Probability

We consider all distinct rational numbers $\frac{p}{q}$ such that $p, q \in\{1,2,3,4,5$, $6\}$ and GCD $(p, q)=1$

Now, we will consider each possible $q$ from 1 to 6 and for each, count the $p \in \{1,2,3,4,5,6\}$ such that g.c.d. $(p, q) =1$

$ \begin{array}{clc} \hline \boldsymbol{q} & \text { Valid } \boldsymbol{p} \text { (C.G.D. }(\boldsymbol{p}, \boldsymbol{q})=\mathbf{1} & \text { Count } \\ \hline 1 & 2,3,4,5,6 & 5 \\ \hline 2 & 1,3,5 & 3 \\ \hline 3 & 1,2,4,5 & 4 \\ \hline 4 & 1,3,5 & 3 \\ \hline 5 & 1,2,3,4,6 & 5 \\ \hline 6 & 1,5 & 2 \\ \hline \text { Total } & & \mathbf{2 2} \\ \hline \end{array} $

So, there are 22 distinct reduced rational numbers of the form $\frac{p}{q}$.

As we know that, a proper fraction is one where $p

Among the 22 reduced fractions, we will now count how many satisfy $p

When, $q=1 ; p=2,3,4,5,6 \rightarrow$ all $p>q \rightarrow$ None proper fraction.

when $q=2, p=1,3,5 \rightarrow$ only $p=1 \rightarrow 1$ proper fraction.

When $q=3 ; p=1,2,4,5 \rightarrow p=1,2 \rightarrow 2$ proper fraction.

when $q=4 ; p=1,3,5 \rightarrow p=1,3 \rightarrow 2$ proper fraction when $q=5 ; p=1,2,3,4,6 \rightarrow p=1,2$, $3,4 \rightarrow 4$ proper fraction when $q=6, p=1,5 \rightarrow p=1,5 \rightarrow$ both $<6 \rightarrow 2$ proper fraction

$\therefore$ Total proper fraction

$ =1+2+2+4+2=11 $

So, required probability

$ \begin{aligned} & =\frac{\text { Number of proper fraction }}{\text { Total distinct reduced fractions }} \\ & =\frac{11}{22}=\frac{1}{2} \end{aligned} $

$ \begin{aligned} & \text { } \Sigma x^2 \cdot P(X=x)=(-1)^2 \cdot \frac{1}{3}+0^2 \cdot \frac{1}{6} +1^2 \cdot \frac{1}{6}+2^2 \cdot \frac{1}{3}=1 \cdot \frac{1}{3}+0+1 \times \frac{1}{6}+4 \times \frac{1}{3} =\frac{11}{6} \\ & \begin{aligned} E(X)=\Sigma x \cdot P(X)=(-1) \cdot \frac{1}{3}+0 \cdot \frac{1}{6}+1 & \times \frac{1}{6} +2 \times \frac{1}{3}=\frac{1}{2} \end{aligned} \\ & \begin{aligned} \therefore \operatorname{var}(X) & =E\left(X^2\right)-(E(X))^2 \\ & =\frac{11}{6}-\frac{1}{4}=\frac{19}{12} \end{aligned} \end{aligned} $

Hence, $6 \cdot \Sigma x^2 \cdot P(X=x)-\operatorname{var}(X)$

$ =6 \times \frac{11}{6}-\frac{19}{12}=\frac{132-19}{12}=\frac{113}{12} $

Average number of accidents per weeks $(\lambda)=5$

We want to find the probability of at most one accident i.e.

$ P(X \leq 1)=P(0)+P(1) . $

As we know that, poisson probability formula,

$ \begin{aligned} P(X & =k)=\frac{e^{-\lambda} \lambda^k}{k!} \\ \therefore \quad & P(0)=\frac{e^{-5} \cdot 5^0}{0!}=e^{-5} \\ P(1) & =\frac{e^{-5} \cdot 5^1}{1!}=5 e^{-5} \\ \therefore P(X \leq 1) & =e^{-5}(1+5)=6 e^{-5}=\frac{6}{e^5} \end{aligned} $

Total number of outcomes

$ =2^8=256=n(S) $

Favourable outcomes $n(E)=10$

$ \Rightarrow P(E)=\frac{n(E)}{n(S)}=\frac{10}{256}=\frac{5}{128} $

Total number of ways

$ ={ }^{12} C_3=\frac{12!}{3!\times 9!}=220 $

Number of ways to draw exactly 2 red balls and 1 non-red ball

$ ={ }^4 C_2 \times{ }^8 C_1=6 \times 8=48 $

Number of ways to draw exactly 2 green balls and 1 non-green ball

$ \begin{aligned} & ={ }^5 C_2 \times{ }^7 C_1 \\ & =\frac{5!}{2!\times 3!} \times \frac{7!}{1!6!}=10 \times 7=70 \end{aligned} $

Number of ways to choose 2 white balls from 3.

$ ={ }^3 C_2 $

Number of ways to draw exactly 2 white balls and 1 non-white ball

$ \begin{aligned} & ={ }^3 C_2 \times{ }^9 C_1=\frac{3!}{2!\times 1!} \times \frac{9!}{1!8!} \\ & =3 \times 9=27 \end{aligned} $

Total number of ways to draw exactly 2 balls of the same colour

$ =48+70+27=145 $

$\therefore$ Required probability $=\frac{145}{220}=\frac{29}{44}$

We want to find the chance that the selected girl comes from $F_2$.

Step 1: Find the chance of picking each family.
There are three families, so the chance of picking any one family ($F_1$, $F_2$, or $F_3$) is $\frac{1}{3}$.

Step 2: Find the chance of picking a girl from each family.

• For $F_1$ (2 boys, 1 girl): Probability of picking a girl = $\frac{1}{3}$
• For $F_2$ (1 boy, 2 girls): Probability of picking a girl = $\frac{2}{3}$
• For $F_3$ (1 boy, 1 girl): Probability of picking a girl = $\frac{1}{2}$

Step 3: Find the total probability of picking a girl (from any family).

To get the total chance of picking a girl, add up the chance for each family:

$ P(G) = P(G|F_1) \cdot P(F_1) + P(G|F_2) \cdot P(F_2) + P(G|F_3) \cdot P(F_3) $

$ = \frac{1}{3} \cdot \frac{1}{3} + \frac{2}{3} \cdot \frac{1}{3} + \frac{1}{2} \cdot \frac{1}{3} $

$ = \frac{1}{9} + \frac{2}{9} + \frac{1}{6} = \frac{1+2+3}{18} = \frac{6}{12} = \frac{1}{2} $

Step 4: Use Bayes’ Theorem to find the chance that the girl is from $F_2$.
Bayes’ Theorem tells us: $ P(F_2|G) = \frac{P(G|F_2) \cdot P(F_2)}{P(G)} $ $ = \frac{\frac{2}{3} \cdot \frac{1}{3}}{\frac{1}{2}} $ $ = \frac{\frac{2}{9}}{\frac{1}{2}} = \frac{2}{9} \times 2 = \frac{4}{9} $

Final Answer:
The probability that the selected girl is from $F_2$ is $\frac{4}{9}$.

The probability of transferring a white ball from $A$ to $B=\frac{4}{5}$

The probability of transferring a black ball from $A$ to $B=\frac{1}{5}$

The probability of drawing a black ball from $C$ after transferring a white ball from $A$ to $B$

$ =\frac{4}{6} \cdot \frac{3}{6}+\frac{2}{6} \cdot \frac{4}{6}=\frac{12}{36}+\frac{8}{36}=\frac{20}{36} $

The probability of drawing a black ball from $C$ after transferring a black ball from $A$ to $B$

$ =\frac{3}{6} \cdot \frac{3}{6}+\frac{3}{6} \cdot \frac{4}{6}=\frac{9}{36}+\frac{12}{36}=\frac{21}{36} $

The total probability of drawing a black ball from $C$ is

$ =\frac{4}{5} \cdot \frac{20}{36}+\frac{1}{5} \cdot \frac{21}{36}=\frac{101}{180} $

$ \begin{aligned} & \text { } \sum_{k=0}^{\infty} P(X=k)=1 \\ & \Rightarrow \sum_{K=0}^{\infty} \frac{2^{-k}(3 k+1)}{2^C}=1 \\ & \Rightarrow \frac{1}{2^C}\left(3 \sum_{k=0}^{\infty} k \cdot 2^{-k}+\sum_{k=0}^{\infty} 2^{-k}\right)=1 \\ & \Rightarrow \frac{1}{2^C}\left(3 \cdot \frac{2^{-1}}{\left(1-2^{-1}\right)^2}+\frac{1}{1-2^{-1}}\right)=1 \\ & \Rightarrow \frac{1}{2^C}\left(\frac{\frac{3}{2}}{\frac{1}{4}}+\frac{1}{\frac{1}{2}}\right)=1 \end{aligned} $

$ \begin{aligned} & \Rightarrow \frac{1}{2^c}(6+2)=1 \Rightarrow \frac{8}{2^c}=1 \\ & \Rightarrow 2^c=8 \Rightarrow 2^c=2^3 \\ & \Rightarrow c=3 \\ & \begin{aligned} P(X \leq c)=P(X \leq 3) & =P(X=0)+P(X=1) +P(X=2)+P(X=3) \end{aligned} \end{aligned} $

$ \begin{aligned} & \begin{aligned} =\frac{1}{2^3} & +\frac{2^{-1} \times 4}{2^3}+\frac{2^{-2} \times 7}{2^3}+\frac{2^{-3} \times 10}{2^3} \\ & =\frac{1}{2^3}+\frac{4}{2^4}+\frac{7}{2^5}+\frac{10}{2^6} \\ & =\frac{2^3+16+14+10}{2^6} \\ & =\frac{8+16+14+10}{2^6}=\frac{48}{64}=\frac{3}{4} \end{aligned} \\ & P(X \leq c)=\frac{3}{4}=\frac{c}{4} \end{aligned} $

We know that $n = 4$ (there are 4 trials) and $P(X=0) = \frac{16}{81}$ (the chance of getting zero successes is $\frac{16}{81}$).

In a binomial distribution, $p$ is the chance of success each time, and $q$ is the chance of failure. $q = 1 - p$.

The formula to find the probability of getting $k$ successes is: $P(X=k) = {^n C_k} p^k q^{n-k}$.

Step 1: Find $q$

$P(X=0) = {^4 C_0} p^0 q^4$
${^4 C_0} = 1$ and $p^0 = 1$, so:
$P(X=0) = 1 \cdot 1 \cdot q^4 = q^4$
We know $q^4 = \frac{16}{81}$.

To find $q$, take the fourth root of both sides:
$q = \frac{2}{3}$ because $(\frac{2}{3})^4 = \frac{16}{81}$.

Step 2: Find $p$

Since $q = \frac{2}{3}$,
$p = 1 - \frac{2}{3} = \frac{1}{3}$.

Step 3: Find $P(X=4)$

$P(X=4)$ is the probability of getting 4 successes out of 4 tries.
$P(X=4) = {^4 C_4} p^4 q^0$
${^4 C_4} = 1$ and $q^0 = 1$,
So:
$P(X=4) = 1 \cdot \left(\frac{1}{3}\right)^4 \cdot 1 = \left(\frac{1}{3}\right)^4$

Calculate $(\frac{1}{3})^4 = \frac{1}{81}$.

Final Answer:

$P(X=4) = \frac{1}{81}$

$ \begin{aligned} &\text { Given that, }\\ &\begin{aligned} P(A) & =\frac{2}{3}, \quad P(B)=\frac{3}{4} \\ P(\text { not } A)=P\left(A^{\prime}\right) & =1-\frac{2}{3}=\frac{1}{3} \\ P\left(B^{\prime}\right)=1-\frac{3}{4} & =\frac{1}{4} \\ P(A \text { or } B) \text { or both }) & =1-P\left(A^{\prime}\right) P\left(B^{\prime}\right) \\ & =1-\frac{1}{3} \times \frac{1}{4}=\frac{11}{12} \end{aligned} \end{aligned} $

Given set $\{1,2,3,4,5,6,7\}$

Two number can be taken from set $=49$ ways.

Now, for $l x^2+m x+1>0$

$ m^2-4 l<0 \Rightarrow m^2<4 l $

If $l=1 \Rightarrow m^2<4 \Rightarrow m=1$

$ \begin{aligned} & l=2 \Rightarrow m^2<8 \Rightarrow m=1,2 \\ & l=3 \Rightarrow m^2<12 \Rightarrow m=1,2,3 \\ & l=4 \Rightarrow m^2<16 \Rightarrow m=1,2,3 \\ & l=5 \Rightarrow m^2<20 \Rightarrow m=1,2,3,4 \\ & l=6 \Rightarrow m^2<24 \Rightarrow m=1,2,3,4 \\ & l=7 \Rightarrow m^2<28 \Rightarrow m=1,2,3,4,5 \end{aligned} $

$\therefore$ Required probability $=\frac{22}{49}=\frac{22}{7^2}$

$ \begin{aligned} &\text { Given that, }\\ &\begin{aligned} & \quad P(A \text { wins })=0.6 \\ & P(A \text { loses })=0.3 \\ & P(\text { game draw })=0.1 \\ & P(\text { A win at least two game }) \\ & \Rightarrow \quad P(W W W)+P(W W L)+P(W L W) \\ & \quad+P(L W W)+P(W W D) \\ & \quad+P(D W W)+P(W D W) \end{aligned} \end{aligned} $

$ \begin{aligned} & =(0.6)^3+3(0.6)^2(0.3)+3(0.6)^2(0.1) \\ & =(0.6)^2(0.6+0.9+0.3) \\ & =0.36(1.8)=0.648=\frac{648}{1000}=\frac{81}{125} \end{aligned} $

Given that

$U r n U_1$ contains 5 red, 3 white, 2 black balls

$U r n U_2$ contains 4 red, 4 white, 2 black balls

$U r n U_3$ contains 3 red, 4 white, 3 black balls

Total number of balls $=30$

Non-black balls = 23

Total probability of selecting one $\operatorname{Ur} n=\frac{1}{3}$

Probability of getting not a black balls from $U_1$ or $U_2$ and $U_3$.

$ =\frac{1}{3}\left(\frac{8}{10}+\frac{8}{10}+\frac{7}{10}\right) $

$\therefore$ Required probability $=\frac{1}{3} \times \frac{23}{10}=\frac{23}{30}$

$ \begin{aligned} &\text { We know, } \Sigma P_i=1\\ &\begin{aligned} & \therefore \quad 3 k+5 k+3 k^2+4 k^2+k+3 k^2=1 \\ & \Rightarrow \quad 10 k^2+9 k-1=0 \\ & \Rightarrow \quad(10 k-1)(k+1)=0 \\ & \Rightarrow \quad k=\frac{1}{10} \\ & \quad k \neq-1 \\ & P(X \leq 2)=P(X=0)+P(X=2)+P(X=3) \\ & =3 k+5 k+3 k^2=\frac{8}{10}+\frac{3}{100}=\frac{83}{100} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { } \text { Variance }=2=\lambda\\ &\sigma=2 \end{aligned} $

$ \begin{aligned} & P(X=k)=\frac{e^{-\lambda} \lambda^k}{k!} \\ & \begin{aligned} P(X \geq 3) & =1-P(X=0)-P(X=1) \\ -P(X & =2) \\ & =1-\left(\frac{e^{-2} 2^0}{0!}+\frac{e^{-2} 2^1}{1!}+\frac{e^{-2} 2^2}{2!}\right) \\ & =1-5 e^{-2}=\frac{e^2-5}{e^2} \end{aligned} \end{aligned} $

Given, $P(A)=\frac{2}{5}, P(B)=\frac{3}{4}$

$P($ Problem is solved $)=1-P($ Problem is not solve)

$ \begin{aligned} P(A \cup B) & =1-P\left(A^{\prime} \cap B^{\prime}\right) \\ & =1-P(A) \cdot P(B) \\ & =1-\frac{3}{5} \times \frac{1}{4}=\frac{20-3}{20}=\frac{17}{20} \end{aligned} $

$A=$ getting a sum $>14$

$B=$ sum is multiple of 3

$ \begin{aligned} & n(A)=14 \quad P(A)=\frac{14}{216} \\ & \begin{array}{l} n(B)=78 \quad P(B)=\frac{78}{216} \\ n(A \cap B)=11 P(A \cap B)=\frac{11}{216} \\ \begin{aligned} \because P(A & \cap \bar{B})+P(\bar{A} \cap B) \\ & =P(A)+P(B)-2 P(A \cap B) \\ & =\frac{14}{216}+\frac{78}{216}-\frac{11 \times 2}{216} \\ & =\frac{92-22}{216}=\frac{70}{216}=\frac{35}{108} \end{aligned} \end{array} \end{aligned} $

Let $E_1$ mean the bulb was made by unit $A$.

Let $E_2$ mean the bulb was made by unit $B$.

Let $E_3$ mean the bulb was made by unit $C$.

Let $E$ mean the bulb is defective.

The chance a bulb comes from each unit is:
$P\left(E_1\right) = \frac{25}{100}$ (from $A$), $P\left(E_2\right) = \frac{35}{100}$ (from $B$), and $P\left(E_3\right) = \frac{40}{100}$ (from $C$).

The chance a bulb is defective from each unit is:
$P\left(\frac{E}{E_1}\right) = \frac{5}{100}$, $P\left(\frac{E}{E_2}\right) = \frac{4}{100}$, $P\left(\frac{E}{E_3}\right) = \frac{2}{100}$.

To find how likely it is a defective bulb came from unit $B$, use Bayes’ theorem:

$ P\left(\frac{E_2}{E}\right) = \frac{P\left(E_2\right) P\left(\frac{E}{E_2}\right)}{P\left(E_1\right) P\left(\frac{E}{E_1}\right) + P\left(E_2\right) P\left(\frac{E}{E_2}\right) + P\left(E_3\right) P\left(\frac{E}{E_3}\right) } $

Now plug in the numbers:

$ P\left(\frac{E_2}{E}\right) = \frac{\frac{35}{100} \times \frac{4}{100}}{\frac{25}{100} \times \frac{5}{100} + \frac{35}{100} \times \frac{4}{100} + \frac{40}{100} \times \frac{2}{100}} $

The top ($35 \times 4 = 140$) and the bottom adds up ($25 \times 5 = 125$, $35 \times 4 = 140$, $40 \times 2 = 80$) so:

$ P\left(\frac{E_2}{E}\right) = \frac{140}{125 + 140 + 80} $

$ P\left(\frac{E_2}{E}\right) = \frac{140}{345} = \frac{28}{69} $

$ \begin{array}{ccccc} \hline \boldsymbol{X}_{\boldsymbol{i}} & \boldsymbol{P}_{\boldsymbol{i}} & \boldsymbol{X}_{\boldsymbol{i}} \boldsymbol{P}_{\boldsymbol{i}} & \boldsymbol{X}_{\boldsymbol{i}}{ }^{\boldsymbol{2}} & \boldsymbol{P}_{\boldsymbol{i}} \boldsymbol{X}_{\boldsymbol{i}}{ }^{\boldsymbol{2}} \\ \hline 1 & \alpha & \alpha & 1 & \alpha \\ \hline 2 & \alpha & 2 \alpha & 4 & 4 \alpha \\ \hline 3 & \alpha & 3 \alpha & 9 & 9 \alpha \\ \hline 4 & \beta & 4 \beta & 16 & 16 \beta \\ \hline 5 & \beta & 5 \beta & 25 & 25 \beta \\ \hline 6 & 0.3 & 1.8 & 36 & 10.8 \\ \hline \end{array} $

$ \begin{array}{ll} \because & \Sigma P_i=1 \\ & 3 \alpha+2 \beta+0.3=1 \\ & 3 \alpha+2 \beta=0.7 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \because & \mu=\Sigma X_i P_i \\ & 6 \alpha+9 \beta+1.8=4.2 \\ & 2 \alpha+3 \beta=0.8 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $

Solving Eqs. (i) and (ii) we get,

$ \begin{aligned} \operatorname{var}(x) & =\sigma^2=\Sigma P_i X_i^2-(\mu)^2 \\ & \sigma^2+\mu^2=\Sigma P_i X_i^2 \\ & =14 \alpha+41 \beta+10.8 \\ & =1.4+8.2+10.8=20.4 \end{aligned} $

$X=$ Students get distinction

$ \begin{gathered} p=\frac{2}{3}, q=1-p=\frac{1}{3}, n=5 \\ P(X \geq 3)=P(X=3)+P(X=4)+P(X=5)\end{gathered} $

We know that, $P(X=r)={ }^n C_r p^r q^{n-r}$

$ \begin{aligned} & P(X=3)={ }^5 C_3\left(\frac{2}{3}\right)^3\left(\frac{1}{3}\right)^2=10 \times \frac{8}{3^5}=\frac{80}{3^5} \\ & P(X=4)={ }^5 C_4\left(\frac{2}{3}\right)^4 \cdot \frac{1}{3}=\frac{5 \times 16}{3^5}=\frac{80}{3^5} \\ & P(X=5)={ }^5 C_5\left(\frac{2}{3}\right)^5=\frac{32}{3^5} \\ & P(X \geq 3)=\frac{192}{3^5}=\frac{64}{3^4}=\frac{64}{81} \end{aligned} $

Given, $P(A \cup B)=\frac{3}{4} P(A \cap B)=\frac{1}{4}$

$ P(\bar{A})=\frac{2}{3} \Rightarrow P(A)=\frac{1}{3} $

$ (\because P(B)=P(A \cup B)+P(A \cap B)-P(A)) $

Now, $P(B)=\frac{2}{3}$

Now, $P(\bar{A} \cap B)=P(B)-P(A \cap B)$

$ =\frac{2}{3}-\frac{1}{4}=\frac{8-3}{12}=\frac{5}{12} $

$A=$ Both cards are black

$B=$ At least one card is a face card.

$ P(A)=\frac{26}{52} \times \frac{25}{51}=\frac{650}{2652} $

$\because 20$ Black card are not a face card.

$\because P$ (two black non-face card)

$ =\frac{20}{52} \times \frac{19}{51}=\frac{380}{2652} $

$P(B)=P($ two black card $)$

$-P$ (two non-face card)

$ =\frac{650}{2652}-\frac{380}{2652}=\frac{270}{2652} $

$\because P\left(\frac{B}{A}\right)=\frac{\frac{270}{2652}}{\frac{650}{2652}}=\frac{270}{650}=\frac{27}{65}$

The person tells the truth 3 out of every 4 times.

Let’s define:
- $A$: The die really shows six.
- $B$: The die does not show six.
- $E$: The person says the die shows six.

The chance of rolling a six when throwing a die, $P(A)$, is $\frac{1}{6}$.

The chance of not rolling a six, $P(B)$, is $\frac{5}{6}$.

If a six really comes up, the chance the person will say “six” (tells the truth) is $P(E|A) = \frac{3}{4}$.

If a six did NOT come up, the chance the person wrongly says “six” (tells a lie) is $P(E|B) = \frac{1}{4}$.

We want to find: What is the chance the die REALLY shows six if the person says it is a six? This is $P(A|E)$.

Use Bayes’ Theorem:
$ P(A|E) = \frac{P(A) \cdot P(E|A)}{P(A)\cdot P(E|A) + P(B)\cdot P(E|B)} $

Put in the values:
$ P(A|E) = \frac{\frac{1}{6} \times \frac{3}{4}}{\frac{1}{6} \times \frac{3}{4} + \frac{5}{6} \times \frac{1}{4}} $

Solve:
$ P(A|E) = \frac{\frac{3}{24}}{\frac{3}{24} + \frac{5}{24}} = \frac{3}{8} $

Let $E_1=$ Employee be a man $E_2=$ Employee be a woman $E=$ Employee be a technical assistants

$ \begin{aligned} Given\,\,& \mathrm{n}\left(E_1\right)=\frac{70}{100}=\frac{7}{10}, P\left(E_2\right)=\frac{3}{10} \\ & P\left(\frac{E}{E_1}\right)=\frac{3}{10} \\ & P\left(\frac{E}{E_2}\right)=\frac{15}{100} \end{aligned} $

$ \begin{aligned} P\left(\frac{E_1}{E}\right) & =\frac{P\left(E_1\right) P\left(\frac{E}{E_1}\right)}{P\left(E_1\right) P\left(\frac{E}{E_1}\right)+P\left(E_2\right) P\left(\frac{E}{E_2}\right)} \\ & =\frac{\frac{7}{10} \times \frac{3}{10}}{\frac{7}{10} \times \frac{3}{10}+\frac{3}{10} \times \frac{15}{100}} \\ & =\frac{210}{210+45}=\frac{210}{255} \\ P\left(\frac{E_1}{E}\right) & =\frac{42}{51}=\frac{14}{17} \end{aligned} $

Given, $P(X=x)=K \cdot \frac{2^{2 x+1}}{(2 x+1)!}$

$ \begin{aligned} & \because \quad \sum P(X=x)=1 \\ & \Rightarrow \sum_{x=0}^{\infty} k \cdot \frac{2^{2 x+1}}{(2 x+1)!}=1 \\ & \Rightarrow 2 k \sum_{x=0}^{\infty} \frac{4^x}{(2 x+1)!}=1 \end{aligned} $

We know that $\sin h x=\sum_{n=0}^{\infty} \frac{x^{2 n+1}}{(2 n+1)!}$

$ \begin{aligned} \frac{\sin h x}{x} & =\sum_{n=0}^{\infty} \frac{x^{2 n}}{(2 n+1)!} \\ \because \quad \sum_{x=0}^{\infty} \frac{4^x}{(2 x+1)!} & =\frac{\sin h(2)}{2} \\ 2 k \cdot \frac{\sin h(2)}{2} & =1 \\ k & =\frac{1}{\sin h 2}=\operatorname{cosech} 2 \end{aligned} $

Given, mean - variance $=1$

$ \begin{aligned} & n p-n p q=1 \\ & n p(1-q)=1 \\ & n p^2=1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

and given $2 P(X=2)=3 P(X=1)$

$ \begin{aligned} 2 \cdot{ }^n C_2 p^2 q^{n-2} & =3 \cdot{ }^n C_1 p q^{n-1} \\ 2 \cdot \frac{n(n-1)}{2} p & =3 \cdot n \cdot q \\ (n-1) p & =3 q=3(1-p) \\ p(n+2) & =3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Solving Eqs. (i) and (ii), we get

$ p=\frac{1}{2}, q=\frac{1}{2} \text { and } n=4 $

Now, $P(X>1)=1-P(X=0)-P(X=1)$

$ \begin{aligned} & =1-4\left(\frac{1}{2}\right)^4-\left(\frac{1}{2}\right)^4 \\ & =1-\frac{5}{16}=\frac{11}{16} \end{aligned} $

$ \because \quad n^2 P(X>1)=16 \cdot \frac{11}{16}=11 $

To get a sum of 9 ,possible dice combinations are

$ \begin{array}{ll} (1,3,6) & -6 \text { ways } \\ (1,3,5) & -5 \text { ways } \\ (1,4,4) & -3 \text { ways } \\ (2,3,5) & -3 \text { ways } \\ (2,3,4) & -6 \text { ways } \\ (3,3,3) & -1 \text { way } \end{array} $

Total ways $=6+6+3+3+6+1=25$

Total possible outcomes when rolling three dice $=6 \times 6 \times 6=216$.

Number of favourable outcomes $=25$

∴ Probability of getting a sum of 9 in a single throw of three fair dice $=\frac{25}{216}$

We know that

Mean, $\mu=n p=10\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Variance, $\sigma^2=n p(1-p)=6\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Dividing Eq. (ii) by Eq. (i),

$ \begin{gathered} (1-p)=\frac{6}{10}=0.6 \\ p=0.4 \end{gathered} $

Substituting $p=0.4 \mathrm{in}$ Eq. (i),

$ \begin{aligned} n \times 0.4 & =10 \\ n & =25 \end{aligned} $

The median of a binomial distribution is typically approximated by

$ M \approx[n p] $

Substituting values

$ M=[25 \times 0.4]=[10] $

Thus, the median is 10 .

Total number of ways in which four digit number can be found $=4!=24$

So, exhaustive number of cases $=24$

Now, the number is divisible by 5 if at units place we have 5.

∴ Unit place can be filled in one way the remaining 3 places can be filled with other digits in 3! ways.

So, the number of ways of forming the number divisible by 5 is $3!=6$

So, the required probability $=\frac{6}{24}=\frac{1}{4}$

Let's denote the outcomes of the three rolls as $X_1$, $X_2$, and $X_3$, where $X_i$ represents the number obtained in the $i^{\text{th}}$ roll. We are looking for the probability that:

$X_2 > X_1 \text{ and } X_3 > X_2$

The total number of outcomes when rolling a dice three times is:

$6^3 = 216$

Let's count the favorable outcomes. For each satisfying outcome, we must ensure that both inequalities are adhered to. Consider the possible sequences where every subsequent roll's number is higher than the previous roll's number. These sequences are:

• (1, 2, 3)
• (1, 2, 4)
• (1, 2, 5)
• (1, 2, 6)
• (1, 3, 4)
• (1, 3, 5)
• (1, 3, 6)
• (1, 4, 5)
• (1, 4, 6)
• (1, 5, 6)
• (2, 3, 4)
• (2, 3, 5)
• (2, 3, 6)
• (2, 4, 5)
• (2, 4, 6)
• (2, 5, 6)
• (3, 4, 5)
• (3, 4, 6)
• (3, 5, 6)
• (4, 5, 6)

Clearly, there are 20 such favorable outcomes. Hence, the probability is given by:

$ \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{20}{216} = \frac{5}{54} $

Therefore, the correct option is:

Option A: $\frac{5}{54}$

To solve this problem, we use Bayes' theorem. Let's define the events:

• $X$: Selecting bag $X$
• $Y$: Selecting bag $Y$
• $Z$: Selecting bag $Z$
• $A$: Drawing a one-rupee coin

We are given that a bag is selected at random, so the probabilities for choosing any of the bags are:

$P(X) = P(Y) = P(Z) = \frac{1}{3}$

Next, we need the probability of drawing a one-rupee coin from each bag:

• From bag $X$: $P(A|X) = \frac{5}{5+4} = \frac{5}{9}$
• From bag $Y$: $P(A|Y) = \frac{4}{4+5} = \frac{4}{9}$
• From bag $Z$: $P(A|Z) = \frac{3}{3+6} = \frac{3}{9} = \frac{1}{3}$

We need to find the probability that the coin came from bag $Y$ given that a one-rupee coin was drawn, i.e., we need $P(Y|A)$. Using Bayes' theorem:

$P(Y|A) = \frac{P(A|Y) \cdot P(Y)}{P(A)}$

To find $P(A)$, the total probability of drawing a one-rupee coin can be calculated as follows:

$P(A) = P(A|X) \cdot P(X) + P(A|Y) \cdot P(Y) + P(A|Z) \cdot P(Z)$

Substituting the values:

$P(A) = \left(\frac{5}{9} \cdot \frac{1}{3}\right) + \left(\frac{4}{9} \cdot \frac{1}{3}\right) + \left(\frac{1}{3} \cdot \frac{1}{3}\right)$

Calculating the above, we get:

$P(A) = \frac{5}{27} + \frac{4}{27} + \frac{1}{9}$

Note that $\frac{1}{9} = \frac{3}{27}$, so:

$P(A) = \frac{5}{27} + \frac{4}{27} + \frac{3}{27} = \frac{12}{27} = \frac{4}{9}$

Now, substituting back into Bayes' theorem:

$P(Y|A) = \frac{\left(\frac{4}{9}\right) \cdot \left(\frac{1}{3}\right)}{\frac{4}{9}} = \frac{4}{9} \cdot \frac{1}{3} \cdot \frac{9}{4} = \frac{1}{3}$

Hence, the probability that the coin came from bag $Y$ is:

Option B: $\frac{1}{3}$

Take two numbers as $a$ and $b$

$a+b=24$

For product to be maximum

$\begin{aligned} & \frac{a+b}{2} \geq \sqrt{a b} \\ & 144>a b \end{aligned}$

Maximum product is 144

Now, $a b \geq \frac{3}{4} \cdot 144=108$

Sample space $=\{(23,1),(22,2), \ldots\}$

Integer points on line in shaded region

$\begin{aligned} & \{(6,18),(7,17),(8,16), \ldots(18,6)\} \\ & P(E)=\frac{n(E)}{n(S)}=\frac{13}{23}=\frac{m}{n} \Rightarrow n-m=10 \end{aligned}$

We have 3 letters and 5 addresses, where 3 letters are posted to exactly 2 addresses. First, we will select 2 addresses in ${ }^5 C_2$ ways.

Now, 3 letters can be posted to exactly 2 addresses in 6 ways.

$\begin{aligned} \therefore \text { Probability }=\frac{{ }^5 C_2 \times 6}{5^3} & \\ & =\frac{60}{125}=\frac{12}{25} \end{aligned}$

$\begin{aligned} & P(\text { standard automobile from } A)=\frac{6}{10} \times \frac{8}{10}=\frac{12}{25} \\ & P(\text { standard automobile from } B)=\frac{4}{10} \times \frac{9}{10}=\frac{9}{25} \end{aligned}$

$\begin{aligned} & \text { Required Probability } \frac{\frac{9}{25}}{\frac{12}{25}+\frac{9}{25}} \\ & P=\frac{9}{21}=\frac{3}{7} \end{aligned}$

So, $126 P=126 \times \frac{3}{7}=54$

Equation is $a x^2+b x+c=0$

$\mathrm{D}>0$ [for roots to be real & distinct]

$\Rightarrow b^2-4 a c>0$

For $b<2$ no value of $a$ & $c$ are possible

$\begin{aligned} & \text { For } b=3 \Rightarrow a c<\frac{9}{4} \\ & (a, c) \in\{(1,1),(1,2),(2,1)\} \Rightarrow 3 \text { cases } \end{aligned}$

For $b=4 \Rightarrow a c<4$

$(a, c) \in\{(1,1),(1,2),(2,1),(3,1),(1,3)\} \Rightarrow 5 \text { cases }$

For $b=5 \Rightarrow a c<\frac{25}{4}$

$\begin{aligned} & (a, c) \in\{(1,1),(1,2),(2,1),(3,1),(1,3),(2,2), \\ & (4,1),(1,4),(3,2),(2,3),(5,1),(1,5),(1,6), \\ & (6,1)\}=14 \text { cases } \end{aligned}$

For $b=6 \Rightarrow a c<9$

$\begin{aligned} & (a, c) \in\{(1,1),(1,2),(2,1),(3,1),(1,3),(2,2), \\ & (4,1),(1,4),(3,2),(2,3),(5,1),(1,5),(1,6), \\ & (6,1),(2,4),(4,2)\}=16 \text { cases } \end{aligned}$

Total cases $=3+5+14+16=38$ cases

$\Rightarrow$ Probability, $p=\frac{38}{216}$

$\Rightarrow 216 p=38$

Given quadratic equation is

$a x^2+b x+c=0 \text { where } a, b, c \in\{1,2,3, \ldots, 8\}$

For repeated roots,

$\begin{aligned} & b^2-4 a c=0 \\ & \Rightarrow b^2=4 a c \end{aligned}$

$\Rightarrow a c$ must be a perfect square

$(a, c) \in\{(1,1),(1,4),(2,2),(2,8),(3,3),(4,1),(4,4),(5,5),(6,6),(7,7),(8,2),(8,8)\}$

Corresponding $b$ must lie in set $\{1,2,3, \ldots 8\}$

$\begin{aligned} & (a, b, c) \in\{(1,2,1),(1,2,4),(2,4,2),(2,8,8) \\ & (3,6,3),(4,4,1),(4,8,4),(8,8,2)\} \\ & \therefore \text { probability }=\frac{8}{8^3} \\ & =\frac{1}{64} \\ & \end{aligned}$

$\begin{aligned} & \text { Mean }=\sum x_i P\left(x_i\right) \\ & \frac{46}{9}=4 a+4 a+4 b+12 b+24 b \\ & \frac{46}{9}=8 a+40 b \\ & \frac{23}{9}=4 a+20 b \\ & 36 a+180 b=23 \quad \text{.... (1)} \end{aligned}$

Sum of probability is 1

$\Rightarrow 4 a+6 b=1 \quad \text{... (2)}$

$\begin{aligned} & \text { Solving (1) and (2) } \\ & a=\frac{1}{12}, b=\frac{1}{9} \\ & \sigma^2=\sum x_i^2 P\left(x_i\right)-\left(\sum x_i P\left(x_i\right)\right)^2 \\ & =4 \times 2 a+16(a+b)+36(2 b)+64(3 b)-\left(\frac{46}{9}\right)^2 \\ & =8(a+2(a+b)+9 b+24 b)-\left(\frac{46}{9}\right)^2 \\ & =8(3 a+35 b)-\left(\frac{46}{9}\right)^2 \\ & =8\left(\frac{3}{12}+\frac{35}{9}\right)-\left(\frac{46}{9}\right)^2 \\ & =8\left(\frac{149}{36}\right)-\left(\frac{46}{9}\right)^2=\frac{566}{81} \\ \end{aligned}$

Let's denote the events as follows :

Let $ A_1, A_2, $ and $ A_3 $ be the events that urns A, B, and C are chosen, respectively.

Let $ B $ be the event that a black ball is drawn.

We need to find the probability that the chosen urn is A given that a black ball is drawn, which is $ P(A_1|B) $.

Using Bayes' theorem, we have :

$ P(A_1|B) = \frac{P(B|A_1)P(A_1)}{P(B)} $

First, we calculate each term individually :

1. The probability of choosing any urn, since they are chosen at random :

$ P(A_1) = P(A_2) = P(A_3) = \frac{1}{3} $

2. The probability of drawing a black ball from each urn :

Urn A: $ P(B|A_1) = \frac{5}{12} $

Urn B: $ P(B|A_2) = \frac{7}{12} $

Urn C: $ P(B|A_3) = \frac{6}{12} = \frac{1}{2} $

3. The total probability of drawing a black ball, $ P(B) $ :

$ P(B) = P(B|A_1)P(A_1) + P(B|A_2)P(A_2) + P(B|A_3)P(A_3) $

$ P(B) = \frac{5}{12} \cdot \frac{1}{3} + \frac{7}{12} \cdot \frac{1}{3} + \frac{1}{2} \cdot \frac{1}{3} $

$ P(B) = \frac{5}{36} + \frac{7}{36} + \frac{6}{36} $

$ P(B) = \frac{18}{36} = \frac{1}{2} $

Now, substitute these into Bayes' theorem :

$ P(A_1|B)= \frac{P(B|A_1)P(A_1)}{P(B)} $

$ P(A_1|B) = \frac{\frac{5}{12} \cdot \frac{1}{3}}{\frac{1}{2}} $

$ P(A_1|B) = \frac{\frac{5}{36}}{\frac{1}{2}} $

$ P(A_1|B) = \frac{5}{18} $

Therefore, the probability that the black ball drawn is from urn A is option C :

$ \frac{5}{18} $

We are given that the probability of Ajay not appearing in the JEE exam is $\mathrm{p}=\frac{2}{7}$, and the probability that both Ajay and Vijay will appear in the exam is $\mathrm{q}=\frac{1}{5}$.

We are asked to find the probability that Ajay will appear in the exam and Vijay will not. Let's denote this probability as $\mathrm{r}$.

To find $\mathrm{r}$, we need to use the concept of complementary events. The probability that Ajay will appear in the exam is the complement of the probability that he will not appear. So,

$ P(\text{Ajay appears}) = 1 - P(\text{Ajay does not appear}) = 1 - \mathrm{p} = 1 - \frac{2}{7} = \frac{5}{7}. $

The event that both Ajay and Vijay appear in the exam is independent of the event that only Ajay appears (and Vijay does not). Therefore, we can express the probability that only Ajay will appear (and Vijay will not) as the difference of Ajay appearing minus both Ajay and Vijay appearing, because the probability of both appearing ($\mathrm{q}$) is included in the probability of Ajay appearing:

$ \mathrm{r} = P(\text{Ajay appears}) - P(\text{Both Ajay and Vijay appear}) = \frac{5}{7} - \frac{1}{5}. $

To subtract these two fractions, we need a common denominator, which would be $35$ in this case. So,

$ \mathrm{r} = \frac{5}{7} \cdot \frac{5}{5} - \frac{1}{5} \cdot \frac{7}{7} = \frac{25}{35} - \frac{7}{35} = \frac{25 - 7}{35} = \frac{18}{35}. $

Therefore, the probability that Ajay will appear in the exam and Vijay will not appear is $\mathrm{r} = \frac{18}{35}$, which corresponds to Option D.

To solve this problem, we need to first determine the probability of getting a head (H) and the probability of getting a tail (T).

Since a head is twice as likely to occur as a tail, we can denote the probability of getting a tail as $ P(T) = p $ and the probability of getting a head as $ P(H) = 2p $.

These probabilities must sum to 1 because those are the only two possible outcomes for each coin toss :

$ P(H) + P(T) = 1 $

$ 2p + p = 1 $

$ 3p = 1 $

$ p = \frac{1}{3} $

Therefore, the probability of getting a tail (T) is $ P(T) = \frac{1}{3} $ and the probability of getting a head (H) is $ P(H) = 2 \times \frac{1}{3} = \frac{2}{3} $.

Now to find the probability of getting two tails and one head, we need to consider the different sequences in which this can occur. There are three unique sequences: TTH, THT, and HTT.

The probability of each sequence is found by multiplying the probabilities of each individual event since each coin toss is independent:

$ P(TTH) = P(T) \times P(T) \times P(H) = \left(\frac{1}{3}\right)^2 \times \frac{2}{3} = \frac{1}{9} \times \frac{2}{3} = \frac{2}{27} $

$ P(THT) = P(T) \times P(H) \times P(T) = \frac{1}{3} \times \frac{2}{3} \times \frac{1}{3} = \frac{2}{27} $

$ P(HTT) = P(H) \times P(T) \times P(T) = \frac{2}{3} \times \left(\frac{1}{3}\right)^2 = \frac{2}{27} $

The overall probability of getting two tails and one head in any order is the sum of these individual probabilities :

$ P(2T1H) = P(TTH) + P(THT) + P(HTT) = \frac{2}{27} + \frac{2}{27} + \frac{2}{27} = \frac{6}{27} $

Simplifying this expression gives us: $ P(2T1H) = \frac{6}{27} = \frac{2}{9} $

Therefore, the correct answer is :

Option B : $\frac{2}{9}$

3 bad apples, 15 good apples.

Let $\mathrm{X}$ be no of bad apples

$\begin{aligned} & \text { Then } \mathrm{P}(\mathrm{X}=0)=\frac{{ }^{15} \mathrm{C}_2}{{ }^{18} \mathrm{C}_2}=\frac{105}{153} \\ & \mathrm{P}(\mathrm{X}=1)=\frac{{ }^3 \mathrm{C}_1 \times{ }^{15} \mathrm{C}_1}{{ }^{18} \mathrm{C}_2}=\frac{45}{153} \\ & \mathrm{P}(\mathrm{X}=2)=\frac{{ }^3 \mathrm{C}_2}{{ }^{18} \mathrm{C}_2}=\frac{3}{153} \\ & \mathrm{E}(\mathrm{X})=0 \times \frac{105}{153}+1 \times \frac{45}{153}+2 \times \frac{3}{153}=\frac{51}{153} \\ & =\frac{1}{3} \\ & \operatorname{Var}(\mathrm{X})=\mathrm{E}\left(\mathrm{X}^2\right)-(\mathrm{E}(\mathrm{X}))^2 \\ & =0 \times \frac{105}{153}+1 \times \frac{45}{153}+4 \times \frac{3}{153}-\left(\frac{1}{3}\right)^2 \\ & =\frac{57}{153}-\frac{1}{9}=\frac{40}{153} \end{aligned}$

To solve this problem, we need to calculate the probability of two independent events occurring in succession: the first marble drawn is red, and the second marble drawn is white. Since the drawing is with replacement, the number of marbles of each color remains the same for both draws.

The total number of marbles in the box is the sum of red, white, blue, and orange marbles:

The probability of drawing a red marble in the first draw is the number of red marbles divided by the total number of marbles:

Since the marble is replaced, the probability of drawing a white marble in the second draw remains as the number of white marbles divided by the total number of marbles:

The probability of both independent events occurring in succession (drawing a red marble first and then a white marble) is the product of their individual probabilities:

Now, let's calculate this probability:

$ P(\text{First is red and second is white}) = \frac{10 \times 30}{75 \times 75} = \frac{300}{5625}= \frac{4}{75}. $

Therefore, the correct answer is

$\mathrm{E}_1: \mathrm{A}$ is selected

$\mathrm{E}_2: \mathrm{B}$ is selected

$\mathrm{E}$ : white ball is drawn

$\mathrm{P}\left(\mathrm{E}_1 / \mathrm{E}\right)=$

$\begin{aligned} & \frac{P(E) \cdot P\left(E / E_1\right)}{P\left(E_1\right) \cdot P\left(E / E_1\right)+P\left(E_2\right) \cdot P\left(E / E_2\right)}=\frac{\frac{1}{2} \times \frac{3}{10}}{\frac{1}{2} \times \frac{3}{10}+\frac{1}{2} \times \frac{3}{5}} \\ & =\frac{3}{3+6}=\frac{1}{3} \end{aligned}$

If $x=0, y=6,7,8,9,10$

If $x=1, y=7,8,9,10$

If $x=2, y=8,9,10$

If $x=3, y=9,10$

If $x=4, y=10$

If $x=5, y=$ no possible value

Total possible ways $=(5+4+3+2+1) \times 2$

$=30$

Required probability $=\frac{30}{11 \times 11}=\frac{30}{121}$

Given set $=\{1,2,3, \ldots \ldots . .50\}$

$\mathrm{P}(\mathrm{A})=$ Probability that number is multiple of 4

$\mathrm{P(B)}=$ Probability that number is multiple of 6

$\mathrm{P}(\mathrm{C})=$ Probability that number is multiple of 7

Now,

$\mathrm{P}(\mathrm{A})=\frac{12}{50}, \mathrm{P}(\mathrm{B})=\frac{8}{50}, \mathrm{P}(\mathrm{C})=\frac{7}{50}$

again

$\begin{aligned} & \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{4}{50}, \mathrm{P}(\mathrm{B} \cap \mathrm{C})=\frac{1}{50}, \mathrm{P}(\mathrm{A} \cap \mathrm{C})=\frac{1}{50} \\ & \mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})=0 \end{aligned}$

Thus

$\begin{aligned} P(A & \cup B \cup C)=\frac{12}{50}+\frac{8}{50}+\frac{7}{50}-\frac{4}{50}-\frac{1}{50}-\frac{1}{50}+0 \\ & =\frac{21}{50} \end{aligned}$

Required probability $=$

$\begin{aligned} & \frac{5}{6} \times \frac{1}{6}+\left(\frac{5}{6}\right)^3 \times \frac{1}{6}+\left(\frac{5}{6}\right)^5 \times \frac{1}{6}+\ldots . . \\ & =\frac{1}{6} \times \frac{\frac{5}{6}}{1-\frac{25}{36}}=\frac{5}{11} \end{aligned}$

$\frac{{ }^6 \mathrm{C}_4}{{ }^{15} \mathrm{C}_4} \times \frac{{ }^9 \mathrm{C}_4}{{ }^{11} \mathrm{C}_4}=\frac{3}{715}$

Hence option (3) is correct.