Probability

Total number of possible words in the word 'CURVE' is

$ \begin{aligned} & \text { 2-letter words }={ }^5 P_2=\frac{5!}{(5-2)!}=5 \times 4=20 \\ & \text { 3-letter words }={ }^5 P_3=5 \times 4 \times 3=60 \\ & \text { 4-letter words }={ }^5 P_4=5 \times 4 \times 3 \times 2=120 \\ & \text { 5-letter words } \\ & ={ }^5 P_5=5 \times 4 \times 3 \times 2 \times 1=120 \end{aligned} $

So, total words $=20+60+120+120$

$ =320 $

$\therefore$ P(3-letter words)

$ \begin{aligned} & =\frac{\text { Number of } 3 \text { - letter words }}{\text { Total words }} \\ & =\frac{60}{320}=\frac{3}{16} \end{aligned} $

Total ways of choosing 3 numbers

$ \begin{aligned} & ={ }^{30} C_3=\frac{30 \times 29 \times 28}{3 \times 2 \times 1} \\ & =4060 \end{aligned} $

The number of sets of three consecutive numbers

$ \begin{aligned} & \{(1,2,3),(2,3,4) \ldots(28,29,30)\} \\ & \Rightarrow 28 \text { sets. } \end{aligned} $

So, the number of sets that are not consecutive numbers $=4060-28=4032$

Hence, required probability

$ =\frac{4032}{4060}=\frac{144}{145} $

$P(\bar{A})=0.3, P(B)=0.4, P(A \cap \bar{B})=0.5$

So, $P(A)=1-P(\bar{A})=0.7$,

$ P(\bar{B})=1-0.4=0.6 $

Now, $P(A \cap B)$

$ \because \quad P(A)=P(A \cap B)+P(A \cap \bar{B}) $

$ \begin{aligned} & \Rightarrow \quad 0.7=P(A \cap B)+0.5 \\ & \Rightarrow \quad P(A \cap B)=0.2 \\ & \text { then, } P(B / A \cup \bar{B})=\frac{P(B \cap(A \cup \bar{B}))}{P(A \cup \bar{B})} \\ & =\frac{P((B \cap A) \cup(B \cap \bar{B})}{P(A)+P(\bar{B})-P(A \cap \bar{B})} \\ & =\frac{P(A \cap B)}{0.7+0.6-0.5}=\frac{0.2}{0.8}=0.25 \end{aligned} $

• Probability that A will get the job, $ P(A) = 0.8 $

• Probability that B will get the job, $ P(B) = 0.7 $

We are asked to find the probability that at least one of them will get the job.

Step 1: Formula for "at least one"

$ P(\text{at least one}) = P(A \cup B) = P(A) + P(B) - P(A \cap B) $

Step 2: If events $A$ and $B$ are independent:

$ P(A \cap B) = P(A) \times P(B) $

$ P(A \cap B) = 0.8 \times 0.7 = 0.56 $

Step 3: Substitute values

$ P(A \cup B) = 0.8 + 0.7 - 0.56 = 0.94 $

✅ Final Answer: 0.94

Correct Option: (B) 0.94

$ \begin{aligned} & \text { } \because P(x=k)={ }^n C_k p^{n-k} q^k \\ & \text { Here, } p=q=1 / 2 \\ & \text { and } 2 P(X=5)=P(X=4)+P(X=6) \\ & \Rightarrow 2^n C_5\left(\frac{1}{2}\right)^{n-5}\left(\frac{1}{2}\right)^5={ }^n C_4\left(\frac{1}{2}\right)^{n-4}\left(\frac{1}{2}\right)^4 +{ }^n C_6\left(\frac{1}{2}\right)^{n-6}\left(\frac{1}{2}\right)^6 \end{aligned} $

$ \begin{aligned} & \Rightarrow 2^n C_5\left(\frac{1}{2}\right)^n={ }^n C_4\left(\frac{1}{2}\right)^n+{ }^n C_6\left(\frac{1}{2}\right)^n \\ & \Rightarrow 2\left[\frac{n!}{5!(n-5)!}\right]=\frac{n!}{4!(n-4)!}+\frac{n!}{6!(n-6)!} \\ & \Rightarrow \frac{2}{5!(n-5)}=\frac{1}{4!}+\frac{1}{6!(n-6)(n-5)} \\ & \Rightarrow \frac{2}{5(n-5)}=1+\frac{1}{30(n-6)(n-9)} \\ & \Rightarrow \frac{2}{5(n-5)}=\frac{30\left(n^2-11 n+30\right)+1}{30(n-6)(n-9)} \\ & \Rightarrow 12(n-6)=30\left(n^2-11 n+30\right)+1 \\ & \Rightarrow n^2-21 n+98=0 \Rightarrow(n-14)(n-7)=0 \end{aligned} $

So, $n=7$ or 14

Hence, largest value of $n=14$

$ \begin{array}{clllcc} \hline \boldsymbol{X}=x_i & -2 & -1 & 0 & 1 & 2 \\ \hline \boldsymbol{P}\left(\boldsymbol{X}=x_i\right) & k^2 / 3 & k^2 & 2 k^2 / 3 k / 2 & k / 2 \\ \hline \end{array} $

$ \begin{aligned} & \because \quad \sum_{i=1}^n P\left(X=x_i\right)=1 \\ & \Rightarrow \quad \frac{k^2}{3}+k^2+\frac{2 k^2}{3}+k / 2+k / 2=1 \\ & \Rightarrow \quad 2 k^2+k-1=0 \\ & \Rightarrow \quad 2 k^2+2 k-k-1=0 \\ & \Rightarrow \quad(2 k-1)(k+1)=0 \\ & \text { so, } k=-1,1 / 2 \\ & \text { for } k=1 / 2 \\ & k^2 / 3=1 / 12>0, k^2=1 / 4>0 \\ & \frac{2 k^2}{3}=1 / 6>0 \end{aligned} $

$ k / 2=1 / 4>0 $

and sum of

$1 / 12+1 / 4+1 / 6+1 / 4+1 / 4=1$

thus, $k=1 / 2$ is a valid solution

Now, $E(X)=\sum x_i P\left(X=x_i\right)$

$ \begin{aligned} &=-2 \times 1 / 12+(-1) \times 1 / 4+0 \times 1 / 6 +1 \times 1 / 4+2 \times 1 / 4 \\ &=-1 / 6-1 / 4+1 / 4+1 / 2 \end{aligned} $

$ =1 / 2-1 / 6=\frac{3-1}{6}=1 / 3 $

Let $A=$ first student get qualified in the exam and $B=$ second students get qualified in the exam $\because P(A)=\frac{1}{4}$ and $P(B)=\frac{2}{5}$ and $A, B$ are independent events.

$ \begin{aligned} P(A \cup B) & =1-P(\overline{A \cup B}) \\ & =1-P(\bar{A} \cap \bar{B}) \\ & =1-P(\bar{A}) \cdot P(\bar{B}) \\ & =1-(1-P(A))(1-P(B)) \\ & =1-\left(1-\frac{1}{4}\right)\left(1-\frac{2}{5}\right) \\ & =1-\frac{3}{4} \times \frac{3}{5}=\frac{11}{20} \end{aligned} $

$P$ (exactly one or $A$ or $B$ occurs) $=P$ (exactly one $B$ or $C$ occurs) $=P$ (exactly one of $C$ or $A$ occurs) $=\frac{1}{4}$

$\therefore \quad P(A \cap B \cap C)=\frac{1}{16}$

$ \begin{aligned} & P(A)+P(B)-2 P(A \cap B)=\frac{1}{4} \\ & P(B)+P(C)-2 P(B \cap C)=\frac{1}{4} \\ & P(C)+P(A)-2 P(C \cap A)=\frac{1}{4} \end{aligned} $

On adding Eqs. (i), (ii) and (iii), we get

$ \begin{aligned} & 2[P(A)+P(B)+P(C)-P(A \cap B) \\ & -P(B \cap C)-P(C \cap A)]=\frac{3}{4} \end{aligned} $

$ \begin{aligned} &\text {}\\ &\begin{aligned} Now, \,\, & P(A \cup B \cup C)=P(A)+P(B)+P(C) \\ & -P(A \cap B)-P(B \cap C) \\ & -P(C \cap A)+P(A \cap B \cap C) \\ & =\frac{3}{8}+\frac{1}{16}=\frac{6+1}{16}=\frac{7}{16} \end{aligned} \end{aligned} $

Probability out of 3 balls 2 are black and one red.

$ \begin{aligned} & =\frac{(1 R)_P(2 B)_Q+(1 B)_P(1 R \cdot 1 B)_Q}{(1 \text { ball })_P \cdot(2 \text { balls })_Q} \\ & =\frac{{ }^4 C_1 \cdot{ }^6 C_2+{ }^5 C_1 \cdot{ }^3 C_1 \cdot{ }^6 C_1}{{ }^9 C_1 \cdot{ }^9 C_2} \\ & =\frac{4 \times \frac{6 \times 5}{2 \times 1}+5 \times 3 \times 6}{9 \times \frac{9 \times 8}{2 \times 1}} \\ & =\frac{60+90}{36 \times 9}=\frac{150}{36 \times 9}=\frac{25}{54} \end{aligned} $

Let, $A=$ Watching TV

$B=$ Read book and $E=$ Student sleep

$ \begin{aligned}Now,\,\, & v, P(A)=\frac{4}{5} \text { and } P(B)=1-\frac{4}{5}=\frac{1}{5} \\ & P\left(\frac{E}{A}\right)=\frac{3}{4} \Rightarrow P\left(\frac{E}{B}\right)=\frac{1}{4} \end{aligned} $

$ \begin{aligned}Now,\,\, P\left(\frac{A}{E}\right) & =\frac{P(A) \cdot P\left(\frac{E}{A}\right)}{P(A) \cdot P\left(\frac{E}{A}\right)+P(B) \cdot P\left(\frac{E}{B}\right)} \\ & =\frac{\frac{4}{5} \times \frac{3}{4}}{\frac{4}{5} \times \frac{3}{4}+\frac{1}{4} \times \frac{1}{5}}=\frac{12}{13} \end{aligned} $

We have, $P(X=k)=\frac{1}{n}$

$ P(X=1)=P(X=2) \ldots, P(X=n)=\frac{1}{n} $

Now, variance

$ \begin{aligned} & \Sigma P_i x_i^2-\left(\Sigma P_i x_i\right)^2 \\ & \frac{1}{n}\left(1^2+2^2+\ldots+n^2\right)-\left(\frac{1}{n}(1+2+\ldots+n)\right)^2 \\ & \frac{(n+1)}{6}(2 n+1)-\left(\frac{n+1}{2}\right)^2 \\ & (n+1)\left(\frac{2 n+1}{6}-\frac{n+1}{4}\right) \\ & (n+1)\left(\frac{4 n+2-3 n-3}{12}\right) \\ & \quad=\frac{n^2-1}{12} \end{aligned} $

$p=$ Probability of detect an enemy plane $=\frac{1}{10}$

and $q=$ Probability of not detect an enemy plane $=1-\frac{1}{10}=\frac{9}{10}$.

$ \begin{aligned} & n=4 \\ P(X \geq 2)= & 1-P(X=0)-P(X=1) \\ = & 1-{ }^4 C_0 P^0 q^4-{ }^4 C_1 p^1 q^3 \\ = & 1-\left(\frac{9}{10}\right)^4-4\left(\frac{1}{10}\right)\left(\frac{9}{10}\right)^3 \\ = & \frac{10000-6561-2916}{104} \\ = & \frac{523}{10000}=0.0523 \end{aligned} $

Total ways of distributing 5 identical samples among 12 house, with atmost one sample $={ }^{12} \mathrm{C}_5$

Number of ways to select 5 house with no two consecutive

$ \begin{gathered} ={ }^{n-r+1} C_r={ }^{12-5+1} C_5={ }^8 C_5 \\ \therefore \text { Required probability }=\frac{{ }^8 C_5}{{ }^{12} C_6}=\frac{\frac{8!}{3!5}}{\frac{12!}{5!7}} \\ \quad=\frac{8!7!}{12!3!}=\frac{7 \times 6 \times 5 \times 4}{12 \times 11 \times 10 \times 9}=\frac{7}{99} \end{gathered} $

Step 1: Show what we know.

$A$ and $B$ are independent, so $P(A \cap B) = P(A) \cdot P(B) = \frac{1}{6}$.

The chance that neither happens is $P(\bar{A} \cap \bar{B}) = (1 - P(A)) \cdot (1 - P(B)) = \frac{1}{3}$.

Step 2: Set up equations.

From above, we have two equations:

• $P(A) \cdot P(B) = \frac{1}{6}$
• $(1 - P(A)) \cdot (1 - P(B)) = \frac{1}{3}$

Step 3: Expand the second equation.

Expand $(1 - P(A)) \cdot (1 - P(B)) = 1 - P(A) - P(B) + P(A) \cdot P(B) = \frac{1}{3}$.

Step 4: Replace $P(A) \cdot P(B)$.

Since $P(A) \cdot P(B) = \frac{1}{6}$, substitute:

• $1 - P(A) - P(B) + \frac{1}{6} = \frac{1}{3}$

• $1 + \frac{1}{6} - \frac{1}{3} = P(A) + P(B)$
• $1 + \frac{1}{6} - \frac{2}{6} = P(A) + P(B)$
• $1 - \frac{1}{6} = P(A) + P(B)$
• $\frac{5}{6} = P(A) + P(B)$

Step 5: Build a quadratic equation.

Let $P(A) = x$. Then $P(B) = \frac{5}{6} - x$.

Also, $x \cdot P(B) = \frac{1}{6}$.

So, $x \left( \frac{5}{6} - x \right) = \frac{1}{6}$

$5x/6 - x^2 = 1/6$

Multiply both sides by 6: $5x - 6x^2 = 1$

$6x^2 - 5x + 1 = 0$

Step 6: Solve the quadratic equation.

$x = \frac{5 \pm \sqrt{25-24}}{12}$

$x = \frac{5 \pm 1}{12}$, so $x = \frac{6}{12} = \frac{1}{2}$ and $x = \frac{4}{12} = \frac{1}{3}$

Step 7: Choose the correct values using $P(A) > P(B)$.

Since $P(A) > P(B)$, $P(A) = \frac{1}{2}$ and $P(B) = \frac{1}{3}$.

Step 1: Understand events A and B

Event $A$ means the sum that comes from the two dice is a prime number.

Event $B$ means the sum is more than 8.

$\bar{B}$ means the sum is 8 or less.

Step 2: List all possible sums when rolling two dice

When rolling two dice, the possible sums go from 2 to 12.

Step 3: Find all ways to get a prime sum ($A$)

The prime numbers between 2 and 12 are: 2, 3, 5, 7, and 11.

We need to count all pairs of dice rolls that add up to one of these numbers.

Step 4: Find the pairs for $A \cap \bar{B}$

We only want the pairs where the sum is a prime number and also less than or equal to 8.

The prime numbers that are 8 or less are: 2, 3, 5, and 7.

Let’s list all pairs for each:

Sum = 2: (1,1)
Sum = 3: (1,2), (2,1)
Sum = 5: (1,4), (2,3), (3,2), (4,1)
Sum = 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)

Counting all these, there are 13 pairs.

Step 5: Total number of possible outcomes

There are 36 possible combinations when rolling two dice (6 sides on the first die times 6 sides on the second die).

Step 6: Write the probability

The probability of both $A$ happening and $\bar{B}$ (sum is a prime less than or equal to 8) happening is:

$P(A \cap \bar{B})=\frac{13}{36}$

$\Sigma_1=$ In the family of 8 , men and women are in equal number $\Sigma_i=$ In the family of 8 , men and women are not in equal number.

Total cases $=9$ for each case $P\left(\Sigma_1\right)=1 / 9 A=4$ persons are selected in which ${ }^2$ are men and 2 are women.

$ \begin{aligned} P\left(\Sigma_1 / A\right) & =\frac{P(A) \cdot P\left(A / \Sigma_1\right)}{P(A) \cdot P\left(A / \Sigma_1\right)+P\left(\Sigma_i\right) \cdot P\left(A / \Sigma_i\right)} \\ & =\frac{\frac{1}{9} \cdot \frac{{ }^4 C_2 \cdot{ }^4 C_2}{{ }^8 C_4}}{\frac{1}{9} \cdot \frac{{ }^4 C_2 \cdot{ }^4 C_2}{{ }^8 C_4}+\frac{1}{9} \frac{\left({ }^2 C_2 \cdot{ }^6 C_2+{ }^3 C_2 \cdot{ }^5 C_2\right)_2}{{ }^8 C_4}} \\ & =\frac{1}{1+\frac{2(15+30)}{36}}=\frac{1}{1+\frac{45}{18}}=\frac{1}{1+\frac{5}{2}}=\frac{2}{7} \end{aligned} $

$ \begin{aligned} &\text { We have, } n=6\\ &\begin{aligned} & \text { Mean - Variance }=27 / 8 \\ & \quad\Rightarrow n p-n p q=\frac{27}{8} \Rightarrow n p(1-q)=\frac{27}{8} \\ & \quad\Rightarrow n p^2=\frac{27}{8} \Rightarrow p^2=\frac{27}{8 \times 6} \Rightarrow p^2=\frac{9}{16} \Rightarrow p=\frac{3}{4} \\ & \begin{aligned} \therefore P(X \leq 2) & =P(X=0)+P(X=1)+P(X=2) \\ & ={ }^6 C_0\left(\frac{3}{4}\right)^0\left(\frac{1}{4}\right)^6+{ }^6 C_1\left(\frac{3}{4}\right)^1\left(\frac{1}{4}\right)^5+{ }^6 C_2\left(\frac{3}{4}\right)^2 \cdot\left(\frac{1}{4}\right)^2 \\ & =\frac{1}{4^6}(1+18+135)=\frac{154}{4^6} \end{aligned} \end{aligned} \end{aligned} $

The mean of a binomial distribution is $\mu = n p$.

The variance is $\sigma^2 = n p q$ where $q = 1 - p$.

Given $\mu = 2\sigma^2$, so $n p = 2(n p q)$. This means $n p = 2 n p q$.

Dividing both sides by $n p$ (assuming $n p \ne 0$), we get $1 = 2q$, so $q = \frac{1}{2}$.

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

Also, we are told that $\mu + \sigma^2 = 3$. Substitute the formulas to get: $n p + n p q = 3$.

Factor $n p$ to get $n p (1+q) = 3$.

Now, substitute $p = \frac{1}{2}$ and $q = \frac{1}{2}$ into the equation: $n \left(\frac{1}{2}\right)\left(1+\frac{1}{2}\right) = 3$.

Rewrite $1+\frac{1}{2} = \frac{3}{2}$, so $n \left(\frac{1}{2}\right)\left(\frac{3}{2}\right) = 3$.

This becomes $n \cdot \frac{3}{4} = 3$. Divide both sides by $\frac{3}{4}$: $n = 3 \div \frac{3}{4} = 3 \times \frac{4}{3} = 4$.

The distribution is $B(4, \frac{1}{2})$.

We need to find $P(X \leq 3)$.

For a binomial distribution, $P(X \leq 3) = 1 - P(X = 4)$, because the only other possible value for $X$ is $4$ when $n = 4$.

$P(X=4) = {4 \choose 4}\left(\frac{1}{2}\right)^4 = 1 \times \left(\frac{1}{16}\right) = \frac{1}{16}$

Therefore, $P(X \leq 3) = 1 - \frac{1}{16} = \frac{15}{16}$.

Total number of fruits in one basket $=5$ apples +7 oranges $=12$

Probability of picking an apple from Ist basket $=\frac{5}{12}=P\left(A_1\right)$ Probability of picking an orange from Ist basket $=\frac{7}{12}=P\left(B_1\right)$ Total number of fruits in second baskd $=4$ apples +8 oranges $=12$ Probability of picking an apples from IInd basket, $P\left(A_2\right)=\frac{4}{12}$

Probability of picking an oranges from IInd basket $P\left(B_2\right)=\frac{8}{12}$

Probability of picking an apple from Ist and an orange from IInd basket

$ =\frac{5}{12} \times \frac{8}{12}=\frac{40}{144}=\frac{10}{36}=\frac{5}{18} $

Probability of picking an apple from IInd basket and an orange from Ist basket $=\frac{4}{12} \times \frac{7}{12}=\frac{28}{144}$

So, total probability

$ =\frac{40}{144}+\frac{28}{144}=\frac{68}{144}=\frac{17}{36} $

Probability of drawing a black queen first $=\frac{2}{52}$

Remaining black face card $=5$

Probability of drawing a black face card after a black queen $=\frac{5}{51}$

Now, probability of red queen first $=\frac{2}{52}$

Probability of a black face card after a red queen $=\frac{6}{51}$

So, total probability $=\frac{2}{52} \times \frac{5}{51}+\frac{2}{52} \times \frac{6}{51}$

$ =\frac{10}{2652}+\frac{12}{2652}=\frac{22}{2652}=\frac{11}{1326} $

Let $D$ be the event that an item is defective, and $N$ be the event that an item is not defective.

So, $P(D)=0.3$ and $P(N)=1-0.3=0.7$

Probability of an accurate result

$ =\frac{8}{10}=0.8=P\left(R_D / D\right) $

Probability of an inaccurate result

$ \begin{aligned} & =1-0.8=0.2=P\left(R_N / D\right) \\ & \text { So, } P\left(D \cap R_N\right)=P\left(R_N / D\right) \times P(D) \\ & =0.2 \times 0.3=0.06 \\ & \begin{aligned} P\left(N \cap R_N\right) & =P\left(R_N / N\right) \times P(N) \\ & =0.8 \times 0.7=0.56 \end{aligned} \end{aligned} $

$ \begin{aligned} 0, P\left(R_N\right) & =P\left(D \cap R_N\right)+P\left(N \cap R_N\right) \\ & =0.06+0.56=0.62 \\ P\left(D / R_N\right) & =\frac{P\left(D \cap R_N\right)}{P\left(R_N\right)}=\frac{0.06}{0.62}=\frac{6}{62}=\frac{3}{31} \end{aligned} $

$\therefore$ The probability of an item is actually defective given device reports it as not defective is $\frac{3}{31}$.

Let $A, B$ and $C$ be the event that a student is from section $A, B$ and $C$.

So, $P(A)=0.2, P(B)=0.3, P(C)=0.5$

Now, section A : 20 girls, 30 boys, total 50 students

Section B : 40 girls, 20 boys, total 60 students

Section C : 10 girls, 30 boys, total 40 students

$ \begin{aligned} P(G / A) & =\frac{20}{50}=0.4, P(G / B)=\frac{40}{60}=\frac{2}{3} \\ P(G / C) & =\frac{10}{40}=0.25 \\ P(G) & =P(G / A) \cdot P(A)+P(G / B) \cdot P(B)+P(G / C) \cdot P(C) \\ & =0.4 \times 0.2+\frac{2}{3} \times 0.3+0.25 \times 0.5 \\ & =0.08+0.2+0.125 \\ & =0.405 \end{aligned} $

Now, $P(G \cap A)=P(G / A) \cdot P(A)$

$ =0.4 \times 0.2=0.08 $

Using Bayes' theorem,

$ \begin{aligned} & P(A / G)=\frac{P(A \cap G)}{P(G)} \\ & =\frac{0.08}{0.405}=\frac{80}{405}=\frac{16}{81} \end{aligned} $

So, the probability that a randomly selected girl belongs to section $A$ is $\frac{16}{81}$.

$ \begin{array}{|l|l|l|l|l|} \hline X=x_i & -1 & 0 & 1 & 2 \\ \hline P\left(X=x_i\right) & k^3 & 2 k^3+k & 4 k-10 k^2 & 4 k-1 \\ \hline \end{array} $

$ \begin{aligned} & \sum_{i=-1}^2 P\left(X=x_i\right)=1 \\ \Rightarrow & k^3+2 k^3+k+4 k-10 k^2+4 k-1=1 \\ \Rightarrow & 3 k^3-10 k^2+9 k-2=0 \\ \Rightarrow & (k-1)\left(3 k^2-7 k+2\right)=0 \\ \Rightarrow & 3(k-1)(k-2)\left(k-\frac{1}{3}\right)=0 \end{aligned} $

So, $k=1,2$ or $\frac{1}{3}$

$ k \neq 1,2 $

For $k=\frac{1}{3^{\prime}}$,

$ \begin{aligned} \mu & =-1 \cdot\left(\frac{1}{3}\right)^3+1 \cdot\left(\frac{4}{3}-\frac{10}{9}\right)+2 \cdot\left(\frac{4}{3}-1\right) \\ & =\frac{-1}{27}+0+\frac{12-10}{9}+2 \cdot \frac{1}{3} \\ & =\frac{-1}{27}+\frac{2}{9}+\frac{2}{3}=\frac{-1+6+18}{27}=\frac{23}{27} \end{aligned} $

So, $\mu=\frac{23}{27}$

Mean, $\mu=\frac{16}{5}$, variance, $\sigma^2=\frac{48}{25}$

We know that, $\mu=n p$ and $\sigma^2=n p(1-p)$

So, $\frac{16}{5}=n p$ and $\frac{48}{25}=n p(1-p)$

$ \begin{aligned} & \Rightarrow \quad \frac{16}{5}(1-p)=\frac{48}{25} \\ & \Rightarrow \quad 1-p=\frac{48}{25} \times \frac{5}{16}=\frac{3}{5} \end{aligned} $

$ \Rightarrow \quad p=1-\frac{3}{5}=\frac{2}{5}, 1-p=\frac{3}{5} $

Now, $n p=\frac{16}{5}$

$ \begin{aligned} & \Rightarrow \quad n \cdot \frac{2}{5}=\frac{16}{5} \\ & \Rightarrow \quad n=8 \end{aligned} $

$ \begin{aligned} & \text { Now, } \\ & \begin{array}{r} P(X \leq 2)=P(X=0)+P(X=1)+P(X=2) \\ =\binom{8}{0} P^0(1-p)^8+\binom{8}{1} p^1(1-p)^7 +\binom{8}{2} p^2(1-p)^6 \end{array} \\ & =1 \cdot\left(\frac{2}{5}\right)^0 \cdot\left(\frac{3}{5}\right)^8+8 \cdot \frac{2}{5} \cdot\left(\frac{3}{5}\right)^7 +28 \cdot\left(\frac{2}{5}\right)^2 \cdot\left(\frac{3}{5}\right)^6 \end{aligned} \text { } \text { } $

$ \begin{aligned} & =\left(\frac{3}{5}\right)^8+\frac{16 \cdot 3^7}{5^8}+\frac{28 \cdot 4 \cdot 3^6}{5^8} \\ & =\frac{3^8+16 \cdot 3^7+28 \cdot 4 \cdot 3^6}{5^8} \\ & =\frac{3^6\left(3^2+16 \cdot 3+112\right)}{5^8}=\frac{3^6(169)}{5^8} \end{aligned} $

Total childrens $=15(8 B+7 G)$

Total ways of choosing 3 childrens out of 15 is

$ ={ }^{15} C_3=\frac{15 \times 14 \times 13}{3 \times 2 \times 1}=455 $

Case I Choosing 1 boy and 2 girls

$ ={ }^8 C_1 \cdot{ }^7 C_2=\frac{8 \times 7 \times 6}{2}=168 $

Case II Choosing 2 boys and 1 girl

$ ={ }^8 C_2 \cdot{ }^7 C_1=\frac{8 \times 7}{2} \times 7=196 $

Thus, required probability

$ =\frac{168+196}{455}=\frac{364}{455}=\frac{4}{5} $

Total people $=14(9 \mathrm{M}+5 \mathrm{~W})$

Total ways to choose 4 peoples from 14 is

$ ={ }^{14} C_4 $

Total ways to choose all men is $={ }^9 C_4$

So, total ways to choose atleast one women

$ \begin{aligned} & ={ }^{14} C_4-{ }^9 C_4=\frac{14 \times 13 \times 12 \times 11}{4 \times 3 \times 2 \times 1} -\frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} \\ & =1001-126=875 \end{aligned} $

Hence, required probability

$ =\frac{875}{1001}=\frac{125}{143} $

Total outcomes $=\{1,2,3,4,5,6\}=6$

Prime number $=\{2,3,5\}=3$

So, $P(A)=3 / 6=1 / 2$

and even numbers $=\{2,4,6\}=3$

$\therefore P(B)=3 / 6=1 / 2$

Thus, $P(\bar{B})=1-P(B)=1-1 / 2=1 / 2$

Total sample for 2 throws $=6 \times 6=36$

Then, $P(A / \bar{B})=\frac{P(A \cap \bar{B})}{P(\bar{B})}$

First throw is prime $=\{2,3,5\}$

Second throw is not even $=\{1,3,5\}$

Pairs where first is in $\{2,3,5\}$ and second is in $\{1,3,5\}$

$ \begin{aligned} &=3 \times 3=9 \text { outcomes }\\ &\text { So, } P(A \cap \bar{B})=9 / 36=1 / 4\\ &\text { Therefore, } P(A / \bar{B})=\frac{(1 / 4)}{(1 / 2)}=1 / 2 \end{aligned} $

Let $R_k$ be the event that are exactly $k$ red balls is in bag, where

$ k=1,2,3,4,5 $

So, $P\left(R_k\right)=\frac{1}{5}$ for $k=1,2,3,4,5$

Let $A$ be the event that a red ball is drawn.

$ \begin{aligned} & \begin{aligned} P\left(R_1 / A\right) & =\frac{P\left(R_1 \cap A\right)}{P(A)} \\ \because \quad P\left(A / R_1\right) & =\frac{1}{5} \\ \text { and } P\left(R_1 \cap A\right) & =P\left(R_1\right) \cdot P\left(A / R_1\right) \\ & =\frac{1}{5} \cdot \frac{1}{5}=\frac{1}{25} \end{aligned} \\ & \text { } \begin{aligned}and\,\, & P(A)=\sum_{k=0}^5 P\left(R_k\right) \cdot P\left(A / R_k\right) \\ &=\frac{1}{5} \sum_{k=0}^5 \frac{k}{5}=\frac{1}{25}(1+2+3+4+5) \\ &=\frac{1}{25} \times 15=\frac{15}{25} \\ & \text { Hence, } P\left(R_1 / A\right)=\frac{P\left(R_1 \cap A\right)}{P(A)}=\frac{\frac{1}{25}}{\frac{15}{25}}=\frac{1}{15} \end{aligned} \end{aligned} $

$\because P(X=r)={ }^n C_r p^r(1-p)^{n-r}$

$ \begin{array}{ll} \text { and } & P(X=3)=P(X=6) \\ \Rightarrow & { }^9 C_3 p^3(1-p)^6={ }^9 C_6 p^6(1-p)^3 \\ \Rightarrow & (1-p)^3=p^3 \\ \Rightarrow & \frac{1-p}{p}=1 \\ \Rightarrow & p=1 / 2 \end{array} $

So, $P(X<3)=P(X=0)+P(X=1)+P(X=2)$

$ \begin{aligned} & ={ }^9 C_0\left(\frac{1}{2}\right)^9+{ }^9 C_1\left(\frac{1}{2}\right)^9+{ }^9 C_2\left(\frac{1}{2}\right)^9 \\ & =\frac{1}{512}+\frac{9}{512}+\frac{36}{512}=\frac{46}{512}=23 / 256 \end{aligned} $

We can choose three squares in a diagonal line parallel to $B D$ in the $\triangle A B D$. Clearly, three squares in $\triangle A B D$ and in a diagonal line parallel to $B D$ can be chosen in

$ { }^3 C_3+{ }^4 C_3+{ }^5 C_3+{ }^6 C_3+{ }^7 C_3+{ }^8 C_3 \text { ways } $

Similarly, in $\triangle B C D$ the squares can be chosen parallel to $B D$ in an equal number of ways.

Hence, the total number of ways in which three squares can be chosen in a diagonal line parallel to $B D$ is

$ =2\left({ }^3 C_3+{ }^4 C_3+{ }^5 C_3+{ }^6 C_3+{ }^7 C_3\right)+{ }^8 C_3 $

( $\because B D$ is common to both the triangles) Similarly, squares can be chosen in a diagonal line parallel to $A C$ and hence the total number of favourable ways

$ \begin{aligned} =4\left({ }^3 C_3+{ }^4 C_3+{ }^5 C_3\right. & \left.+{ }^6 C_3+{ }^7 C_3\right) +2 \cdot{ }^8 C_3=392 \end{aligned} $

Hence, the required probability

$ =\frac{392}{{ }^{64} C_3}=\frac{392 \times 6}{64 \times 63 \times 62}=\frac{7}{744} $

Total number of ways to choose 4 shoes from $8={ }^8 C_4=70$

We want to choose 4 shoes that none of them form a pair.

Firstly, choose 4 different pairs from the 4 available pairs $={ }^4 C_4=1$

For each of the 4 pairs, 2 choices (left or right) $=2^4=16$.

Hence, total number of ways to choose 4 shoes with no pair $=1 \times 2^4=16$.

Thus, number of ways with atleast one pair $=70-16=54$.

$\therefore$ Required probability $=\frac{54}{70}=\frac{27}{35}$

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} $