Questions and Answers (Board Exam)

Mastering Board Exam Probability

In Board Exams, the evaluation focuses heavily on the logical progression of your steps. For probability, it is vital to explicitly define your events (for example, "Let $E_1$ be the event that …") and state the specific theorem you are applying (for example, "By Bayes' Theorem …"). This section provides 30 key questions in a board-exam style to help you understand the presentation format required for full marks.

Question 1 [CBSE Style]

If $P(A) = 0.4,\; P(B) = 0.8$ and $P(B\mid A) = 0.6$, find $P(A \cup B)$.

Solution: Let us first use the definition of conditional probability: $$P(B\mid A) = \frac{P(A \cap B)}{P(A)}$$ Therefore, $$P(A \cap B) = P(B\mid A)\,P(A) = 0.6 \times 0.4 = 0.24$$ Now apply the addition theorem: $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$ Substituting the values, $$P(A \cup B) = 0.4 + 0.8 - 0.24 = 0.96$$

Answer: $$\boxed{0.96}$$

Question 2 [CBSE Style]

Ten cards numbered 1 to 10 are placed in a box, mixed thoroughly, and one card is drawn randomly. If it is known that the number on the drawn card is more than 3, what is the probability that it is an even number?

Solution: Let $$A = \{\text{the drawn number is even}\}$$ and $$B = \{\text{the drawn number is greater than }3\}$$ Then, $$B = \{4,5,6,7,8,9,10\}$$ so that $$n(B)=7$$ Also, $$A \cap B = \{4,6,8,10\}$$ Hence, $$n(A \cap B)=4$$ Therefore, by conditional probability, $$P(A\mid B)=\frac{n(A\cap B)}{n(B)}=\frac{4}{7}$$

Answer: $$\boxed{\frac{4}{7}}$$

Question 3 [ISC Style]

Probability of solving a specific problem independently by A and B are $\frac12$ and $\frac13$ respectively. If both try to solve the problem independently, find the probability that the problem is solved.

Solution: Let $$P(A)=\frac12, \qquad P(B)=\frac13$$ Since the attempts are independent, the probability that neither solves the problem is $$P(A' \cap B') = P(A')P(B')$$ Now, $$P(A')=1-\frac12=\frac12, \qquad P(B')=1-\frac13=\frac23$$ Thus, $$P(A' \cap B')=\frac12 \times \frac23 = \frac13$$ Therefore, the probability that the problem is solved is $$1-P(A' \cap B')=1-\frac13=\frac23$$

Answer: $$\boxed{\frac23}$$

Question 4 [CBSE Style]

A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag, which is found to be red. Find the probability that the ball is drawn from the first bag.

Solution: Let $$E_1=\{\text{Bag I is chosen}\}, \qquad E_2=\{\text{Bag II is chosen}\}$$ Since a bag is selected at random, $$P(E_1)=P(E_2)=\frac12$$ Let $$A=\{\text{a red ball is drawn}\}$$ Then, $$P(A\mid E_1)=\frac48=\frac12, \qquad P(A\mid E_2)=\frac28=\frac14$$ By Bayes' theorem, $$P(E_1\mid A)=\frac{P(E_1)P(A\mid E_1)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)}$$ Substituting, $$P(E_1\mid A)=\frac{\frac12\cdot\frac12}{\frac12\cdot\frac12+\frac12\cdot\frac14} =\frac{\frac14}{\frac14+\frac18} =\frac{\frac14}{\frac38} =\frac23$$

Answer: $$\boxed{\frac23}$$

Question 5 [CBSE Style]

A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.

Solution: Let $$E=\{\text{die actually shows 6}\}, \qquad E'=\{\text{die does not show 6}\}$$ Then, $$P(E)=\frac16, \qquad P(E')=\frac56$$ Let $$A=\{\text{man reports that the die shows 6}\}$$ Now, $$P(A\mid E)=\frac34$$ because if the die is actually 6, he reports 6 when he speaks truth. Also, $$P(A\mid E')=\frac14$$ because if the die is not 6, he reports 6 only when he lies. Applying Bayes' theorem, $$P(E\mid A)=\frac{P(E)P(A\mid E)}{P(E)P(A\mid E)+P(E')P(A\mid E')}$$ So, $$P(E\mid A)=\frac{\frac16\cdot\frac34}{\frac16\cdot\frac34+\frac56\cdot\frac14} =\frac{\frac18}{\frac18+\frac{5}{24}} =\frac{\frac18}{\frac{8}{24}} =\frac38$$

Answer: $$\boxed{\frac38}$$

Question 6 [State Board Style]

From a lot of 30 bulbs, which includes 6 defectives, a sample of 4 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs.

Solution: Let $X$ denote the number of defective bulbs obtained in 4 draws. Since drawing is done with replacement, the trials are independent.

Probability of getting a defective bulb is $$p=\frac{6}{30}=\frac15$$ Hence, $$q=1-p=\frac45$$ Therefore, $$X \sim B\left(4,\frac15\right)$$ Using the binomial formula, $$P(X=r)=\binom{4}{r}\left(\frac15\right)^r\left(\frac45\right)^{4-r}, \qquad r=0,1,2,3,4$$ Now, $$P(X=0)=\binom40\left(\frac15\right)^0\left(\frac45\right)^4=\frac{256}{625}$$ $$P(X=1)=\binom41\left(\frac15\right)^1\left(\frac45\right)^3=4\cdot\frac{64}{625}=\frac{256}{625}$$ $$P(X=2)=\binom42\left(\frac15\right)^2\left(\frac45\right)^2=6\cdot\frac{16}{625}=\frac{96}{625}$$ $$P(X=3)=\binom43\left(\frac15\right)^3\left(\frac45\right)^1=4\cdot\frac{4}{625}=\frac{16}{625}$$ $$P(X=4)=\binom44\left(\frac15\right)^4=\frac{1}{625}$$ Thus the distribution is: $$\begin{array}{c|ccccc} X & 0 & 1 & 2 & 3 & 4 \\\hline P(X) & \frac{256}{625} & \frac{256}{625} & \frac{96}{625} & \frac{16}{625} & \frac{1}{625} \end{array}$$

Answer: $$\boxed{P(X=0)=\frac{256}{625},\; P(X=1)=\frac{256}{625},\; P(X=2)=\frac{96}{625},\; P(X=3)=\frac{16}{625},\; P(X=4)=\frac{1}{625}}$$

Question 7 [CBSE Style]

Find the mean and variance of the number of tails in three tosses of a fair coin.

Solution: Let $X$ be the number of tails in 3 tosses of a fair coin. Then, $$X \sim B\left(3,\frac12\right)$$ So, $$n=3, \qquad p=\frac12, \qquad q=\frac12$$ For a binomial distribution, $$E(X)=np, \qquad Var(X)=npq$$ Thus, $$E(X)=3\cdot\frac12=\frac32$$ and $$Var(X)=3\cdot\frac12\cdot\frac12=\frac34$$

Answer: $$\boxed{\text{Mean}=\frac32, \qquad \text{Variance}=\frac34}$$

Question 8 [CBSE Style]

The probability of hitting a target is $\frac14$. If a person fires 7 times, find the probability that he hits the target at least once.

Solution: Let $X$ be the number of hits in 7 shots. Then, $$X \sim B\left(7,\frac14\right)$$ Hence, $$p=\frac14, \qquad q=\frac34$$ We need $$P(X\ge 1)=1-P(X=0)$$ Now, $$P(X=0)=\left(\frac34\right)^7=\frac{2187}{16384}$$ Therefore, $$P(X\ge 1)=1-\frac{2187}{16384}=\frac{14197}{16384}$$

Answer: $$\boxed{\frac{14197}{16384}}$$

Question 9 [ISC Style]

An urn contains 5 white, 7 red and 8 black balls. If 4 balls are drawn one by one with replacement, what is the probability that at least one is white?

Solution: Total number of balls is $$5+7+8=20$$ Probability of drawing a white ball is $$p=\frac{5}{20}=\frac14$$ Therefore, the probability of not drawing a white ball is $$q=1-\frac14=\frac34$$ If 4 draws are made with replacement, then $$P(\text{at least one white})=1-P(\text{no white in all 4 draws})$$ So, $$P(\text{at least one white})=1-\left(\frac34\right)^4 =1-\frac{81}{256} =\frac{175}{256}$$

Answer: $$\boxed{\frac{175}{256}}$$

Question 10 [CBSE Style]

Find the probability distribution of the sum of numbers obtained in two throws of a die.

Solution: When a die is thrown twice, the total number of equally likely ordered outcomes is $$6 \times 6 = 36$$ Let $X$ denote the sum of the two numbers. Then $$X \in \{2,3,4,5,6,7,8,9,10,11,12\}$$ The corresponding frequencies are: $$\begin{aligned} P(X=2)&=\frac{1}{36} \\ P(X=3)&=\frac{2}{36} \\ P(X=4)&=\frac{3}{36} \\ P(X=5)&=\frac{4}{36} \\ P(X=6)&=\frac{5}{36} \\ P(X=7)&=\frac{6}{36} \\ P(X=8)&=\frac{5}{36} \\ P(X=9)&=\frac{4}{36} \\ P(X=10)&=\frac{3}{36} \\ P(X=11)&=\frac{2}{36} \\ P(X=12)&=\frac{1}{36} \end{aligned}$$ Thus the distribution table is $$\begin{array}{c|ccccccccccc} X & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\\hline P(X) & \frac1{36} & \frac2{36} & \frac3{36} & \frac4{36} & \frac5{36} & \frac6{36} & \frac5{36} & \frac4{36} & \frac3{36} & \frac2{36} & \frac1{36} \end{array}$$

Answer: $$\boxed{P(X=2)=\frac1{36},\; P(X=3)=\frac2{36},\; \dots,\; P(X=12)=\frac1{36}}$$

Question 11 [CBSE Style]

If $E$ and $F$ are independent events, then show that $E$ and $F'$ are also independent.

Solution: Since $E$ and $F$ are independent, $$P(E \cap F)=P(E)P(F)$$ Also, $$P(E)=P(E\cap F)+P(E\cap F')$$ Therefore, $$P(E\cap F')=P(E)-P(E\cap F)$$ Substitute the value of $P(E\cap F)$: $$P(E\cap F')=P(E)-P(E)P(F)$$ Take $P(E)$ common: $$P(E\cap F')=P(E)[1-P(F)]$$ But, $$1-P(F)=P(F')$$ Hence, $$P(E\cap F')=P(E)P(F')$$ So $E$ and $F'$ are independent.

Answer: $$\boxed{\text{Hence proved that } E \text{ and } F' \text{ are independent.}}$$

Question 12 [CBSE Style]

Bag A contains 3 red and 5 black balls, Bag B contains 2 red and 3 black balls. One ball is transferred from Bag A to Bag B and then a ball is drawn from Bag B. The drawn ball is found to be red. Find the probability that the transferred ball was black.

Solution: Let $$E_1=\{\text{red ball transferred from A to B}\}, \qquad E_2=\{\text{black ball transferred from A to B}\}$$ Then, $$P(E_1)=\frac38, \qquad P(E_2)=\frac58$$ Let $$A=\{\text{red ball drawn from Bag B}\}$$ If a red ball is transferred, Bag B becomes: $$3R,\;3B$$ Hence, $$P(A\mid E_1)=\frac36=\frac12$$ If a black ball is transferred, Bag B becomes: $$2R,\;4B$$ Hence, $$P(A\mid E_2)=\frac26=\frac13$$ By Bayes' theorem, $$P(E_2\mid A)=\frac{P(E_2)P(A\mid E_2)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)}$$ Thus, $$P(E_2\mid A)=\frac{\frac58\cdot\frac13}{\frac38\cdot\frac12+\frac58\cdot\frac13} =\frac{\frac{5}{24}}{\frac{3}{16}+\frac{5}{24}}$$ Taking LCM $48$ in the denominator, $$\frac{3}{16}+\frac{5}{24}=\frac{9}{48}+\frac{10}{48}=\frac{19}{48}$$ So, $$P(E_2\mid A)=\frac{5}{24}\cdot\frac{48}{19}=\frac{10}{19}$$

Answer: $$\boxed{\frac{10}{19}}$$

Question 13 [CBSE Style]

Given that the two numbers appearing on throwing two dice are different, find the probability that the sum of the numbers on the dice is 4.

Solution: Let $$B=\{\text{numbers on the two dice are different}\}$$ There are 36 total outcomes in throwing two dice. Out of these, 6 outcomes have equal numbers: $$(1,1),(2,2),\dots,(6,6)$$ Hence, $$n(B)=36-6=30$$ Let $$A=\{\text{sum of numbers is }4\}$$ Then, $$A=\{(1,3),(2,2),(3,1)\}$$ Since the numbers are different, we exclude $(2,2)$. Thus, $$A\cap B=\{(1,3),(3,1)\}$$ So, $$n(A\cap B)=2$$ Therefore, $$P(A\mid B)=\frac{n(A\cap B)}{n(B)}=\frac{2}{30}=\frac{1}{15}$$

Answer: $$\boxed{\frac{1}{15}}$$

Question 14 [ISC Style]

A laboratory blood test is 99% effective in detecting a certain disease when it is, in fact, present. However, the test also yields a false positive for 0.5% of healthy people. If 0.1% of the population has the disease, what is the probability that a person has the disease given that his test is positive?

Solution: Let $$E_1=\{\text{person has the disease}\}, \qquad E_2=\{\text{person is healthy}\}$$ Then, $$P(E_1)=0.001, \qquad P(E_2)=0.999$$ Let $$A=\{\text{test result is positive}\}$$ Given, $$P(A\mid E_1)=0.99, \qquad P(A\mid E_2)=0.005$$ By Bayes' theorem, $$P(E_1\mid A)=\frac{P(E_1)P(A\mid E_1)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)}$$ Substituting the values, $$P(E_1\mid A)=\frac{0.001\times 0.99}{0.001\times 0.99+0.999\times 0.005}$$ $$=\frac{0.00099}{0.00099+0.004995} =\frac{0.00099}{0.005985} =\frac{22}{133}$$ Hence, $$P(E_1\mid A)\approx 0.1654$$

Answer: $$\boxed{\frac{22}{133} \approx 0.1654}$$

Question 15 [CBSE Style]

Two cards are drawn successively with replacement from a well-shuffled pack of 52 cards. Find the probability distribution of the number of aces.

Solution: Let $X$ be the number of aces obtained in two draws with replacement. Then, $$X \in \{0,1,2\}$$ Probability of drawing an ace in one draw is $$p=\frac{4}{52}=\frac{1}{13}$$ Therefore, $$q=1-p=\frac{12}{13}$$ So, $$X \sim B\left(2,\frac{1}{13}\right)$$ Now, $$P(X=0)=\binom20\left(\frac{1}{13}\right)^0\left(\frac{12}{13}\right)^2=\frac{144}{169}$$ $$P(X=1)=\binom21\left(\frac{1}{13}\right)\left(\frac{12}{13}\right)=\frac{24}{169}$$ $$P(X=2)=\binom22\left(\frac{1}{13}\right)^2=\frac{1}{169}$$ Thus the probability distribution is: $$\begin{array}{c|ccc} X & 0 & 1 & 2 \\\hline P(X) & \frac{144}{169} & \frac{24}{169} & \frac{1}{169} \end{array}$$

Answer: $$\boxed{P(X=0)=\frac{144}{169},\; P(X=1)=\frac{24}{169},\; P(X=2)=\frac{1}{169}}$$

Question 16 [State Board Style]

In a series of 10 Bernoulli trials, if $P(\text{Success}) = p$, and $P(X=1) = P(X=2)$, find $p$.

Solution: Since $X$ follows a binomial distribution with parameters $n=10$ and success probability $p$, $$P(X=1)=\binom{10}{1}p q^9$$ and $$P(X=2)=\binom{10}{2}p^2 q^8$$ Given that $$P(X=1)=P(X=2)$$ we get $$\binom{10}{1}p q^9=\binom{10}{2}p^2 q^8$$ That is, $$10pq^9 = 45p^2 q^8$$ Cancelling the common factors $pq^8$, $$10q=45p$$ Since $$q=1-p$$ we have $$10(1-p)=45p$$ $$10=55p$$ $$p=\frac{10}{55}=\frac{2}{11}$$

Answer: $$\boxed{\frac{2}{11}}$$

Question 17 [CBSE Style]

Two independent events A and B are such that $P(A \cup B) = 0.6$ and $P(A) = 0.2$. Find $P(B)$.

Solution: Let $$P(B)=x$$ Since A and B are independent, $$P(A\cap B)=P(A)P(B)=0.2x$$ Now use the addition theorem: $$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$ So, $$0.6=0.2+x-0.2x$$ $$0.6=0.2+0.8x$$ $$0.4=0.8x$$ $$x=0.5$$ Hence, $$P(B)=0.5$$

Answer: $$\boxed{0.5}$$

Question 18 [CBSE Style]

A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all three oranges are good, the box is approved. Find the probability that a box containing 15 oranges out of which 12 are good will be approved.

Solution: A box will be approved only if all the three selected oranges are good. The probability is therefore $$\frac{12}{15}\times\frac{11}{14}\times\frac{10}{13}$$ Simplifying, $$\frac{12}{15}=\frac45$$ Hence, $$P(\text{approved})=\frac45\times\frac{11}{14}\times\frac{10}{13}$$ Canceling common factors, $$=\frac{44}{91}$$

Answer: $$\boxed{\frac{44}{91}}$$

Question 19 [ISC Style]

A random variable $X$ has PMF: $P(X=0)=k$, $P(X=1)=2k$, $P(X=2)=3k$. Find $P(X < 2)$.

Solution: Since total probability is 1, $$P(X=0)+P(X=1)+P(X=2)=1$$ Thus, $$k+2k+3k=1$$ $$6k=1$$ $$k=\frac16$$ Now, $$P(X<2)=P(X=0)+P(X=1)=k+2k=3k$$ Hence, $$P(X<2)=3\cdot\frac16=\frac12$$

Answer: $$\boxed{\frac12}$$

Question 20 [CBSE Style]

A pair of dice is thrown. If the sum is 8, find the probability that one of the numbers is 5.

Solution: Let $$B=\{\text{sum is }8\}$$ Then, $$B=\{(2,6),(3,5),(4,4),(5,3),(6,2)\}$$ so that $$n(B)=5$$ Let $$A=\{\text{one of the numbers is }5\}$$ Among the outcomes in $B$, the favorable ones are $$A\cap B=\{(3,5),(5,3)\}$$ Hence, $$n(A\cap B)=2$$ Therefore, $$P(A\mid B)=\frac{2}{5}$$

Answer: $$\boxed{\frac25}$$

Question 21 [Corrected for consistency]

Given $P(A)=\frac12$, $P(B)=\frac{7}{12}$ and $P(A' \cup B')=\frac{17}{24}$, are A and B independent?

Solution: Using De Morgan's law, $$A' \cup B' = (A \cap B)'$$ Therefore, $$P(A' \cup B') = 1 - P(A \cap B)$$ Given that $$P(A' \cup B')=\frac{17}{24}$$ we obtain $$\frac{17}{24}=1-P(A \cap B)$$ Hence, $$P(A \cap B)=1-\frac{17}{24}=\frac{7}{24}$$ Now, $$P(A)P(B)=\frac12\times\frac{7}{12}=\frac{7}{24}$$ Since $$P(A\cap B)=P(A)P(B)$$ A and B are independent.

Answer: $$\boxed{\text{Yes, A and B are independent.}}$$

Question 22 [CBSE Style]

Find the variance of the number of heads in two tosses of a coin.

Solution: Let $X$ denote the number of heads obtained in two tosses of a fair coin. Then, $$X \sim B\left(2,\frac12\right)$$ Thus, $$n=2, \qquad p=\frac12, \qquad q=\frac12$$ For a binomial distribution, $$Var(X)=npq$$ So, $$Var(X)=2\cdot\frac12\cdot\frac12=\frac12$$

Answer: $$\boxed{\frac12}$$

Question 23 [State Board Style]

In a factory, Machines A, B, C produce 50%, 30%, 20% of items. Defective rates are 1%, 2%, 3% respectively. If an item is defective, find the probability that it was produced by machine C.

Solution: Let $$E_1=\{\text{item produced by A}\}, \quad E_2=\{\text{item produced by B}\}, \quad E_3=\{\text{item produced by C}\}$$ Then, $$P(E_1)=0.5, \qquad P(E_2)=0.3, \qquad P(E_3)=0.2$$ Let $$D=\{\text{item is defective}\}$$ Given, $$P(D\mid E_1)=0.01, \qquad P(D\mid E_2)=0.02, \qquad P(D\mid E_3)=0.03$$ By Bayes' theorem, $$P(E_3\mid D)=\frac{P(E_3)P(D\mid E_3)}{P(E_1)P(D\mid E_1)+P(E_2)P(D\mid E_2)+P(E_3)P(D\mid E_3)}$$ Substituting, $$P(E_3\mid D)=\frac{0.2\cdot 0.03}{0.5\cdot 0.01+0.3\cdot 0.02+0.2\cdot 0.03}$$ $$=\frac{0.006}{0.005+0.006+0.006}=\frac{0.006}{0.017}=\frac{6}{17}$$

Answer: $$\boxed{\frac{6}{17}}$$

Question 24 [CBSE Style]

If $P(A\mid B) > P(A)$, show that $P(B\mid A) > P(B)$.

Solution: Given, $$P(A\mid B)>P(A)$$ Using the definition of conditional probability, $$\frac{P(A\cap B)}{P(B)} > P(A)$$ Multiplying both sides by $P(B)$, $$P(A\cap B) > P(A)P(B)$$ Now divide both sides by $P(A)$: $$\frac{P(A\cap B)}{P(A)} > P(B)$$ But, $$\frac{P(A\cap B)}{P(A)}=P(B\mid A)$$ Therefore, $$P(B\mid A)>P(B)$$

Answer: $$\boxed{\text{Hence proved that } P(B\mid A)>P(B).}$$

Question 25 [CBSE Style]

Find the probability of getting exactly two 6's in 5 throws of a die.

Solution: Let $X$ be the number of times 6 appears in 5 throws. Then, $$X \sim B\left(5,\frac16\right)$$ Thus, $$p=\frac16, \qquad q=\frac56$$ We need $$P(X=2)=\binom52\left(\frac16\right)^2\left(\frac56\right)^3$$ Now, $$\binom52=10$$ Therefore, $$P(X=2)=10\cdot\frac{1}{36}\cdot\frac{125}{216} =\frac{1250}{7776} =\frac{625}{3888}$$

Answer: $$\boxed{\frac{625}{3888}}$$

Question 26 [CBSE Style]

Events A and B are such that $P(A)=\frac12$, $P(A \cup B)=\frac35$ and $P(B)=p$. Find $p$ if A and B are independent.

Solution: Since A and B are independent, $$P(A\cap B)=P(A)P(B)=\frac12 p$$ Now use the addition theorem: $$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$ Thus, $$\frac35 = \frac12 + p - \frac12 p$$ So, $$\frac35 - \frac12 = \frac12 p$$ $$\frac{6-5}{10}=\frac12 p$$ $$\frac1{10}=\frac12 p$$ Hence, $$p=\frac{1}{5}$$

Answer: $$\boxed{\frac15}$$

Question 27 [ISC Style]

Find the mean of the number of heads in three tosses of a fair coin.

Solution: Let $X$ denote the number of heads in 3 tosses of a fair coin. Then, $$X \sim B\left(3,\frac12\right)$$ For a binomial distribution, $$E(X)=np$$ Thus, $$E(X)=3\cdot\frac12=\frac32$$

Answer: $$\boxed{\frac32}$$

Question 28 [CBSE Style]

A card is lost from a pack of 52. Two cards drawn from the remainder are both spades. Find the probability that the lost card was a spade.

Solution: Let $$E_1=\{\text{lost card is a spade}\}, \qquad E_2=\{\text{lost card is not a spade}\}$$ Then, $$P(E_1)=\frac{13}{52}=\frac14, \qquad P(E_2)=\frac34$$ Let $$A=\{\text{the two drawn cards are both spades}\}$$ If the lost card is a spade, then 12 spades remain among 51 cards. Hence, $$P(A\mid E_1)=\frac{\binom{12}{2}}{\binom{51}{2}}=\frac{66}{1275}$$ If the lost card is not a spade, then 13 spades remain among 51 cards. Hence, $$P(A\mid E_2)=\frac{\binom{13}{2}}{\binom{51}{2}}=\frac{78}{1275}$$ By Bayes' theorem, $$P(E_1\mid A)=\frac{P(E_1)P(A\mid E_1)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)}$$ Therefore, $$P(E_1\mid A)=\frac{\frac14\cdot 66}{\frac14\cdot 66+\frac34\cdot 78} =\frac{66}{66+234} =\frac{66}{300} =\frac{11}{50}$$

Answer: $$\boxed{\frac{11}{50}}$$

Question 29 [CBSE Style]

If the mean of a binomial distribution is 20 and the standard deviation is 4, find $n$.

Solution: For a binomial distribution, $$np=20$$ Also, the standard deviation is 4, so $$\sqrt{npq}=4$$ Squaring both sides, $$npq=16$$ Now divide by $np=20$: $$q=\frac{16}{20}=0.8$$ Hence, $$p=1-q=1-0.8=0.2$$ Using $np=20$, $$n\cdot 0.2=20$$ $$n=100$$

Answer: $$\boxed{100}$$

Question 30 [CBSE Style]

If $P(A)=\frac{7}{13},\; P(B)=\frac{9}{13}$ and $P(A \cap B)=\frac{4}{13}$, find $P(A'\mid B)$.

Solution: First find $$P(A\mid B)=\frac{P(A\cap B)}{P(B)}$$ So, $$P(A\mid B)=\frac{\frac{4}{13}}{\frac{9}{13}}=\frac49$$ Now, $$P(A'\mid B)=1-P(A\mid B)=1-\frac49=\frac59$$

Answer: $$\boxed{\frac59}$$