Probability

$ \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. 1.$x \in A$ and $x \notin B$

2. 2.$x \notin A$ and $x \in B$

3. 3.$x \notin A$ and $x \notin B$

4. 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

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 $

Let the number of Mathematics books chosen be $a$ and the number of Physics books chosen be $b$.

Since a total of 6 books are selected, we have:

$a + b = 6$

The random variable is defined as the absolute difference between these two numbers:

$X = |a - b|$

Case 1: $X = 6$ This happens if all 6 books are Mathematics books and none are Physics books, or vice versa. But since only 6 Maths and 5 Physics books exist, only $a=6$, $b=0$ is possible.

$P(X = 6) = \frac{^6C_6 \times ^5C_0}{^{11}C_6} = \frac{1}{^{11}C_6}$

Case 2: $X = 4$ This happens if $(a,b) = (5,1)$ or $(a,b) = (1,5)$.

$P(X = 4) = \frac{^6C_5 \times ^5C_1 + ^6C_1 \times ^5C_5}{^{11}C_6} = \frac{36}{^{11}C_6}$

Case 3: $X = 2$ This happens if $(a,b) = (4,2)$ or $(a,b) = (2,4)$.

$P(X = 2) = \frac{^6C_4 \times ^5C_2 + ^6C_2 \times ^5C_4}{^{11}C_6} = \frac{225}{^{11}C_6}$

Case 4: $X = 0$ This happens if $(a,b) = (3,3)$.

$P(X = 0) = \frac{^6C_3 \times ^5C_3}{^{11}C_6} = \frac{200}{^{11}C_6}$

Mean of the random variable $X$:

$\alpha = \sum X \cdot P(X)$

Substituting the values:

$\alpha = \frac{6(1) + 4(36) + 2(225) + 0(200)}{^{11}C_6}$

$\alpha = \frac{100}{77}$

Hence,

$77\alpha = 100$

Total number of ways to select any $3$ different dates from $31$ days is

$ \binom{31}{3} $

Now we count the number of selections in which the three dates are in an increasing A.P.

Let the three dates be

$ a-d,\ a,\ a+d $

where $d \ge 1$ is an integer.

So we need all triples of the form

$ x,\ x+r,\ x+2r $

with

$ x \ge 1,\quad x+2r \le 31 $

For a fixed common difference $r$, the first term $x$ can be chosen from

$ 1,2,3,\dots,31-2r $

So number of A.P.s with common difference $r$ is

$ 31-2r $

Now $r$ can take values

$ 1,2,3,\dots,15 $

because

$ x+2r \le 31 \Rightarrow 2r \le 30 \Rightarrow r \le 15 $

Hence total favourable selections are

$ \sum_{r=1}^{15} (31-2r) $

This is

$ 15\cdot 31 - 2\sum_{r=1}^{15} r $

$ = 465 - 2\cdot \frac{15\cdot 16}{2} $

$ = 465 - 240 $

$ = 225 $

Therefore,

$ P=\frac{225}{\binom{31}{3}} $

Now,

$ \binom{31}{3}=\frac{31\cdot 30\cdot 29}{6}=4495 $

So,

$ P=\frac{225}{4495} $

Divide numerator and denominator by $5$:

$ P=\frac{45}{899} $

Thus,

$ a=45,\quad b=899 $

Therefore,

$ a+b=45+899=944 $

So the required answer is

$ \boxed{944} $

Let the outcomes of the 8 tosses be denoted by $X_1, X_2, X_3, X_4, X_5, X_6, X_7, X_8$.

The condition “exactly 4 heads in the first six tosses” means the block $(x_1, x_2, x_3, x_4, x_5, x_6)$ contains 4 heads.

The condition “exactly 3 heads in the last five tosses” means the block $(x_4, x_5, x_6, x_7, x_8)$ contains 3 heads.

Notice that these two blocks overlap in the common tosses $(x_4, x_5, x_6)$.

Let $k$ be the number of heads in the common tosses $(x_4, x_5, x_6)$.

Then:

• In $(x_1, x_2, x_3)$ we must have $4-k$ heads (so that the first six have total 4 heads).

• In $(x_7, x_8)$ we must have $3-k$ heads (so that the last five have total 3 heads).

Now we count the number of valid outcome sequences for each possible value of $k$.

$k=1$ ∴ ways to overlap = ${}^3C_1$

Here, we choose:

• ${}^3C_3$ ways to put $4-k=3$ heads in $(x_1, x_2, x_3)$,

• ${}^3C_1$ ways to put $k=1$ head in $(x_4, x_5, x_6)$,

• ${}^2C_2$ ways to put $3-k=2$ heads in $(x_7, x_8)$.

${}^3C_3 \times {}^3C_1 \times {}^2C_2 = 3 \text{ ways}$

$k=2$

Here, we choose:

• ${}^3C_2$ ways to put $4-k=2$ heads in $(x_1, x_2, x_3)$,

• ${}^3C_2$ ways to put $k=2$ heads in $(x_4, x_5, x_6)$,

• ${}^2C_1$ ways to put $3-k=1$ head in $(x_7, x_8)$.

${}^3C_2 \times {}^3C_2 \times {}^2C_1 = 18$

$k=3$

Here, we choose:

• ${}^3C_1$ ways to put $4-k=1$ head in $(x_1, x_2, x_3)$,

• $1$ way to put $k=3$ heads in $(x_4, x_5, x_6)$ (all heads),

• $1$ way to put $3-k=0$ heads in $(x_7, x_8)$ (all tails).

${}^3C_1 \times 1 = 3$

Total favourable outcomes $= 3+18+3 = 24$.

Since each full sequence of 8 tosses has probability $\left(\frac{1}{2}\right)^8$, we get:

Total ways $24\left(\frac{1}{2}\right)^8$

So $p = 24\left(\frac{1}{2}\right)^8 = \frac{24}{2^8} = \frac{24}{256} = \frac{3}{32}$, hence $96p = 96 \cdot \frac{3}{32} = 9$.

$96 \times \frac{24}{28} = \frac{6 \times 24}{16} = 9$

For the quadratic expression

$ ax^2 + 2\sqrt{2}\,bx + c $

to be positive for all $x \in \mathbb{R}$, we use the NCERT condition for a quadratic $Ax^2+Bx+C$ to satisfy

$ Ax^2+Bx+C > 0 \quad \text{for all } x \in \mathbb{R}, $

which is:

1. $A>0$

2. Discriminant $<0$

Here,

$ A=a,\quad B=2\sqrt{2}\,b,\quad C=c $

Since $a \in \{1,2,3,4\}$, we always have

$ a>0 $

So we only need

$ D = B^2 - 4AC < 0 $

Now,

$ D=(2\sqrt{2}\,b)^2 - 4ac = 8b^2 - 4ac $

Thus,

$ 8b^2 - 4ac < 0 $

Dividing by $4$:

$ 2b^2 - ac < 0 $

So the required condition is

$ ac > 2b^2 $

Now $a,b,c \in \{1,2,3,4\}$, so total number of choices is

$ 4 \times 4 \times 4 = 64 $

We count favourable triples $(a,b,c)$.

Case 1: $b=1$

Then

$ ac > 2 $

Now $a,c \in \{1,2,3,4\}$. Total pairs $(a,c)$ are $16$.

We exclude pairs with $ac \le 2$:

• $(1,1)$ gives $ac=1$

• $(1,2)$ gives $ac=2$

• $(2,1)$ gives $ac=2$

So favourable pairs:

$ 16-3=13 $

Case 2: $b=2$

Then

$ ac > 8 $

Possible products greater than $8$ are obtained from:

• $(3,3)$ gives $9$

• $(3,4)$ gives $12$

• $(4,3)$ gives $12$

• $(4,4)$ gives $16$

So favourable pairs = $4$.

Case 3: $b=3$

Then

$ ac > 18 $

But maximum possible value of $ac$ is

$ 4 \cdot 4 = 16 $

So no favourable pair.

Case 4: $b=4$

Then

$ ac > 32 $

Again impossible, since maximum $ac=16$.

So no favourable pair.

Hence total favourable triples are

$ 13+4+0+0 = 17 $

Therefore, the required probability is

$ \frac{17}{64} $

Thus,

$ m=17,\quad n=64 $

and

$ m+n=17+64=81 $

So the answer is

$ \boxed{81} $

$P\left(\frac{\text { Lost }_{\text {(spade })}}{\mathrm{n} \text { cards are spade }}\right)$

$\begin{aligned} & =\frac{P\left(\frac{n_s}{L_s}\right) P\left(L_s\right)}{P\left(\frac{n_s}{L_s}\right) P\left(L_s\right)+P\left(\frac{n_s}{\bar{L}_s}\right) P\left(\bar{L}_s\right)} \\ & =\frac{\frac{{ }^{12} C_n}{{ }^{51} C_n} \times \frac{1}{4}}{\frac{{ }^{12} C_n}{{ }^{51} C_n} \times \frac{1}{4}+\frac{3}{4} \times \frac{{ }^{13} C_n}{{ }^{51} C_n}}=\frac{1}{1+3 \cdot \frac{{ }^{13} C_n}{{ }^{12} C_n}}=\frac{13-n}{52-n} \\ & \Rightarrow \frac{13-n}{52-n}=\frac{11}{50} \\ & \Rightarrow n=2 \end{aligned}$

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 $m=7,\; n=2470 \Rightarrow m+n=2477.$

$H_1$: 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.

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)

Total for $a = 1 = 1 + 2 + 4 = 7$

• 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)

Total for $a = 2 = 1 + 2 = 3$

• 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)

Total for $a = 3 = 1$

• 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 $a = 4 = 1$

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.

$\begin{aligned} & \bar{X}=\sum X p(X) \\ & \bar{Y}=\sum Y p(Y) \\ & P(X=3)=P(Y=0)=\frac{{ }^5 C_3 \cdot C_0}{{ }^9 C_3}=\frac{{ }^5 C_2}{{ }^9 C_3}=\frac{5}{42} \\ & P(X=2)=P(Y=1)=\frac{{ }^5 C_2 \cdot C_1}{{ }^9 C_3}=\frac{10}{21} \\ & P(X=1)=P(Y=2)=\frac{{ }^5 C_1 \cdot C_2}{{ }^9 C_3}=\frac{5}{14} \\ & P(X=0)=P(Y=3)=\frac{{ }^5 C_0 \cdot C_3}{{ }^9 C_3}=\frac{4}{84}=\frac{1}{21} \\ & \bar{X}=3 \times \frac{5}{42}+2 \times \frac{10}{21}+\frac{5}{14}+0 \times \frac{1}{21}=\frac{15+40+15}{42}=\frac{70}{42} \\ & \bar{Y}=0 \times \frac{5}{42}+1 \times \frac{10}{21}+2 \times \frac{5}{14}+3 \times \frac{1}{21}=\frac{20+30+6}{42}=\frac{56}{42} \\ & 7 \bar{X}+4 \bar{Y}=17 \end{aligned}$

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)

$\begin{aligned} & \frac{9}{12} \cdot \frac{8}{11} \cdot \frac{7}{10} \cdot \frac{6}{9} \cdot \frac{5}{8}=\frac{21}{132}=\frac{7}{44} \\ & P(X=1)=5\left[\frac{9 \cdot 8 \cdot 7 \cdot 6 \cdot 3}{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8}\right]=\frac{21}{44} \\ & P(X=2)=5\left[\frac{9 \cdot 8 \cdot 7 \cdot 3 \cdot 2}{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8}\right]=\frac{14}{44} \\ & P(X=3)=5\left[\frac{3 \cdot 2 \cdot 1 \cdot 9 \cdot 8}{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8}\right]=\frac{2}{44} \\ & P(X=4)=0 \\ & P(X=5)=0 \end{aligned}$

$\begin{aligned} & \sigma_x^2=\sum X^2 P(x)-\left(\sum x P(x)\right)^2 \\ & =\frac{95}{44}-\left(\frac{55}{44}\right)^2 \\ & =\frac{4180-3025}{1936}=\frac{1155}{1936}=\frac{105}{176}=\frac{m}{n} \\ & =n-m=71 \end{aligned}$

$\begin{aligned} & \mu=\sum x P(x)=0+\frac{5}{12}+\frac{10}{12}+\frac{3}{12}=\frac{3}{2} \\ & \sigma^2=\sum(x-\mu) P(x)=\sum\left(x-\frac{3}{2}\right)^2 P(x) \\ & =\frac{9}{4} \times \frac{1}{12}+\frac{1}{4} \times \frac{5}{12}+\frac{1}{4} \times \frac{5}{12}+\frac{9}{4} \times \frac{1}{12}=\frac{7}{12} \end{aligned}$

$\Rightarrow \sigma^2 \cdot 96=8 \times 7=56$

$\begin{aligned} & x+y=10 \\ & A=x-y \\ & P(|A|<2) \text { is } P \\ & \Rightarrow|A|=2,1,0 \Rightarrow A=0,1,-1,2,-2 \\ & \Rightarrow x=\frac{10+A}{2} \Rightarrow A \in \text { even as } x \in \text { integer } \\ & \Rightarrow A=0,-2,2 \\ & \Rightarrow P(|A| \leq 2)=P(A=0)+P(A=-2)+P(A=2) \end{aligned}$

(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

$\begin{aligned} & \text { Similarly, } P(A=2)={ }^{10} C_6\left(\frac{1}{3}\right)^6\left(\frac{2}{3}\right)^4 \\ & \Rightarrow P(|A| \leq 2) 3^9=3\left({ }^{10} C_5 \cdot 2^5+{ }^{10} C_4 \cdot 2^6+{ }^{10} C_6 \cdot 2^4\right) \\ & =8288 \end{aligned}$

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, $\frac{b+c}{a} = 12$.

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 equation of line is $\mathrm{y=mx+c}$

$\sum_{x=0}^4(m x+c)=1 \Rightarrow 10 m+5 c=1 \Rightarrow 2 m+c=\frac{1}{5}\quad \text{.... (i)}$

$\begin{aligned} & \text { mean }=\sum x_i P_i=\sum_{i=0}^4\left(m x_i+c\right) \cdot x_i=30 m+10 c=\frac{5}{2} \\ & \therefore 3 m+c=\frac{1}{4} \ldots(2) \\ & \text { from (1) and (2) } \mathrm{m}=\frac{1}{20}, \mathrm{c}=\frac{1}{10} \\ & \Sigma \mathrm{P}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}^2=\sum_{\mathrm{i}=0}^4\left(\mathrm{mx}_{\mathrm{i}}+\mathrm{c}\right) \mathrm{x}_1^2 \\ & =\sum_{\mathrm{i}=0}^4\left(\mathrm{mx}_{\mathrm{i}}^3+c \mathrm{x}_{\mathrm{i}}^2\right) \Rightarrow 100 \mathrm{~m}+30 \mathrm{c}(\text { Now putting } \mathrm{m} \text { and } \mathrm{c}) \\ & \Rightarrow \Sigma \mathrm{P}_{\mathrm{i}}^2=5+3=8 \\ & \text { Variance }=\Sigma \mathrm{P}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}^2-\left(\Sigma \mathrm{P}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}\right)^2=8-\left(\frac{5}{2}\right)^2=\frac{7}{4} \\ & \therefore 24 \alpha=42 \end{aligned}$

$p = {6 \over {36}} = {1 \over 6}$

$q = {{{}^6{C_1} \times {}^5{C_1} \times {{4!} \over {3!}}} \over {{6^4}}} = {{120} \over {1296}} = {5 \over {54}}$

${p \over q} = {{{1 \over 6}} \over {{5 \over {54}}}} = {{54} \over {6 \times 5}} = {9 \over 5} = {m \over n}$

$m + n = 14$

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)

Total number of ways of selecting 2 persons $={ }^{49} \mathrm{C}_2$

Number of ways in which 2 friends are selected $=6 \times 7 \times 2=84$

$7 \mathrm{P}=\frac{84 \times 2}{49 \times 48} \times 7=\frac{1}{2}$

$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}}$

$100{\sigma ^2} = 56.$

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 $d = - {1 \over 3}$, $E(x) = {{6 + 3 \times - {1 \over 3}} \over 4} = {5 \over 4}$

$\therefore$ $60E(x) = 60 \times {5 \over 4} = 75$

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

Total number of numbers from given

Condition = n(s) = 2⁶.

Every required number is of the form

A = 7 . (10^a₁ + 10^a₂ + 10^a₃ + .......) + 111111

Here 111111 is always divisible by 21.

$\therefore$ If A is divisible by 21 then

10^a₁ + 10^a₂ + 10^a₃ + ....... 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 = ${{22} \over {{2^6}}} = p$

$\therefore$ ${{11} \over {32}} = p$

$\therefore$ $96p = 33$

Student guesses only two wrong. So there are three possibilities.

(i) Student guesses both wrong from 1^st section

(ii) Student guesses both wrong from 2^nd 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}}}}$

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.

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

$=\mathrm{P}(\mathrm{A} \cap \overline{\mathrm{B}} \cap \overline{\mathrm{C}})=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\overline{\mathrm{B}}) \cdot \mathrm{P}(\overline{\mathrm{C}})$.