Solved Examples

Important Theorems and Formulas Recap

1. Conditional Probability:

$$P(A\mid B)=\frac{P(A\cap B)}{P(B)}, \qquad P(B)\neq 0$$

2. Multiplication Theorem:

• For dependent events: $$P(A\cap B)=P(A)\,P(B\mid A)$$
• For independent events: $$P(A\cap B)=P(A)\,P(B)$$

3. Theorem of Total Probability

If $E_1,E_2,\dots,E_n$ is a partition of the sample space $S$, then for any event $A$, $$P(A)=\sum_{j=1}^{n} P(E_j)\,P(A\mid E_j)$$

4. Bayes' Theorem

$$P(E_i\mid A)=\frac{P(E_i)\,P(A\mid E_i)}{\sum_{j=1}^{n} P(E_j)\,P(A\mid E_j)}$$

5. Random Variable Parameters

• Mean (Expectation): $$E(X)=\sum x_i p_i$$
• Variance: $$Var(X)=E(X^2)-[E(X)]^2$$

6. Binomial Distribution $B(n,p)$

• Probability mass function: $$P(X=r)=\binom{n}{r}p^r q^{n-r}, \qquad q=1-p$$
• Mean: $$E(X)=np$$
• Variance: $$Var(X)=npq$$

Example 1: Basic Conditional Probability

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

Solution: Using the definition of conditional probability, $$P(A\mid B)=\frac{P(A\cap B)}{P(B)}$$ Substituting the given values, $$P(A\mid B)=\frac{\frac{4}{13}}{\frac{9}{13}}=\frac{4}{13}\cdot\frac{13}{9}=\frac{4}{9}$$

Answer: $$\boxed{\frac{4}{9}}$$

Example 2: Family Probability

A family has two children. What is the probability that both are boys given that at least one of them is a boy?

Solution: Assume that each child is equally likely to be a boy or a girl, and the births are independent.

The sample space is $$S=\{BB, BG, GB, GG\}$$ Let $$A=\{BB\} \quad \text{(both are boys)}$$ and $$B=\{BB,BG,GB\} \quad \text{(at least one is a boy)}$$ Then, $$A\cap B=\{BB\}$$ Hence, $$P(A\mid B)=\frac{n(A\cap B)}{n(B)}=\frac{1}{3}$$

Answer: $$\boxed{\frac{1}{3}}$$

Example 3: Two Dice Sum

A die is thrown twice and the sum of the numbers appearing is observed to be 6. What is the conditional probability that the number 4 has appeared at least once?

Solution: Let $$B=\{\text{sum is }6\}$$ Then the possible ordered outcomes are $$B=\{(1,5),(2,4),(3,3),(4,2),(5,1)\}$$ So, $$n(B)=5$$ Let $$A=\{\text{the number }4\text{ appears at least once}\}$$ Among the outcomes in $B$, the favorable ones for $A$ are $$A\cap B=\{(2,4),(4,2)\}$$ Thus, $$n(A\cap B)=2$$ Therefore, $$P(A\mid B)=\frac{n(A\cap B)}{n(B)}=\frac{2}{5}$$

Answer: $$\boxed{\frac{2}{5}}$$

Example 4: Independent Events Check

If $P(A)=\frac{3}{5}$ and $P(B)=\frac{1}{5}$, find $P(A\cap B)$ if $A$ and $B$ are independent events.

Solution: For independent events, $$P(A\cap B)=P(A)\,P(B)$$ Hence, $$P(A\cap B)=\frac{3}{5}\cdot\frac{1}{5}=\frac{3}{25}$$

Answer: $$\boxed{\frac{3}{25}}$$

Example 5: Target Hitting

A and B appear for an interview. The probability of A's selection is $\frac{1}{7}$ and B's selection is $\frac{1}{5}$. Find the probability that at least one of them is selected.

Solution: Assuming the selections are independent, $$P(A)=\frac{1}{7}, \qquad P(B)=\frac{1}{5}$$ The probability that at least one is selected is $$1-P(\text{none selected})$$ Now, $$P(A')=1-\frac{1}{7}=\frac{6}{7}, \qquad P(B')=1-\frac{1}{5}=\frac{4}{5}$$ So, $$P(\text{none selected})=P(A'\cap B')=P(A')P(B')=\frac{6}{7}\cdot\frac{4}{5}=\frac{24}{35}$$ Hence, $$P(\text{at least one selected})=1-\frac{24}{35}=\frac{11}{35}$$

Answer: $$\boxed{\frac{11}{35}}$$

Example 6: Total Probability (Bags)

Bag I contains 3 red and 4 black balls while Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags. Find the probability that the ball drawn is red.

Solution: Let $$E_1=\{\text{Bag I is chosen}\}, \qquad E_2=\{\text{Bag II is chosen}\}$$ Since one of the two bags is chosen at random, $$P(E_1)=P(E_2)=\frac{1}{2}$$ Let $$A=\{\text{red ball is drawn}\}$$ Then, $$P(A\mid E_1)=\frac{3}{7}, \qquad P(A\mid E_2)=\frac{5}{11}$$ By the theorem of total probability, $$P(A)=P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)$$ $$=\frac{1}{2}\cdot\frac{3}{7}+\frac{1}{2}\cdot\frac{5}{11}$$ $$=\frac{3}{14}+\frac{5}{22}$$ Taking LCM $154$, $$P(A)=\frac{33+35}{154}=\frac{68}{154}=\frac{34}{77}$$

Answer: $$\boxed{\frac{34}{77}}$$

Example 7: Bayes' Theorem (Classic)

In a factory, machine A produces 60% and machine B produces 40% of items. 2% from A and 1% from B are defective. An item is chosen and found to be defective. Find the probability it was from machine B.

Solution: Let $$E_A=\{\text{item produced by A}\}, \qquad E_B=\{\text{item produced by B}\}$$ Then, $$P(E_A)=0.6, \qquad P(E_B)=0.4$$ Let $$D=\{\text{item is defective}\}$$ Given, $$P(D\mid E_A)=0.02, \qquad P(D\mid E_B)=0.01$$ First compute $P(D)$ using total probability: $$P(D)=P(E_A)P(D\mid E_A)+P(E_B)P(D\mid E_B)$$ $$=(0.6)(0.02)+(0.4)(0.01)=0.012+0.004=0.016$$ Now apply Bayes' theorem: $$P(E_B\mid D)=\frac{P(E_B)P(D\mid E_B)}{P(D)}$$ $$=\frac{(0.4)(0.01)}{0.016}=\frac{0.004}{0.016}=\frac{1}{4}$$

Answer: $$\boxed{\frac{1}{4}}$$

Example 8: Student Knowing/Guessing

A student knows the answer with probability $\frac{3}{4}$ and guesses with probability $\frac{1}{4}$. If he guesses, the probability of being correct is $\frac{1}{4}$. Find the probability he knew the answer given that his answer was correct.

Solution: Let $$E_1=\{\text{student knows the answer}\}, \qquad E_2=\{\text{student guesses}\}$$ Then, $$P(E_1)=\frac{3}{4}, \qquad P(E_2)=\frac{1}{4}$$ Let $$C=\{\text{answer is correct}\}$$ Since a student who knows the answer answers correctly with certainty, $$P(C\mid E_1)=1$$ If he guesses, $$P(C\mid E_2)=\frac{1}{4}$$ By Bayes' theorem, $$P(E_1\mid C)=\frac{P(E_1)P(C\mid E_1)}{P(E_1)P(C\mid E_1)+P(E_2)P(C\mid E_2)}$$ $$=\frac{\frac{3}{4}\cdot 1}{\frac{3}{4}\cdot 1+\frac{1}{4}\cdot\frac{1}{4}}$$ $$=\frac{\frac{3}{4}}{\frac{3}{4}+\frac{1}{16}}$$ $$=\frac{\frac{3}{4}}{\frac{13}{16}}=\frac{3}{4}\cdot\frac{16}{13}=\frac{12}{13}$$

Answer: $$\boxed{\frac{12}{13}}$$

Example 9: Bayes (Card Lost)

A card from a pack of 52 is lost. From the remaining cards, 2 cards are drawn and found to be diamonds. Find the probability that the lost card was a diamond.

Solution: Let $$E_1=\{\text{lost card is a diamond}\}, \qquad E_2=\{\text{lost card is not a diamond}\}$$ Then, $$P(E_1)=\frac{13}{52}=\frac{1}{4}, \qquad P(E_2)=\frac{3}{4}$$ Let $$A=\{\text{the two drawn cards are diamonds}\}$$ If the lost card is a diamond, then 12 diamonds remain among 51 cards, so $$P(A\mid E_1)=\frac{\binom{12}{2}}{\binom{51}{2}}=\frac{66}{1275}$$ If the lost card is not a diamond, then all 13 diamonds remain among 51 cards, so $$P(A\mid E_2)=\frac{\binom{13}{2}}{\binom{51}{2}}=\frac{78}{1275}$$ Applying Bayes' theorem, $$P(E_1\mid A)=\frac{P(E_1)P(A\mid E_1)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)}$$ $$=\frac{\frac{1}{4}\cdot 66}{\frac{1}{4}\cdot 66+\frac{3}{4}\cdot 78}$$ $$=\frac{66}{66+234}=\frac{66}{300}=\frac{11}{50}$$

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

Example 10: Bayes (Insurance)

An insurance company has insured 2000 scooters, 4000 cars, and 6000 trucks. The probability of an accident involving a scooter, car, and truck are 0.01, 0.03, and 0.15 respectively. If one of the insured vehicles meets with an accident, find the probability that it is a scooter.

Solution: Let $$E_1=\{\text{vehicle is a scooter}\}, \quad E_2=\{\text{vehicle is a car}\}, \quad E_3=\{\text{vehicle is a truck}\}$$ Total insured vehicles: $$2000+4000+6000=12000$$ Hence, $$P(E_1)=\frac{2000}{12000}=\frac{1}{6}, \qquad P(E_2)=\frac{1}{3}, \qquad P(E_3)=\frac{1}{2}$$ Let $$A=\{\text{vehicle had an accident}\}$$ Then, $$P(A\mid E_1)=0.01, \qquad P(A\mid E_2)=0.03, \qquad P(A\mid E_3)=0.15$$ By Bayes' theorem, $$P(E_1\mid A)=\frac{P(E_1)P(A\mid E_1)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)+P(E_3)P(A\mid E_3)}$$ $$=\frac{\frac{1}{6}(0.01)}{\frac{1}{6}(0.01)+\frac{1}{3}(0.03)+\frac{1}{2}(0.15)}$$ Now, $$\frac{1}{6}(0.01)=\frac{1}{600}, \qquad \frac{1}{3}(0.03)=0.01, \qquad \frac{1}{2}(0.15)=0.075$$ So, $$P(E_1\mid A)=\frac{\frac{1}{600}}{\frac{1}{600}+0.01+0.075}$$ Converting to denominator $600$, $$\frac{1}{600}+0.01+0.075=\frac{1}{600}+\frac{6}{600}+\frac{45}{600}=\frac{52}{600}$$ Hence, $$P(E_1\mid A)=\frac{1/600}{52/600}=\frac{1}{52}$$

Answer: $$\boxed{\frac{1}{52}}$$

Example 11: Random Variable Distribution

Find the probability distribution of the number of tails in three tosses of a coin.

Solution: Let $X$ denote the number of tails in 3 tosses. Then possible values of $X$ are $$X\in\{0,1,2,3\}$$ Since there are $$2^3=8$$ equally likely outcomes, we get:

For $X=0$: $$P(X=0)=P(HHH)=\frac{1}{8}$$ For $X=1$: $$P(X=1)=P(HHT,HTH,THH)=\frac{3}{8}$$ For $X=2$: $$P(X=2)=P(TTH,THT,HTT)=\frac{3}{8}$$ For $X=3$: $$P(X=3)=P(TTT)=\frac{1}{8}$$

Answer: $$P(X=0)=\frac{1}{8}, \quad P(X=1)=\frac{3}{8}, \quad P(X=2)=\frac{3}{8}, \quad P(X=3)=\frac{1}{8}$$

Example 12: Finding $k$ in Distribution

$P(X=x)=kx^2$ for $x=1,2,3$. Find $k$ and $E(X)$.

Solution: Since total probability is 1, $$\sum P(X=x)=1$$ Thus, $$k(1^2+2^2+3^2)=1$$ $$k(1+4+9)=1$$ $$14k=1$$ $$k=\frac{1}{14}$$ Now, $$E(X)=\sum x\,P(X=x)$$ $$=1\cdot k(1^2)+2\cdot k(2^2)+3\cdot k(3^2)$$ $$=k(1+8+27)=36k$$ $$E(X)=36\cdot \frac{1}{14}=\frac{18}{7}$$

Answer: $$\boxed{k=\frac{1}{14}, \quad E(X)=\frac{18}{7}}$$

Example 13: Mean and Variance Calculation

Find mean and variance for $X$: $\{0,1\}$ with probabilities $\{q,p\}$.

Solution: The random variable takes values 0 and 1 with probabilities $q$ and $p$ respectively, where $$p+q=1$$ Now, $$E(X)=0\cdot q+1\cdot p=p$$ Also, $$E(X^2)=0^2\cdot q+1^2\cdot p=p$$ Therefore, $$Var(X)=E(X^2)-[E(X)]^2$$ $$=p-p^2=p(1-p)=pq$$

Answer: $$\boxed{\text{Mean}=p, \quad \text{Variance}=pq}$$

Example 14: Binomial (Exactly $r$)

A fair coin is tossed 10 times. Find the probability of exactly 6 heads.

Solution: Here, $$n=10, \quad p=\frac{1}{2}, \quad q=\frac{1}{2}, \quad r=6$$ Using the binomial formula, $$P(X=6)=\binom{10}{6}\left(\frac{1}{2}\right)^6\left(\frac{1}{2}\right)^4$$ $$=\binom{10}{6}\left(\frac{1}{2}\right)^{10}$$ Since, $$\binom{10}{6}=\binom{10}{4}=210$$ Therefore, $$P(X=6)=\frac{210}{1024}=\frac{105}{512}$$

Answer: $$\boxed{\frac{105}{512}}$$

Example 15: Binomial (At least once)

Probability of hitting a target is $\frac{1}{4}$. A man fires 7 times. Find the probability of hitting the target at least once.

Solution: Let success mean hitting the target. Then, $$n=7, \quad p=\frac{1}{4}, \quad q=\frac{3}{4}$$ We need $$P(X\ge 1)$$ Using the complement rule, $$P(X\ge 1)=1-P(X=0)$$ Now, $$P(X=0)=\left(\frac{3}{4}\right)^7=\frac{2187}{16384}$$ Hence, $$P(X\ge 1)=1-\frac{2187}{16384}=\frac{14197}{16384}$$

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

Example 16: Inverse Binomial (Finding $n$)

Mean of $B(n,p)$ is 4 and variance is 3. Find $n$.

Solution: For a binomial distribution, $$np=4$$ and $$npq=3$$ Dividing the second equation by the first, $$q=\frac{3}{4}$$ Therefore, $$p=1-q=1-\frac{3}{4}=\frac{1}{4}$$ Now use $$np=4$$ so, $$n\cdot \frac{1}{4}=4$$ $$n=16$$

Answer: $$\boxed{16}$$

Example 17: Truth Reporting with a Die

A man speaks truth $\frac{3}{4}$ of the time. He reports that the die shows 6. Find the probability that the die actually showed 6.

Solution: Let $$E_1=\{\text{actual outcome is }6\}, \qquad E_2=\{\text{actual outcome is not }6\}$$ Then, $$P(E_1)=\frac{1}{6}, \qquad P(E_2)=\frac{5}{6}$$ Let $$A=\{\text{man reports }6\}$$ If the actual outcome is 6 and he tells the truth, $$P(A\mid E_1)=\frac{3}{4}$$ If the actual outcome is not 6 and he reports 6, then he must be lying: $$P(A\mid E_2)=\frac{1}{4}$$ By Bayes' theorem, $$P(E_1\mid A)=\frac{P(E_1)P(A\mid E_1)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)}$$ $$=\frac{\frac{1}{6}\cdot\frac{3}{4}}{\frac{1}{6}\cdot\frac{3}{4}+\frac{5}{6}\cdot\frac{1}{4}}$$ $$=\frac{3}{3+5}=\frac{3}{8}$$

Answer: $$\boxed{\frac{3}{8}}$$

Example 18: Multiple Targets

A hits with probability $\frac{4}{5}$, B hits with probability $\frac{3}{4}$, and C hits with probability $\frac{2}{3}$. All fire once. Find the probability that exactly two hit the target.

Solution: Exactly two hit in the following mutually exclusive cases: $$AB\,C', \qquad A B' C, \qquad A' B C$$ Hence, $$P(\text{exactly two hit})=P(AB C')+P(A B' C)+P(A' B C)$$ Using independence, $$=\left(\frac{4}{5}\cdot\frac{3}{4}\cdot\frac{1}{3}\right)+\left(\frac{4}{5}\cdot\frac{1}{4}\cdot\frac{2}{3}\right)+\left(\frac{1}{5}\cdot\frac{3}{4}\cdot\frac{2}{3}\right)$$ $$=\frac{12}{60}+\frac{8}{60}+\frac{6}{60}=\frac{26}{60}=\frac{13}{30}$$

Answer: $$\boxed{\frac{13}{30}}$$

Example 19: Drawing without Replacement

An urn contains 5 white and 3 black balls. Two balls are drawn without replacement. Find the probability distribution of the number of black balls drawn.

Solution: Let $X$ be the number of black balls drawn in 2 draws. Then, $$X\in\{0,1,2\}$$ Now,

For $X=0$: $$P(X=0)=\frac{5}{8}\cdot\frac{4}{7}=\frac{20}{56}=\frac{5}{14}$$ For $X=1$: This can happen in two ways: $$WB \quad \text{or} \quad BW$$ So, $$P(X=1)=\frac{5}{8}\cdot\frac{3}{7}+\frac{3}{8}\cdot\frac{5}{7}=\frac{30}{56}=\frac{15}{28}$$ For $X=2$: $$P(X=2)=\frac{3}{8}\cdot\frac{2}{7}=\frac{6}{56}=\frac{3}{28}$$ Thus the distribution is: $$P(X=0)=\frac{5}{14}, \quad P(X=1)=\frac{15}{28}, \quad P(X=2)=\frac{3}{28}$$

Answer: $$\boxed{\left(\frac{5}{14},\;\frac{15}{28},\;\frac{3}{28}\right)}$$

Example 20: Dice (Multiple 6s)

A die is rolled 6 times. Find the probability of getting at most two 6s.

Solution: Let success mean getting a 6. Then, $$n=6, \quad p=\frac{1}{6}, \quad q=\frac{5}{6}$$ We need $$P(X\le 2)=P(X=0)+P(X=1)+P(X=2)$$ Now, $$P(X=0)=\left(\frac{5}{6}\right)^6$$ $$P(X=1)=\binom{6}{1}\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)^5=6\cdot\frac{1}{6}\left(\frac{5}{6}\right)^5=\left(\frac{5}{6}\right)^5$$ $$P(X=2)=\binom{6}{2}\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^4=15\cdot\frac{1}{36}\left(\frac{5}{6}\right)^4$$ Taking $\left(\frac{5}{6}\right)^4$ common, $$P(X\le 2)=\left(\frac{5}{6}\right)^4\left[\frac{25}{36}+\frac{30}{36}+\frac{15}{36}\right]$$ $$=\left(\frac{5}{6}\right)^4\cdot\frac{70}{36}$$ $$=\frac{35}{18}\left(\frac{5}{6}\right)^4$$ Equivalently, $$P(X\le 2)=\frac{4375}{5832}$$

Answer: $$\boxed{\frac{35}{18}\left(\frac{5}{6}\right)^4=\frac{4375}{5832}}$$

JEE Main Standard

Example 21: Sum of Mean and Variance

If for a binomial distribution, the sum of its mean and variance is 1.8 and their product is 0.8, find $n$.

Solution: Let mean $=m$ and variance $=v$. Given, $$m+v=1.8, \qquad mv=0.8$$ So $m$ and $v$ are the roots of $$t^2-1.8t+0.8=0$$ Multiplying by 10, $$10t^2-18t+8=0$$ Dividing by 2, $$5t^2-9t+4=0$$ Factorizing, $$(5t-4)(t-1)=0$$ Hence, $$t=1 \quad \text{or} \quad t=\frac{4}{5}$$ Since for a binomial distribution mean is greater than variance (when $0<p<1$), we take $$m=1, \qquad v=0.8$$ Now, $$np=1, \qquad npq=0.8$$ Dividing, $$q=0.8$$ So, $$p=1-0.8=0.2$$ Using $np=1$, $$n(0.2)=1$$ $$n=5$$

Answer: $$\boxed{5}$$

Example 22: Intersection of Independent Events

If A and B are independent events such that $P(A\cup B)=0.6$ and $P(A)=0.2$, find $P(B)$.

Solution: Let $$P(B)=x$$ Now, $$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$ Since A and B are independent, $$P(A\cap B)=P(A)P(B)=0.2x$$ Therefore, $$0.6=0.2+x-0.2x$$ $$0.4=0.8x$$ $$x=0.5$$ Hence, $$\boxed{0.5}$$

Example 23: Sequence until Success

A die is thrown until a '1' is obtained. Find the probability that '1' is obtained on an even-numbered throw.

Solution: Let success mean getting 1. Then, $$P(S)=\frac{1}{6}, \qquad P(F)=\frac{5}{6}$$ We need the probability that the first success occurs on the 2nd, 4th, 6th, … throw. So, $$P=FS+FFFS+FFFFFS+\cdots$$ Thus, $$P=\left(\frac{5}{6}\right)\left(\frac{1}{6}\right)+\left(\frac{5}{6}\right)^3\left(\frac{1}{6}\right)+\left(\frac{5}{6}\right)^5\left(\frac{1}{6}\right)+\cdots$$ This is an infinite geometric progression with $$a=\frac{5}{36}, \qquad r=\left(\frac{5}{6}\right)^2=\frac{25}{36}$$ Hence, $$P=\frac{a}{1-r}=\frac{\frac{5}{36}}{1-\frac{25}{36}}=\frac{\frac{5}{36}}{\frac{11}{36}}=\frac{5}{11}$$

Answer: $$\boxed{\frac{5}{11}}$$

Example 24: Conditional Odds

In a class, 40% study Math, 25% study Biology, and 15% study both. A student is selected. If he studies Biology, what is the probability that he also studies Math?

Solution: Let $$M=\{\text{student studies Math}\}, \qquad B=\{\text{student studies Biology}\}$$ Given, $$P(M)=0.4, \qquad P(B)=0.25, \qquad P(M\cap B)=0.15$$ We need $$P(M\mid B)$$ Using conditional probability, $$P(M\mid B)=\frac{P(M\cap B)}{P(B)}=\frac{0.15}{0.25}=\frac{3}{5}=0.6$$

Answer: $$\boxed{0.6}$$

Example 25: Expectation of Absolute Difference

Two dice are rolled. Let $X$ be the absolute difference of the numbers obtained. Find $E(X)$.

Solution: The possible values of $X$ are $$0,1,2,3,4,5$$ The number of outcomes corresponding to each value are: $$\begin{aligned} X=0 &\Rightarrow 6 \text{ outcomes} \\ X=1 &\Rightarrow 10 \text{ outcomes} \\ X=2 &\Rightarrow 8 \text{ outcomes} \\ X=3 &\Rightarrow 6 \text{ outcomes} \\ X=4 &\Rightarrow 4 \text{ outcomes} \\ X=5 &\Rightarrow 2 \text{ outcomes} \end{aligned}$$ Since the total number of outcomes is 36, $$E(X)=\frac{0\cdot 6+1\cdot 10+2\cdot 8+3\cdot 6+4\cdot 4+5\cdot 2}{36}$$ $$=\frac{0+10+16+18+16+10}{36}=\frac{70}{36}=\frac{35}{18}$$

Answer: $$\boxed{\frac{35}{18}}$$

Example 26: Balls in Urns (Total Probability)

Urn A contains 2 white and 1 black ball. Urn B contains 1 white and 2 black balls. One ball is transferred from A to B. Then one ball is drawn from B. Find the probability that it is white.

Solution: Let $$E_1=\{\text{white ball transferred from A to B}\}, \qquad E_2=\{\text{black ball transferred from A to B}\}$$ From urn A, $$P(E_1)=\frac{2}{3}, \qquad P(E_2)=\frac{1}{3}$$ Let $$W=\{\text{white ball drawn from urn B}\}$$ If a white ball is transferred, urn B becomes: $$2W, 2B$$ So, $$P(W\mid E_1)=\frac{2}{4}=\frac{1}{2}$$ If a black ball is transferred, urn B becomes: $$1W, 3B$$ So, $$P(W\mid E_2)=\frac{1}{4}$$ By total probability, $$P(W)=P(E_1)P(W\mid E_1)+P(E_2)P(W\mid E_2)$$ $$=\frac{2}{3}\cdot\frac{1}{2}+\frac{1}{3}\cdot\frac{1}{4}$$ $$=\frac{1}{3}+\frac{1}{12}=\frac{5}{12}$$

Answer: $$\boxed{\frac{5}{12}}$$

Example 27: Multiple Independent Events

If A, B, and C are independent events such that $P(A)=0.2$, $P(B)=0.3$, and $P(C)=0.4$, find $P(\text{exactly one occurs})$.

Solution: Exactly one of A, B, C occurs in the following mutually exclusive cases: $$AB'C', \qquad A'BC', \qquad A'B'C$$ Therefore, $$P(\text{exactly one occurs})=P(AB'C')+P(A'BC')+P(A'B'C)$$ Using independence, $$=0.2(1-0.3)(1-0.4)+(1-0.2)0.3(1-0.4)+(1-0.2)(1-0.3)0.4$$ $$=0.2(0.7)(0.6)+0.8(0.3)(0.6)+0.8(0.7)(0.4)$$ $$=0.084+0.144+0.224=0.452$$

Answer: $$\boxed{0.452}$$

Example 28: Variance Property

If $Var(X)=4$, find $Var(3X+5)$.

Solution: Using the variance property, $$Var(aX+b)=a^2Var(X)$$ Here, $$a=3, \qquad b=5$$ So, $$Var(3X+5)=3^2Var(X)=9\cdot 4=36$$

Answer: $$\boxed{36}$$

Example 29: Coin Bias

Two coins, one fair and one double-headed, are placed in a box. One coin is picked at random and tossed. It shows Head. Find the probability that the chosen coin was double-headed.

Solution: Let $$E_F=\{\text{fair coin chosen}\}, \qquad E_D=\{\text{double-headed coin chosen}\}$$ Since one coin is chosen at random, $$P(E_F)=P(E_D)=\frac{1}{2}$$ Let $$H=\{\text{Head is observed}\}$$ Then, $$P(H\mid E_F)=\frac{1}{2}, \qquad P(H\mid E_D)=1$$ By Bayes' theorem, $$P(E_D\mid H)=\frac{P(E_D)P(H\mid E_D)}{P(E_D)P(H\mid E_D)+P(E_F)P(H\mid E_F)}$$ $$=\frac{\frac{1}{2}\cdot 1}{\frac{1}{2}\cdot 1+\frac{1}{2}\cdot\frac{1}{2}}$$ $$=\frac{\frac{1}{2}}{\frac{1}{2}+\frac{1}{4}}=\frac{\frac{1}{2}}{\frac{3}{4}}=\frac{2}{3}$$

Answer: $$\boxed{\frac{2}{3}}$$

Example 30: System Reliability

A system works if component A or component B works. If $P(A)=0.9$ and $P(B)=0.8$, and the components work independently, find the probability that the system works.

Solution: The system fails only if both A and B fail. Thus, $$P(\text{system works})=1-P(A'\cap B')$$ Since the components are independent, $$P(A'\cap B')=P(A')P(B')$$ Now, $$P(A')=1-0.9=0.1, \qquad P(B')=1-0.8=0.2$$ So, $$P(\text{system works})=1-(0.1)(0.2)=1-0.02=0.98$$

Answer: $$\boxed{0.98}$$