{B_1}} \right)P\left( {\overline {{B_2}} } \right)P\left( {\overline {{B_3}} } \right)$
$ \Rightarrow $ x(1 $-$ y)(1 $-$ z) = $\alpha$
Similarly, y(1 $-$ x)(1 $-$ z) = $\beta$
z(1 $-$ x)(1 $-$ y) = $\gamma$
and (1 $-$ x)(1 $-$ y)(1 $-$ z) = p
($\alpha$ $-$ 2$\beta$)p = $\alpha$$\beta$
(x(1 $-$ y)(1 $-$ z) $-$2y(1 $-$ x)(1 $-$ z)) (1 $-$ x)(1 $-$ y)(1 $-$ z) = xy(1 $-$ x)(1 $-$ y)(1 $-$ z)
x $-$ xy $-$ 2y + 2xy = xy
x = 2y ...... (1)
Similarly ($\beta$ $-$ 3$\gamma $)p = 2$\beta$$\gamma $
$ \Rightarrow $ y = 3z .... (2)
From (1) & (2)
x = 6z
Now
${x \over z} = 6$
Explanation:
at least 2 bombs should hit.
P(x > 2) $ \ge $ 0.99
$ \Rightarrow $ 1 - p(x < 2) $ \ge $ 0.99
$ \Rightarrow $ 1 - (p(x = 0) + p(x = 1)) $ \ge $ 0.99
$ \Rightarrow $ 1 - nC0${\left( {{1 \over 2}} \right)^0}{\left( {{1 \over 2}} \right)^n}$ - nC1.${\left( {{1 \over 2}} \right)^1}{\left( {{1 \over 2}} \right)^{n - 1}}$ $ \ge $ 0.99
$ \Rightarrow $ 1 - ${1 \over {{2^n}}}$ - ${n \over {{2^n}}}$ $ \ge $ ${{99} \over {100}}$
$ \Rightarrow $ ${1 \over {100}}$ $ \ge $ ${{n + 1} \over {{2^n}}}$
$ \Rightarrow $ 2n $ \ge $ 100(n + 1)
Now checking for value of n, we get
n = 11
Explanation:
$ \Rightarrow 1 - {\left( {{9 \over {10}}} \right)^n} > {1 \over 4}$
$ \Rightarrow {3 \over 4} > {\left( {{9 \over {10}}} \right)^n} \Rightarrow n \ge 3$
A factory has a total of three manufacturing units, $M_1, M_2$, and $M_3$, which produce bulbs independent of each other. The units $M_1, M_2$, and $M_3$ produce bulbs in the proportions of $2: 2: 1$, respectively. It is known that $20 \%$ of the bulbs produced in the factory are defective. It is also known that, of all the bulbs produced by $M_1, 15 \%$ are defective. Suppose that, if a randomly chosen bulb produced in the factory is found to be defective, the probability that it was produced by $M_2$ is $\frac{2}{5}$.
If a bulb is chosen randomly from the bulbs produced by $M_3$, then the probability that it is defective is __________.
Explanation:
$H_1$: The bulb is produced by unit $M_1$.
$H_2$: The bulb is produced by unit $M_2$.
$H_3$: The bulb is produced by unit $M_3$.
$E$: The bulb is defective.
The unit production proportions and known probabilities are:
$P(H_1) = \frac{2}{5}$, $P(H_2) = \frac{2}{5}$, $P(H_3) = \frac{1}{5}$
$P(E | H_1) = \frac{15}{100} = \frac{3}{20}$
The total probability of a defective bulb is given by:
$ P(E) = P(E | H_1) \cdot P(H_1) + P(E | H_2) \cdot P(H_2) + P(E | H_3) \cdot P(H_3) $
Given:
$ P(E) = 0.20 = \frac{3}{20} \cdot \frac{2}{5} + P(E | H_2) \cdot \frac{2}{5} + P(E | H_3) \cdot \frac{1}{5} $
This simplifies to:
$ 2P(E | H_2) + P(E | H_3) = \frac{7}{10} \quad \text{...(i)} $
Additionally, the probability that a defective bulb is from $M_2$ is given as:
$ P(H_2 | E) = \frac{2}{5} = \frac{P(E | H_2) \cdot \frac{2}{5}}{\frac{1}{5}} $
Solving the equation above gives:
$ P(E | H_2) = \frac{1}{5} \quad \text{...(ii)} $
Using the equations (i) and (ii), we substitute $P(E | H_2)$ into (i):
$ 2 \cdot \frac{1}{5} + P(E | H_3) = \frac{7}{10} $
Solving for $P(E | H_3)$:
$ \frac{2}{5} + P(E | H_3) = \frac{7}{10} $
Converting $\frac{2}{5}$ to $\frac{4}{10}$ gives:
$ \frac{4}{10} + P(E | H_3) = \frac{7}{10} $
$ P(E | H_3) = \frac{7}{10} - \frac{4}{10} = \frac{3}{10} $
Therefore, the probability that a randomly chosen bulb from $M_3$ is defective is 0.30.
Explanation:
3 White
6 Green
$(\mathrm{N}-9)$ Blue
$\begin{aligned} & \text { Given } P\left(W_1 \cap G_2 \cap B_3\right)=\frac{2}{5 N} \\ & \text { and } P\left(B_3 \mid W_1 \cap G_2\right)=\frac{2}{9} \\ & \Rightarrow \frac{P\left(B_3 \cap W_1 \cap G_2\right)}{P\left(W_1 \cap G_2\right)}=\frac{2}{9} \\ & \Rightarrow \frac{2}{5 N} \times \frac{N \times(N-1)}{3 \times 6}=\frac{2}{9} \\ & \Rightarrow N=11 \end{aligned}$
Let $X$ be a random variable, and let $P(X=x)$ denote the probability that $X$ takes the value $x$. Suppose that the points $(x, P(X=x)), x=0,1,2,3,4$, lie on a fixed straight line in the $x y$-plane, and $P(X=x)=0$ for all $x \in \mathbb{R}-\{0,1,2,3,4\}$. If the mean of $X$ is $\frac{5}{2}$, and the variance of $X$ is $\alpha$, then the value of $24 \alpha$ is _____________.
Explanation:
Let equation of line is $\mathrm{y=mx+c}$
| $\mathrm{x}$ | 0 | 1 | 2 | 3 | 4 | $\mathrm{R-\{0,1,2,3,4\}}$ |
|---|---|---|---|---|---|---|
| $\mathrm{P(x)}$ | $\mathrm{c}$ | $\mathrm{m+c}$ | $\mathrm{2m+c}$ | $\mathrm{3m+c}$ | $\mathrm{4m+c}$ | 0 |
$\sum_{x=0}^4(m x+c)=1 \Rightarrow 10 m+5 c=1 \Rightarrow 2 m+c=\frac{1}{5}\quad \text{.... (i)}$
$\begin{aligned} & \text { mean }=\sum x_i P_i=\sum_{i=0}^4\left(m x_i+c\right) \cdot x_i=30 m+10 c=\frac{5}{2} \\ & \therefore 3 m+c=\frac{1}{4} \ldots(2) \\ & \text { from (1) and (2) } \mathrm{m}=\frac{1}{20}, \mathrm{c}=\frac{1}{10} \\ & \Sigma \mathrm{P}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}^2=\sum_{\mathrm{i}=0}^4\left(\mathrm{mx}_{\mathrm{i}}+\mathrm{c}\right) \mathrm{x}_1^2 \\ & =\sum_{\mathrm{i}=0}^4\left(\mathrm{mx}_{\mathrm{i}}^3+c \mathrm{x}_{\mathrm{i}}^2\right) \Rightarrow 100 \mathrm{~m}+30 \mathrm{c}(\text { Now putting } \mathrm{m} \text { and } \mathrm{c}) \\ & \Rightarrow \Sigma \mathrm{P}_{\mathrm{i}}^2=5+3=8 \\ & \text { Variance }=\Sigma \mathrm{P}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}^2-\left(\Sigma \mathrm{P}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}\right)^2=8-\left(\frac{5}{2}\right)^2=\frac{7}{4} \\ & \therefore 24 \alpha=42 \end{aligned}$
Explanation:
Among these 38 elements, let us calculate when element is not divisible by 20
$\therefore p=\frac{38-7}{38} $
$ \therefore 38 p=31$
Explanation:
Number of points having 0 friend $=0$
Number of points having 1 friend $=0$
Number of points having 2 friends $=4$
Number of points having 3 friends $=5 \times 4=20$
Number of points having 4 friends $=49-24=25$
$\mathrm{P}_{\mathrm{i}}=$ Probability that randomly selected points has friends
$\mathrm{P}_0=0$ (0 friends)
$\mathrm{P}_1=0$ (exactly 1 friends)
$\mathrm{P}_2=\frac{{ }^4 \mathrm{C}_1}{{ }^{49} \mathrm{C}_1}=\frac{4}{9}$ (exactly 2 friends)
$\mathrm{P}_3=\frac{{ }^{20} \mathrm{C}_1}{{ }^{49} \mathrm{C}_1}=\frac{20}{49}$ (exactly 3 friends)
$\mathrm{P}_4=\frac{{ }^{25} \mathrm{C}_1}{{ }^{49} \mathrm{C}_1}=\frac{25}{49}$ (exactly 4 friends)
$ \begin{array}{|c|c|c|c|c|c|} \hline x & 0 & 1 & 2 & 3 & 4 \\ \hline P(x) & 0 & 0 & \frac{4}{49} & \frac{20}{49} & \frac{25}{49} \\ \hline \end{array} $$\begin{aligned} & \text { Mean }=E(x)=\sum x_i P_i=0+0+\frac{8}{49}+\frac{60}{49}+\frac{100}{49}=\frac{168}{49} \\\\ & 7(E(x))=\frac{168}{49} \times 7=24\end{aligned}$
Explanation:
Total number of ways of selecting 2 persons $={ }^{49} \mathrm{C}_2$
Number of ways in which 2 friends are selected $=6 \times 7 \times 2=84$
$7 \mathrm{P}=\frac{84 \times 2}{49 \times 48} \times 7=\frac{1}{2}$
190 persons had symptom of fever,
220 persons had symptom of cough,
220 persons had symptom of breathing problem,
330 persons had symptom of fever or cough or both,
350 persons had symptom of cough or breathing problem or both,
340 persons had symptom of fever or breathing problem or both,
30 persons had all three symptoms (fever, cough and breathing problem).
If a person is chosen randomly from these 900 persons, then the probability that the person has at most one symptom is ____________.
Explanation:
Person had symptom of cough $=n(\mathrm{C})=220$
Person had symptom of breathing $=n(\mathrm{~B})=220$
Person had symptom of fever or cough
$ =n(\mathrm{~F} \cup \mathrm{C})=330 $
Person had symptom of cough or breathing
$ =n(\mathrm{C} \cup \mathrm{B})=350 $
Person had symptom of fever or breathing
$ =n(\mathrm{~F} \cup \mathrm{B})=340 $
Person had symptom of fever, cough and breathing $=n(\mathrm{~F} \cap \mathrm{C} \cup \mathrm{B})=30$
So $n(\mathrm{~F} \cap \mathrm{C})=n(\mathrm{~F})+n(\mathrm{C})-n(\mathrm{~F} \cup \mathrm{C})$
$=190+220-330$
$=80$
and
$ n(\mathrm{F} \cap \mathrm{B}) =n(\mathrm{~F})+n(\mathrm{~B})-n(\mathrm{~F} \cup \mathrm{B}) $
$ =190+220-340 $
$ =70 $
and $ n(\mathrm{C} \cap \mathrm{B}) =n(\mathrm{C})+n(\mathrm{~B})-n(\mathrm{C} \cap \mathrm{B}) $
$ =220+220-350 $
$ =90$
Number of people having at most one symptom
$ =70+80+90+480 =720 $
Required probability $=\frac{720}{900}=0.80$
Explanation:
Let E1 = Event that it is a multiple of 3 = {3, 6, 9, ...., 1998}
$\therefore$ n(E1) = 666
and E2 = Event that it is a multiple of 7 = {7, 14, ..., 1995}
$\therefore$ n(E2) = 285
${E_1} \cap {E_2}$ = multiple of 21 = {21, 42, ....., 1995}
$n({E_1} \cap {E_2}) = 95$
$\therefore$ $P({E_1} \cup {E_2}) = P({E_1}) + P({E_2}) - P({E_1} \cap {E_2})$
$P({E_1} \cup {E_2}) = {{666 + 285 - 95} \over {2000}} = {{856} \over {2000}} = p$ (given)
Hence, $500p = 500 \times {{856} \over {2000}} = {{856} \over 4} = 214$
The value of ${{625} \over 4}{p_1}$ is ___________.
Explanation:
$ = 1 - {\left( {{{80} \over {100}}} \right)^3} = {{125 - 64} \over {125}} = {{61} \over {125}}$
So, ${{625{p_1}} \over 4} = {{625} \over 4} \times {{61} \over {125}} = {5 \over 4} \times 61 = 76.25$
The value of ${{125} \over 4}{p_2}$ is ___________.
Explanation:
$ = 1 - {\left( {{{60} \over {100}}} \right)^3} = 1 - {{27} \over {125}} = {{98} \over {125}}$
So, ${{125{p_2}} \over 4} = {{125} \over 4} \times {{98} \over {125}} = {{98} \over 4} = 24.50$
Explanation:
Now, according to the question, let the minimum number of missiles required to fired is n, so
$^n{C_3}{p^3}{q^{n - 3}}{ + ^n}{C_4}{p^4}{q^{n - 4}} + ...{ + ^n}{C_n}{p^n}\, \ge \,0.95$
$ \Rightarrow 1 - \left\{ {^n{C_0}{{\left( {{1 \over 4}} \right)}^n}{ + ^n}{C_1}\left( {{3 \over 4}} \right){{\left( {{1 \over 4}} \right)}^{n - 1}}{ + ^n}{C_2}{{\left( {{3 \over 4}} \right)}^2}{{\left( {{1 \over 4}} \right)}^{n - 2}}} \right\}\, \ge \,0.95$
$ \Rightarrow 1 - {{95} \over {100}}\, \ge \,{1 \over {{4^n}}} + {{3n} \over {{4^n}}} + {{n(n - 1)} \over 2}{9 \over {{4^n}}}$
$ \Rightarrow {{{4^n}} \over {20}}\, \ge \,{{2 + 6n + 9{n^2} - 9n} \over 2}$
$ \Rightarrow 10(9{n^2} - 3n + 2)\, \le \,{4^n}$
Now, at n = 3, LHS = 720, RHS = 64
at n = 4, LHS = 1340, RHS = 256
at n = 5, LHS = 2120, RHS = 1024
at n = 6, LHS = 3080, RHS = 4096
Hence, n = 6 missiles should be fired.
Explanation:
E = {(1, 3), (2, 2), (3, 1), (3, 6), (4, 5), (5, 4), (6, 3)}
and an event F of sum of outputs are prime numbers
(i.e., 2, 3, 5, 7, 11) so
F = {(1, 1), (1, 2), (2, 1), (1, 4), (2, 3), (3, 2), (4, 1), (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1), (5, 6), (6, 5)}
and event T of sum of outputs are odd numbers
(i.e., 3, 5, 7, 9, 11)
T = {(1, 2), (2, 1), (1, 4), (2, 3), (3, 2), (4, 1), (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1), (3, 6), (4, 5), (5, 4), (6, 3), (5, 6), (6, 5)}
Now, required probability $p = P(T/E)$
$ = {{P(T \cap E)} \over {P(E)}}$
where, ${P(T \cap E)}$ = probability of occurring perfect square odd number before prime
$ = \left( {{4 \over {36}}} \right) + \left( {{{14} \over {36}}} \right)\left( {{4 \over {36}}} \right) + {\left( {{{14} \over {36}}} \right)^2}\left( {{4 \over {36}}} \right) + ....\infty $
$ = {{{4 \over {36}}} \over {1 - {{14} \over {36}}}} = {4 \over {22}} = {2 \over {11}}$
and P(E) = probability of occurring perfect square before prime
$ = \left( {{7 \over {36}}} \right) + \left( {{{14} \over {36}}} \right)\left( {{7 \over {36}}} \right) + {\left( {{{14} \over {36}}} \right)^2}\left( {{7 \over {36}}} \right) + ....\infty $
$ = {{{7 \over {36}}} \over {1 - {{14} \over {36}}}} = {7 \over {22}}$
$ \therefore $ $P(T/E) = {{{2 \over {11}}} \over {{7 \over {22}}}} = {4 \over 7} = p \Rightarrow 14p = 8$
E1 = {A$ \in $S : det A = 0} and
E2 = {A$ \in $S : sum of entries of A is 7}.
If a matrix is chosen at random from S, then the conditional probability P(E1 | E2) equals ...............
Explanation:
E1 = {A$ \in $S : det(A) = 0} and
E2 = {A$ \in $S : sum of entries of A is 7}.
For event E2, means sum of entries of matrix A is 7, then we need seven 1s and two 0s.
$ \therefore $ Number of different possible matrices = ${{9!} \over {7!2!}}$
$ \Rightarrow $ n(E2) = 36
For event E1, |A| = 0, both the zeroes must be in same row/column.
$ \therefore $ Number of matrices such that their determinant is zero = $6 \times {{3!} \over {2!}}$ = 18 = $n({E_1} \cap {E_2})$
$ \therefore $ Required probability,
$P\left( {{{{E_1}} \over {{E_2}}}} \right) = {{n({E_1} \cap {E_2})} \over {n({E_2})}}$
$ = {{18} \over {36}}$
$ = {1 \over 2} = 0.50$
Explanation:
Let the coin is tossed $n$ times.
$\because p$ (at least 2 heads) $=1-[p$ (one heads) $+p$ (No heads)$\}$
As we know, by binominal probability theorem the probability of getting $r$ success in $n$ trials with $p$ being the probability of success and $q$ be the probability of failure, is given by ${ }^n \mathrm{C}_r(p)^r(q)^{n-r}$.
Let head be considered as the success and tail be the failure probability of getting head in a toss $=p=\frac{1}{2}$ and probability of getting tail in a toss $=q=\frac{1}{2}$
$\begin{aligned} & \therefore \mathrm{P}(\text { one head })={ }^n \mathrm{C}_1\left(\frac{1}{2}\right)^1\left(\frac{1}{2}\right)^{n-1} \\ & ={ }^n \mathrm{C}_1\left(\frac{1}{2}\right)^n \\ & \Rightarrow \mathrm{P} \text { (one head) }=n\left(\frac{1}{2}\right)^n \\ & \text { Similarly, } \mathrm{P}(\text {No heads})={ }^n \mathrm{C}_0\left(\frac{1}{2}\right)^0\left(\frac{1}{2}\right)^n \\ & \Rightarrow \mathrm{P}(\text { No heads })=\left(\frac{1}{2}\right)^n \quad\left\{\because n_{c_o}=1\right\} \\ & \therefore \mathrm{P} \text { (at least } 2 \text { heads) }=1-\left\{n\left(\frac{1}{2}\right)^n+\left(\frac{1}{2}\right)^n\right\} \end{aligned}$
$\begin{aligned} & \qquad=1-\frac{(n+1)}{2^n} \\ & \because P(\text { at least } 2 \text { heads }) \geq 0.96 \\ & \Rightarrow 1-\frac{(n+1)}{2^n} \geq 0.96 \\ & \Rightarrow 0.04 \geq \frac{n+1}{2^n} \\ & \Rightarrow \frac{1}{25} \geq \frac{n+1}{2^n} \\ & \Rightarrow \frac{2^n}{n+1} \geq 25 \\ & \Rightarrow n \geq 8 \end{aligned}$
So, the minimum number of times a fair coin to be tossed is 8.
Hint:
(i) P (at least 2 heads) $=1-\{\mathrm{P}$ (one head} + $\mathrm{P}$(No heads)}
(ii) Use binomial probability theorem and simplify it.
Then ${{\Pr obability\,\,of\,\,occurrence\,\,of\,\,{E_1}} \over {\Pr obability\,\,of\,\,occurrence\,\,of\,\,{E_3}}}$
Explanation:
Given, three independent events $\mathrm{E}_1, \mathrm{E}_2$ and $\mathrm{E}_3$.
Probability that only
$\mathrm{E}_1 \text { occurs }=\mathrm{P}\left(\mathrm{E}_1 \cap \overline{\mathrm{E}}_2 \cap \overline{\mathrm{E}}_3\right)=\alpha$
Probability that only
$\mathrm{E}_2 \text { occurs }=\mathrm{P}\left(\mathrm{E}_1 \cap \overline{\mathrm{E}}_2 \cap \overline{\mathrm{E}}_3\right)=\beta$
Probability that only
$\mathrm{E}_3 \text { occurs }=P\left(\bar{E}_1 \cap \bar{E}_2 \cap E_3\right)=\gamma$
Probability that none of $E_1, E_2$ or $E_3$ occurs
$=P\left(\overline{\mathrm{E}}_1 \cap \overline{\mathrm{E}}_2 \cap \overline{\mathrm{E}}_3\right)=\rho$
$\begin{aligned} & \text { Let } \mathrm{P}\left(\mathrm{E}_1\right)=x, \mathrm{P}\left(\mathrm{E}_2\right)=y \text { and } \mathrm{P}\left(\mathrm{E}_3\right)=z \\ & \alpha=\mathrm{P}\left(\mathrm{E}_1\right) \cdot \mathrm{P}\left(\overline{\mathrm{E}}_2\right) \cdot \mathrm{P}\left(\overline{\mathrm{E}}_3\right) \\ & \Rightarrow \quad \alpha=x(1-y)(1-z) \quad \text{... (i)}\\ & \beta=\mathrm{P}\left(\overline{\mathrm{E}}_1\right) \cdot \mathrm{P}\left(\mathrm{E}_2\right) \cdot \mathrm{P}\left(\overline{\mathrm{E}}_3\right) \\ & \Rightarrow \quad \beta=(1-x) \cdot(y)(1-z) \quad \text{... (ii)}\\ & \gamma=\mathrm{P}\left(\overline{\mathrm{E}}_1\right) \cdot \mathrm{P}\left(\overline{\mathrm{E}}_2\right) \cdot \mathrm{P}\left(\mathrm{E}_3\right) \\ & \Rightarrow \quad \gamma=(1-x)(1-y) z \quad \text{... (iii)}\\ & \rho=\mathrm{P}\left(\overline{\mathrm{E}}_1\right) \cdot \mathrm{P}\left(\overline{\mathrm{E}}_2\right) \cdot \mathrm{P}\left(\overline{\mathrm{E}}_3\right) \\ & \Rightarrow \quad \rho=(1-x)(1-y)(1-z) \quad \text{... (iv)}\\ \end{aligned}$
Given $(\alpha-2 \beta) \rho=a \beta$ and $(\beta-3 \gamma) \rho=2 b \gamma$
$\begin{aligned} \Rightarrow \quad & {[x(1-y)(1-z)-2(1-x) y(1-z)](1-x) } \\ & (1-y)(1-z) \\ & =x(1-y)(1-z) \cdot(1-x) y(1-z) \text { and } \\ & {[(1-x) y(1-z)-3(1-x)(1-y) z](1-x) } \\ & (1-y)(1-z) \\ & =2(1-x) y(1-z) \cdot(1-x)(1-y) z \\ \Rightarrow \quad & {[x-x y-2 y+2 x y](1-x)(1-y)(1-z)^2 } \\ & =x y(1-x)(1-y)(1-z)^2 \text { and } \\ & {[y-y z-3 z+3 y z](1-x)^2(1-y)(1-z) } \\ & =2 y z(1-x)^2(1-y)(1-z) \\ \Rightarrow \quad & (x-2 y+x y)=x y \text { and }(y-3 z+2 y z)=2 y z \\ \Rightarrow \quad & x=2 y \text { and } y=3 z \\ \Rightarrow \quad & \frac{x}{2}=y=3 z \\ \Rightarrow \quad & \frac{x}{Z}=6 \end{aligned}$
$\Rightarrow \frac{\text { Probability of occurrence of } E_1}{\text { Probability of occurrence of } E_3}=\frac{x}{Z}=6$
Hints :
If $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are independent events, then probability of occurrence of only event A
$=\mathrm{P}(\mathrm{A} \cap \overline{\mathrm{B}} \cap \overline{\mathrm{C}})=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\overline{\mathrm{B}}) \cdot \mathrm{P}(\overline{\mathrm{C}})$.