Q.A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale.

Showing the 12 most recent of 45 on this concept.

1. COMEDK 2026Set 2026-A1 markMCQ
   Q.Vishnu has two jars of marbles, Jar A and Jar B. Jar A contains 3 yellow marbles and 2 green marbles. Jar B contains 4 yellow marbles and 3 green marbles. Vishnu flips a fair coin. If it lands heads, he picks two marbles at random without replacement from Jar A. If it lands tails, he picks two marbles at random with replacement from Jar B. Given that Vishnu picked one yellow and one green marble, what is the probability that they came from Jar B? (A) $\frac{21}{41}$ (B) $\frac{49}{89}$ (C) $\frac{40}{89}$ (D) $\frac{20}{41}$

   ›Reveal solutionSolution

   $P(E\mid A)=\tfrac35$ (without replacement), $P(E\mid B)=\tfrac{24}{49}$ (with replacement); Bayes gives $P(B\mid E)=\dfrac{40}{89}$ — option (C).

   Likelihoods of drawing one yellow and one green ($E$).

   Jar A (3 yellow, 2 green; two draws without replacement):

   $P(E\mid A)=\frac{\binom{3}{1}\binom{2}{1}}{\binom{5}{2}}=\frac{6}{10}=\frac{3}{5}.$

   Jar B (4 yellow, 3 green; two draws with replacement):

   $P(E\mid B)=2\cdot\frac{4}{7}\cdot\frac{3}{7}=\frac{24}{49}.$

   Bayes' theorem with $P(A)=P(B)=\tfrac12$ (the $\tfrac12$ cancels):

   $P(B\mid E)=\frac{P(E\mid B)}{P(E\mid A)+P(E\mid B)}=\frac{\frac{24}{49}}{\frac{3}{5}+\frac{24}{49}}.$

   Combine the denominator over $245$:

   $\frac{3}{5}+\frac{24}{49}=\frac{147+120}{245}=\frac{267}{245}.$

   $P(B\mid E)=\frac{24}{49}\cdot\frac{245}{267}=\frac{24\cdot 5}{267}=\frac{120}{267}=\frac{40}{89}.$

   ✓Final answer

   $P(B\mid E)=\dfrac{40}{89}$, which is option (C).

2. COMEDK 2026Set 2026-M1 markMCQ
   Q.Samhita faces a three-headed dragon. She wins a "Tactical medal" if she manages to defeat exactly one of the three heads. The battle proceeds head-by-head under the following conditions: The probability of defeating the first head is $\frac{\mathbf{1}}{\mathbf{3}}$. After a win: if she defeats a head, the probability of defeating the next head is $\frac{2}{3}$. After a loss: if she fails to defeat a head, the probability of defeating the next head is $\frac{\mathbf{1}}{\mathbf{4}}$. What is the probability that Samhita earns the "Tactical medal"? (A) $\frac{23}{72}$ (B) $\frac{5}{36}$ (C) $\frac{17}{72}$ (D) $\frac{19}{72}$

   ›Reveal solutionSolution

   Summing the three disjoint "exactly one win" paths gives $\dfrac{1}{12}+\dfrac{1}{18}+\dfrac{1}{8}=\dfrac{19}{72}$ — option (D).

   Set up the conditional probabilities. Let $W_i$ mean "defeats head $i$" and $L_i$ mean "fails":

   • $P(W_1) = \dfrac{1}{3}$, so $P(L_1) = \dfrac{2}{3}$.
   • After a win: next-head win probability $=\dfrac{2}{3}$, so next-head loss probability $=\dfrac{1}{3}$.
   • After a loss: next-head win probability $=\dfrac{1}{4}$, so next-head loss probability $=\dfrac{3}{4}$.

   Earning the medal means exactly one of the three heads is defeated. The three disjoint sequences are $W_1L_2L_3$, $L_1W_2L_3$, and $L_1L_2W_3$.

   Path 1 — $W_1 L_2 L_3$ (win, then loss after a win, then loss after a loss):

   $P = \frac{1}{3}\cdot\frac{1}{3}\cdot\frac{3}{4} = \frac{3}{36} = \frac{1}{12}$

   Path 2 — $L_1 W_2 L_3$ (loss, then win after a loss, then loss after a win):

   $P = \frac{2}{3}\cdot\frac{1}{4}\cdot\frac{1}{3} = \frac{2}{36} = \frac{1}{18}$

   Path 3 — $L_1 L_2 W_3$ (loss, then loss after a loss, then win after a loss):

   $P = \frac{2}{3}\cdot\frac{3}{4}\cdot\frac{1}{4} = \frac{6}{48} = \frac{1}{8}$

   Total (common denominator $72$):

   $\frac{1}{12}+\frac{1}{18}+\frac{1}{8} = \frac{6}{72}+\frac{4}{72}+\frac{9}{72} = \frac{19}{72}$

   ✓Final answer

   $P(\text{Tactical medal}) = \dfrac{19}{72}$ — option (D).

3. COMEDK 2026Set 2026-M1 markMCQ
   Q.A teacher has two jars of candy on her desk: Jar 1: Contains 3 Strawberry candies and 2 Orange candies. Jar 2: Contains 1 Strawberry candy and 4 Orange candies. The teacher randomly picks two candies from Jar 1 and drops them into Jar 2. Then, a student reaches into Jar 2 and picks two candies. What is the probability that the student picks two Strawberry candies? (A) $\frac{6}{35}$ (B) $\frac{4}{21}$ (C) $\frac{3}{70}$ (D) $\frac{1}{14}$

   ›Reveal solutionSolution

   Condition on how many strawberries move from Jar 1 to Jar 2, then compute the chance of drawing two strawberries from the now 7-candy Jar 2. Total $=\frac{1}{14}$.

   Setup. Jar 1 has $3$ Strawberry (S) and $2$ Orange (O). Two candies are moved into Jar 2, which started with $1$ S and $4$ O. After the transfer Jar 2 holds $7$ candies. Let $k$ = number of strawberries transferred.

   Transfer probabilities (choosing $2$ of $5$ from Jar 1, $\binom{5}{2}=10$):

   $P(k{=}2)=\frac{\binom{3}{2}}{10}=\frac{3}{10},\quad P(k{=}1)=\frac{\binom{3}{1}\binom{2}{1}}{10}=\frac{6}{10},\quad P(k{=}0)=\frac{\binom{2}{2}}{10}=\frac{1}{10}.$

   Draw two S from Jar 2 (which now has $1+k$ strawberries out of $7$, $\binom{7}{2}=21$):

   • $k=2$: Jar 2 has $3$ S $\Rightarrow \dfrac{\binom{3}{2}}{21}=\dfrac{3}{21}=\dfrac{1}{7}$.
   • $k=1$: Jar 2 has $2$ S $\Rightarrow \dfrac{\binom{2}{2}}{21}=\dfrac{1}{21}$.
   • $k=0$: Jar 2 has $1$ S $\Rightarrow \dfrac{\binom{1}{2}}{21}=0$.

   Total probability.

   $P=\frac{3}{10}\cdot\frac{1}{7}+\frac{6}{10}\cdot\frac{1}{21}+\frac{1}{10}\cdot 0=\frac{3}{70}+\frac{1}{35}=\frac{3}{70}+\frac{2}{70}=\frac{5}{70}=\frac{1}{14}.$

   ✓Final answer

   The probability is $\dfrac{1}{14}$ — option (D).

4. COMEDK 2026Set 2026-M1 markMCQ
   Q.Advika chooses one of three scarves every morning: Red, Blue, or Green. The probability she chooses Red is $20 \%$. The probability she chooses Blue is twice the probability of choosing Red. On the remaining days she wears a Green scarf. Once a scarf is chosen, she decides whether to wear a Hat (H) and Sunglasses (S). These choices are independent of each other but depend on the scarf colour: $\begin{array}{|l|l|l|} \hline \text { Scarf colour } & \mathbf{P ( H )} & \mathbf{P ( S )} \\ \hline \text { Red } & 0.5 & 0.8 \\ \hline \text { Blue } & 0.4 & 0.5 \\ \hline \text { Green } & 0.1 & 0.5 \\ \hline \end{array}$ Advika is spotted outdoors wearing both a Hat and Sunglasses. What is the probability that she is wearing the Red scarf? (A) $\frac{13}{313}$ (B) $\frac{8}{21}$ (C) $\frac{4}{9}$ (D) $\frac{8}{13}$

   ›Reveal solutionSolution

   Bayes' theorem on scarf colour given that both a hat and sunglasses are worn. Priors $P(R)=0.2,\ P(B)=0.4,\ P(G)=0.4$; likelihoods $P(H\cap S\mid\text{colour})=P(H)\,P(S)$. The posterior $P(R\mid H\cap S)=\frac{4}{9}$ — option (C).

   Concept. Hat and sunglasses are independent given the scarf, so $P(H\cap S\mid\text{colour})=P(H\mid\text{colour})\cdot P(S\mid\text{colour})$. Bayes' theorem then reverses the conditioning to give the probability of the scarf colour from the observed accessories.

   Step 1 — Priors.

   $P(R)=20\%=0.2,\qquad P(B)=2P(R)=0.4,\qquad P(G)=1-0.2-0.4=0.4.$

   Step 2 — Likelihood of wearing both accessories for each colour.

   $P(H\cap S\mid R)=0.5\times0.8=0.40,$

   $P(H\cap S\mid B)=0.4\times0.5=0.20,$

   $P(H\cap S\mid G)=0.1\times0.5=0.05.$

   Step 3 — Total probability of both accessories (denominator).

   $P(H\cap S)=(0.2)(0.40)+(0.4)(0.20)+(0.4)(0.05)=0.08+0.08+0.02=0.18.$

   Step 4 — Posterior for Red.

   $P(R\mid H\cap S)=\frac{P(R)\,P(H\cap S\mid R)}{P(H\cap S)}=\frac{0.08}{0.18}=\frac{8}{18}=\frac{4}{9}.$

   ✓Final answer

   $P(\text{Red}\mid H\cap S)=\dfrac{4}{9}$ — option (C).

   ANSWER: C

5. KCET 2026Set UNKNOWN1 markMCQ
   Q.Recent studies suggest that 12% of the world population is left handed. Depending on parents hand usage, the chances of having left handed children are as follows: A: Both parents are left handed, chances of having left handed children = 24% B: Both parents are right handed, chances of having left handed children = 9% C: Father left handed and mother right handed, chances of having left handed children = 17% D: Father right handed and mother left handed, chances of having left handed children = 22% Given $P(A) = P(B) = P(C) = P(D) = 1/4$ and L denotes child is left handed. What is the probability that $P(A|L)$? (A) $\frac{17}{80}$ (B) $\frac{24}{75}$ (C) $\frac{1}{3}$ (D) $\frac{1}{2}$

   ›Reveal solutionSolution

   Use the law of total probability to find $P(L)$, then apply Bayes' theorem to compute $P(A\mid L) = \dfrac{P(L\mid A)\,P(A)}{P(L)}$.

   Step 1 — List the given conditional probabilities

   With $P(A)=P(B)=P(C)=P(D)=\frac{1}{4}$, the conditional probabilities of a left-handed child are:

   $P(L\mid A) = 0.24,\quad P(L\mid B) = 0.09,\quad P(L\mid C) = 0.17,\quad P(L\mid D) = 0.22.$

   Step 2 — Find the total probability of a left-handed child, $P(L)$

   By the law of total probability:

   $P(L) = P(L\mid A)P(A) + P(L\mid B)P(B) + P(L\mid C)P(C) + P(L\mid D)P(D)$

   $P(L) = \frac{1}{4}\left(0.24+0.09+0.17+0.22\right) = \frac{1}{4}(0.72) = 0.18.$

   Step 3 — Apply Bayes' theorem for $P(A\mid L)$

   $P(A\mid L) = \frac{P(L\mid A)\,P(A)}{P(L)} = \frac{0.24 \times \frac{1}{4}}{0.18} = \frac{0.06}{0.18} = \frac{1}{3}.$

   ✓Final answer

   The correct option is (C) — $\frac{1}{3}$.

6. CA Foundation 2026Set jan-20261 markMCQ
   Q.If in a class, 50% of the student study mathematics and science and 70% of the student study mathematics, then the probability of a student studying science given that he/she is already studying mathematics is (A) $\frac{3}{7}$ (B) $\frac{6}{7}$ (C) $\frac{4}{7}$ (D) $\frac{5}{7}$

   ›Reveal solutionSolution

   Conditional probability $P(S\mid M) = \dfrac{P(M \cap S)}{P(M)}$.

   Step 1 — identify the probabilities

   50% study both maths and science, so $P(M \cap S) = 0.5$; 70% study maths, so $P(M) = 0.7$.

   Step 2 — apply the conditional-probability formula

   $P(S \mid M) = \frac{P(M \cap S)}{P(M)} = \frac{0.5}{0.7} = \frac{5}{7}.$

   Watch out

   Divide by the given event's probability: since maths is given, the denominator is $P(M)=0.7$, not $P(S)$ or the total. Dividing the other way ($0.7/0.5$) gives a value above 1, which is impossible for a probability.

   Tip

   "Given that" tells you the denominator. Here it is "given studying mathematics," so put $P(M)$ on the bottom: $0.5/0.7 = 5/7$.

   ✓Final answer

   (D) 5/7

7. CA Foundation 2026Set jan-20261 markMCQ
   Q.If two dice are rolled, then the probability of getting a greater number on the first die than the one on the second, given that the sum should be equal to 7 is (A) $\frac{1}{2}$ (B) $\frac{1}{3}$ (C) $\frac{1}{6}$ (D) $\frac{2}{3}$

   ›Reveal solutionSolution

   Conditional probability on a reduced sample space: $P(A\mid B) = \dfrac{n(A \cap B)}{n(B)}$.

   Step 1 — list the outcomes with sum 7.

   $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1) \Rightarrow n(B) = 6.$

   Step 2 — count first die greater than second, among those.

   $(4,3),(5,2),(6,1)$ → 3 outcomes.

   Step 3 — conditional probability.

   $P = \frac{3}{6} = \frac{1}{2}.$

   Watch out

   The condition "sum = 7" shrinks the sample space to those 6 outcomes — divide by 6, not by the full 36. Using $3/36$ gives $1/12$, which isn't even an option.

   Tip

   None of the sum-7 pairs are ties, so by symmetry "first > second" and "first < second" split the 6 outcomes evenly — the answer is simply half.

   ✓Final answer

   (A) 1/2

8. KCET 2025Set A-11 markMCQ
   Q.If A and B are two non-mutually exclusive events such that $P(A|B) = P(B|A)$, then (A) $A \subset B$ but $A \neq B$ (B) $A = B$ (C) $A \cap B = \phi$ (D) $P(A) = P(B)$

   ›Reveal solutionSolution

   Write both conditional probabilities with the same numerator $P(A \cap B)$, cancel it (legal because the events are not mutually exclusive), and the denominators must be equal.

   Step 1 — Definition of conditional probability.

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

   Note $A \cap B = B \cap A$, so the two fractions share the same numerator.

   Step 2 — Impose the given condition.

   $\frac{P(A\cap B)}{P(B)} = \frac{P(A\cap B)}{P(A)}.$

   Step 3 — Cancel — and see why we are allowed to.

   The events are non-mutually-exclusive, i.e. $A \cap B \ne \phi$ and $P(A \cap B) \ne 0$. (This hypothesis is exactly what makes the cancellation valid — if $P(A\cap B)$ were 0, both sides would be 0 for any $P(A), P(B)$ and nothing would follow.) Dividing both sides by $P(A\cap B)$:

   $\frac{1}{P(B)} = \frac{1}{P(A)} \;\Longrightarrow\; P(A) = P(B).$

   Step 4 — Why the other options are not forced.

   Equality of probabilities does not force equality of sets. Example: toss a fair coin twice; let $A$ = "first toss is head", $B$ = "second toss is head". Then $P(A) = P(B) = 1/2$, $P(A\cap B) = 1/4 \neq 0$, and indeed $P(A|B) = P(B|A) = 1/2$ — yet $A \ne B$ and neither is a subset of the other. So (A) and (B) are not implied. (C) contradicts the given non-mutual-exclusivity.

   ✓Final answer

   The correct option is (D) — $P(A) = P(B)$.

   ANSWER: D

9. KCET 2025Set A-11 markMCQ
   Q.If A and B are two events such that $A \subset B$ and $P(B) \neq 0$, then which of the following is correct? (A) $P(A|B) = \dfrac{P(B)}{P(A)}$ (B) $P(A|B) < P(A)$ (C) $P(A|B) \geq P(A)$ (D) $P(A) = P(B)$

   ›Reveal solutionSolution

   Use $A \subset B \Rightarrow A \cap B = A$, then note that dividing $P(A)$ by $P(B) \le 1$ cannot make it smaller.

   Step 1 — Simplify the intersection.

   If every element of $A$ lies in $B$, then $A \cap B = A$. Hence

   $P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{P(A)}{P(B)}.$

   Step 2 — Compare with $P(A)$.

   Every probability satisfies $0 < P(B) \le 1$ (we are told $P(B) \ne 0$). Therefore $\dfrac{1}{P(B)} \ge 1$, and multiplying the non-negative number $P(A)$ by a factor $\ge 1$ gives

   $P(A\mid B) = P(A)\cdot\frac{1}{P(B)} \;\ge\; P(A).$

   Equality holds exactly when $P(B) = 1$ (or when $P(A) = 0$).

   Intuition: conditioning on $B$ throws away all outcomes outside $B$ — but none of $A$ lies outside $B$. So $A$'s share of the shrunken sample space can only grow.

   Step 3 — Eliminate.

   • (A) $P(A|B) = P(B)/P(A)$ — wrong; the ratio is upside down (and can exceed 1).
   • (B) $P(A|B) < P(A)$ — the inequality points the wrong way.
   • (D) $P(A) = P(B)$ — not implied; $A$ can be a strictly smaller subset, e.g. rolling a die with $A = \{2\}$, $B = \{2,4,6\}$: $P(A) = 1/6$, $P(B) = 1/2$, and $P(A|B) = 1/3 \ge 1/6$. ✓ consistent only with (C).

   ✓Final answer

   The correct option is (C) — $P(A \mid B) \ge P(A)$.

   ANSWER: C

10. KCET 2025Set A-11 markMCQ
    Q.Meera visits only one of the two temples A and B in her locality. Probability that she visits temple A is $\dfrac{2}{5}$. If she visits temple A, $\dfrac{1}{3}$ is the probability that she meets her friend, whereas it is $\dfrac{2}{7}$ if she visits temple B. Meera met her friend at one of the two temples. The probability that she met her at temple B is (A) $\dfrac{7}{16}$ (B) $\dfrac{5}{16}$ (C) $\dfrac{3}{16}$ (D) $\dfrac{9}{16}$

    ›Reveal solutionSolution

    The friend has already been met (the effect); we want the probability of the cause (temple B) — that reversal of conditioning is exactly Bayes' theorem.

    Step 1 — Name the events.

    Let $A$ = "Meera visits temple A", $B$ = "Meera visits temple B", $F$ = "she meets her friend".

    She visits only one of the two temples, so $A$ and $B$ are mutually exclusive and exhaustive:

    $P(A)=\frac25\quad\Longrightarrow\quad P(B)=1-\frac25=\frac35 .$

    Given: $P(F\mid A)=\dfrac13$, $P(F\mid B)=\dfrac27$.

    Step 2 — Why Bayes.

    We are told the outcome ($F$ happened) and asked for the probability of a cause ($B$). Bayes' theorem inverts the conditioning:

    $P(B\mid F)=\frac{P(B)\,P(F\mid B)}{P(A)\,P(F\mid A)+P(B)\,P(F\mid B)} .$

    The denominator is $P(F)$ by the law of total probability — the friend can be met on either branch.

    Step 3 — Compute the two branch probabilities.

    $P(A\cap F)=\frac25\times\frac13=\frac{2}{15},\qquad P(B\cap F)=\frac35\times\frac27=\frac{6}{35} .$

    Step 4 — Total probability of meeting the friend.

    LCM of 15 and 35 is 105:

    $\frac{2}{15}=\frac{14}{105},\qquad \frac{6}{35}=\frac{18}{105},$

    $P(F)=\frac{14}{105}+\frac{18}{105}=\frac{32}{105}.$

    Step 5 — Apply Bayes.

    $P(B\mid F)=\frac{18/105}{32/105}=\frac{18}{32}=\frac{9}{16}.$

    (Check: $P(A\mid F)=\frac{14}{32}=\frac{7}{16}$, and $\frac{9}{16}+\frac{7}{16}=1$ ✓. Note option (A) $\frac{7}{16}$ is the trap — it is the probability for temple A.)

    ✓Final answer

    The correct option is (D) — $\dfrac{9}{16}$.

    ANSWER: D

11. COMEDK 2025Set 2025-A1 markMCQ
    Q.A bag contains $(n+1)$ coins. It is known that one of these coins has a head on both sides, whereas the other coins are fair. One of these coins is selected at random and tossed. If the probability that the toss results in heads is $\frac{7}{12}$, then the value of $n$ is : (A) 5 (B) 3 (C) 2 (D) 4

    ›Reveal solutionSolution

    The key idea is to treat the coin selection as a two‑case partition (two‑headed coin vs. fair coins) and apply the law of total probability. Solving the resulting equation gives $n = 5$, so the correct option is (A).

    ---

    Concept and intuition

    We have a mixed bag: one trick coin that always lands heads, and $n$ fair coins that land heads with probability $\frac12$. When we pick a coin at random and toss it, the overall chance of heads is a weighted average of the two cases. The weight for the trick coin is $\frac{1}{n+1}$ (since there are $n+1$ coins total), and for any fair coin it’s $\frac{n}{n+1}$. The problem gives that overall probability as $\frac{7}{12}$, so we set up an equation and solve for $n$.

    ---

    Step‑by‑step solution

    1. Define the events Let $T$ be the event that the two‑headed coin is selected, and $F$ the event that a fair coin is selected. Since selection is random,

    $P(T) = \frac{1}{n+1}, \qquad P(F) = \frac{n}{n+1}.$

    2. Conditional probabilities for heads If the trick coin is chosen, heads is certain:

    $P(\text{heads} \mid T) = 1.$

    If a fair coin is chosen, the chance of heads is $\frac12$:

    $P(\text{heads} \mid F) = \frac12.$

    3. Apply the law of total probability The overall probability of heads is

    $P(\text{heads}) = P(T) \cdot P(\text{heads}\mid T) + P(F) \cdot P(\text{heads}\mid F).$

    Substituting the values:

    $P(\text{heads}) = \frac{1}{n+1} \cdot 1 + \frac{n}{n+1} \cdot \frac12.$

    4. Simplify the expression

    $P(\text{heads}) = \frac{1}{n+1} + \frac{n}{2(n+1)} = \frac{2 + n}{2(n+1)}.$

    5. Set equal to the given probability The problem states this equals $\frac{7}{12}$:

    $\frac{2 + n}{2(n+1)} = \frac{7}{12}.$

    6. Solve for $n$ Cross‑multiply:

    $12(2 + n) = 7 \cdot 2(n+1) \quad\Rightarrow\quad 24 + 12n = 14n + 14.$

    Bring terms together:

    $24 - 14 = 14n - 12n \quad\Rightarrow\quad 10 = 2n.$

    Hence

    $n = 5.$

    Tip

    A quick sanity check: with $n=5$, there are 6 coins total. The trick coin contributes $\frac16$ to the overall heads probability, and the five fair coins contribute $\frac56 \cdot \frac12 = \frac{5}{12}$. Sum: $\frac{1}{6} + \frac{5}{12} = \frac{2}{12} + \frac{5}{12} = \frac{7}{12}$. Perfect.

    Watch out

    A common mistake is forgetting that the denominator is $n+1$, not $n$. Always count all coins, including the trick coin.

    ---

    ✓Final answer

    The correct option is (A).

    ANSWER: A

12. COMEDK 2025Set 2025-A1 markMCQ
    Q.If for two events $A$ and $B, P(A-B)=\frac{1}{5}$ and $P(A)=\frac{3}{5}$ then $P(B / A)=$ (A) $\frac{2}{3}$ (B) $\frac{1}{2}$ (C) $\frac{3}{5}$ (D) $\frac{2}{5}$

    ›Reveal solutionSolution

    The key is to interpret $P(A-B)$ as $P(A \cap B^c)$ and use the definition of conditional probability. The result is $P(B/A) = \frac{2}{3}$, so option (A) is correct.

    We are asked for $P(B/A)$, the probability of $B$ given $A$. The definition is

    $P(B/A) = \frac{P(A \cap B)}{P(A)}.$

    We know $P(A) = \frac{3}{5}$, so we need $P(A \cap B)$. The given $P(A-B) = \frac{1}{5}$ is the key: $A-B$ means “$A$ and not $B$,” i.e., $A \cap B^c$.

    1. Relate $P(A-B)$ to $P(A \cap B)$ Since $A$ is the union of the disjoint parts “$A$ and $B$” and “$A$ and not $B$,” we have

    $P(A) = P(A \cap B) + P(A \cap B^c).$

    Here $P(A \cap B^c) = P(A-B) = \frac{1}{5}$ and $P(A) = \frac{3}{5}$.

    2. Solve for $P(A \cap B)$

    $\frac{3}{5} = P(A \cap B) + \frac{1}{5} \quad\Rightarrow\quad P(A \cap B) = \frac{3}{5} - \frac{1}{5} = \frac{2}{5}.$

    3. Apply the conditional probability formula

    $P(B/A) = \frac{P(A \cap B)}{P(A)} = \frac{2/5}{3/5} = \frac{2}{3}.$

    Watch out

    A common mistake is to confuse $P(A-B)$ with $P(B^c)$ or to think $P(A-B) = P(A) - P(B)$. Remember: $A-B$ is only the part of $A$ that excludes $B$, not the whole complement of $B$.

    Tip

    Visualize a Venn diagram: $A$ is a circle split into the $B$ overlap and the rest. $P(A-B)$ is the “crescent” of $A$ outside $B$. Subtracting that from $P(A)$ gives the overlap directly.

    ✓Final answer

    The correct option is (A).

    ANSWER: A