Q.An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the other without replacement. What is the probability that both drawn balls are black?

Showing the 12 most recent of 45 on this concept.

1. COMEDK 2024Set 2024-M1 markMCQ
   Q.An urn contains 2 white and 2 black balls. A ball is drawn at random. If it is white it is not replaced into the urn. Otherwise it is replaced along with another ball of the same colour. The process is repeated. The probability that the third ball drawn is black is (A) $\frac{17}{30}$ (B) $\frac{37}{60}$ (C) $\frac{31}{60}$ (D) $\frac{23}{30}$

   ›Reveal solutionSolution

   Conditioning on the first two draws (white -> not replaced; black -> replaced plus one extra black) and using total probability gives $P(\text{3rd black}) = \tfrac{23}{30}$ — option (D).

   Rules. Start with 2 white, 2 black (total 4). Drawing white removes it (whites $-1$, total $-1$). Drawing black puts it back and adds one more black (blacks $+1$, total $+1$).

   First two draws and the composition just before the 3rd draw

   • WW: $\tfrac24\cdot\tfrac13 = \tfrac16$; urn becomes W0, B2 (total 2) $\Rightarrow P(\text{black}) = 1$.
   • WB: $\tfrac24\cdot\tfrac23 = \tfrac13$; urn becomes W1, B3 (total 4) $\Rightarrow P(\text{black}) = \tfrac34$.
   • BW: $\tfrac24\cdot\tfrac25 = \tfrac15$; urn becomes W1, B3 (total 4) $\Rightarrow P(\text{black}) = \tfrac34$.
   • BB: $\tfrac24\cdot\tfrac35 = \tfrac{3}{10}$; urn becomes W2, B4 (total 6) $\Rightarrow P(\text{black}) = \tfrac23$.

   (The four path probabilities sum to $\tfrac16+\tfrac13+\tfrac15+\tfrac{3}{10} = 1$.)

   Total probability

   $P(\text{3rd black}) = \tfrac16(1) + \tfrac13\cdot\tfrac34 + \tfrac15\cdot\tfrac34 + \tfrac{3}{10}\cdot\tfrac23$

   $= \frac{10}{60} + \frac{15}{60} + \frac{9}{60} + \frac{12}{60} = \frac{46}{60} = \frac{23}{30}$

   ✓Final answer

   $P(\text{third ball is black}) = \dfrac{23}{30}$ — option (D).

   ANSWER: D

2. COMEDK 2021Set 2021-B1 markMCQ
   Q.An urn contains 5 red and 5 black coloured balls. A ball is picked at random, its colour noted and then is put back into the urn. Now, additional two balls of the same colour are put into the urn and then a ball is drawn at random. What is the probability that the ball now chosen is red. (A) 7/24 (B) 1/2 (C) 5/8 (D) 2/3

   ›Reveal solutionSolution

   By symmetry the second-draw red probability averages to $\tfrac12$.

   Start: 5 red, 5 black (10 total). First ball is drawn, replaced, then 2 more of the same colour are added, so the urn always has 12 balls before the second draw.

   • First ball red ($P=\tfrac{5}{10}=\tfrac12$): urn now 7R, 5B ⇒ $P(\text{red})=\tfrac{7}{12}$.
   • First ball black ($P=\tfrac12$): urn now 5R, 7B ⇒ $P(\text{red})=\tfrac{5}{12}$.

   Total probability:

   $\frac12\cdot\frac{7}{12}+\frac12\cdot\frac{5}{12}=\frac{7+5}{24}=\frac{12}{24}=\frac12.$

   ✓Final answer

   The correct option is (B) — 1/2

3. KCET 2021Set A-11 markMCQ
   Q.Given that A and B are two events such that $P(B) = \dfrac{3}{5}$, $P(A/B) = \dfrac{1}{2}$ and $P(A \cup B) = \dfrac{4}{5}$, then $P(A) =$ (A) $\dfrac{3}{10}$ (B) $\dfrac{1}{2}$ (C) $\dfrac{1}{5}$ (D) $\dfrac{3}{5}$

   ›Reveal solutionSolution

   Turn the conditional probability into $P(A\cap B)$, then substitute into the addition theorem and solve for $P(A)$.

   Step 1 — Extract $P(A\cap B)$ from the conditional probability.

   By definition,

   $P(A/B)=\frac{P(A\cap B)}{P(B)}\qquad (P(B)\neq0).$

   Rearranging (the multiplication theorem):

   $P(A\cap B)=P(A/B)\cdot P(B)=\frac{1}{2}\times\frac{3}{5}=\frac{3}{10}.$

   This is the key move — the conditional probability is not directly usable in the union formula, but $P(A\cap B)$ is.

   Step 2 — Apply the addition theorem.

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

   The intersection is subtracted because the elements common to $A$ and $B$ would otherwise be counted twice.

   Step 3 — Substitute the known values.

   $\frac{4}{5}=P(A)+\frac{3}{5}-\frac{3}{10}.$

   Step 4 — Solve for $P(A)$.

   $P(A)=\frac{4}{5}-\frac{3}{5}+\frac{3}{10}=\frac{1}{5}+\frac{3}{10}=\frac{2}{10}+\frac{3}{10}=\frac{5}{10}=\frac{1}{2}.$

   Consistency check. With $P(A)=\tfrac12$, $P(B)=\tfrac35$, $P(A\cap B)=\tfrac{3}{10}$: note $P(A)P(B)=\tfrac12\cdot\tfrac35=\tfrac{3}{10}=P(A\cap B)$, so $A$ and $B$ are independent — entirely consistent with $P(A/B)=\tfrac12=P(A)$. And $P(A\cup B)=\tfrac12+\tfrac35-\tfrac3{10}=\tfrac{5+6-3}{10}=\tfrac{8}{10}=\tfrac45$. ✓

   ✓Final answer

   The correct option is (B) — $\dfrac{1}{2}$.

   ANSWER: B

4. COMEDK 2025Set 2025-E1 markMCQ
   Q.A pot contains 5 red and 2 green balls. A ball is drawn at random from this pot. If a drawn ball is green, then a red ball is added to the pot. If a drawn ball is red, then a green ball is added to the pot, while the original ball drawn is not replaced in the pot. Now a second ball is drawn at random from the pot, what is the probability that the second ball drawn is a red ball? (A) $\frac{12}{49}$ (B) $\frac{32}{49}$ (C) $\frac{3}{7}$ (D) $\frac{27}{49}$

   ›Reveal solutionSolution

   Condition on the first draw: $P(\text{2nd red}) = \tfrac{2}{7}\cdot\tfrac{6}{7} + \tfrac{5}{7}\cdot\tfrac{4}{7} = \tfrac{32}{49}$.

   Start: $5$ red, $2$ green, total $7$. The first ball is not replaced, and a ball of the other colour is added.

   Case 1 — first ball is green. $P = \dfrac{2}{7}$.

   Remove that green ($5R,1G$), then add a red $\Rightarrow 6R,1G$ (total $7$).

   $P(\text{2nd red}\mid \text{green first}) = \frac{6}{7}$

   Case 2 — first ball is red. $P = \dfrac{5}{7}$.

   Remove that red ($4R,2G$), then add a green $\Rightarrow 4R,3G$ (total $7$).

   $P(\text{2nd red}\mid \text{red first}) = \frac{4}{7}$

   Total probability:

   $P(\text{2nd red}) = \frac{2}{7}\cdot\frac{6}{7} + \frac{5}{7}\cdot\frac{4}{7} = \frac{12}{49} + \frac{20}{49} = \frac{32}{49}$

   ✓Final answer

   $P(\text{second ball red}) = \dfrac{32}{49}$ — option (B).

5. COMEDK 2023Set 2023-E1 markMCQ
   Q.Bag A contains 3 white and 2 red balls. Bag B contains only 1 white ball. A fair coin is tossed. If head appears then 1 ball is drawn at random from bag A and put into bag B. However if tail appears then 2 balls are drawn at random from bag A and put into bag B. Now one ball is drawn at random from bag B. Given that the drawn ball from B is white, the probability that head appeared on the coin is (A) $\frac{23}{30}$ (B) $\frac{12}{23}$ (C) $\frac{11}{23}$ (D) $\frac{19}{30}$

   ›Reveal solutionSolution

   Computing $P(\text{white drawn}\mid H)=\frac45$ and $P(\text{white}\mid T)=\frac{11}{15}$, Bayes' theorem gives $P(H\mid\text{white})=\frac{12}{23}$.

   Head (transfer 1 ball from A$(3W,2R)$ to B$(1W)$, then draw from B's 2 balls):

   $P(\text{white})=\tfrac35(1)+\tfrac25\left(\tfrac12\right)=\tfrac35+\tfrac15=\tfrac45.$

   Tail (transfer 2 balls from A to B$(1W)$, then draw from B's 3 balls):

   $P(WW)=\tfrac{3}{10}\Rightarrow P(\text{white})=1,\quad P(WR)=\tfrac{6}{10}\Rightarrow \tfrac23,\quad P(RR)=\tfrac{1}{10}\Rightarrow \tfrac13.$

   $P(\text{white}\mid T)=\tfrac{3}{10}+\tfrac{6}{10}\cdot\tfrac23+\tfrac{1}{10}\cdot\tfrac13=\tfrac{3}{10}+\tfrac{4}{10}+\tfrac{1}{30}=\tfrac{22}{30}=\tfrac{11}{15}.$

   With $P(H)=P(T)=\tfrac12$:

   $P(H\mid\text{white})=\frac{\tfrac12\cdot\tfrac45}{\tfrac12\cdot\tfrac45+\tfrac12\cdot\tfrac{11}{15}}=\frac{\tfrac45}{\tfrac45+\tfrac{11}{15}}=\frac{\tfrac{12}{15}}{\tfrac{23}{15}}=\frac{12}{23}.$

   ✓Final answer

   The correct option is (B) — $\frac{12}{23}$

6. 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

7. 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).

8. COMEDK 2024Set 2024-E1 markMCQ
   Q.Suppose we have three cards identical in form except that both sides of the first card are coloured red, both sides of the second are coloured black, and one side of the third card is coloured red and the other side is coloured black. The three cards are mixed and a card is picked randomly. If the upper side of the chosen card is coloured red, what is the probability that the other side is coloured black. (A) $\frac{1}{6}$ (B) $\frac{1}{2}$ (C) 0 (D) $\frac{1}{3}$

   ›Reveal solutionSolution

   This is a classic conditional probability problem (often called the "three cards" or "Bertrand's box" variant). The key is that seeing a red side updates the probability space to only the red sides, and among those, only one belongs to the mixed card. The answer is 1/3.

   We are asked: given that the visible side is red, what is the probability that the other side is black? This is not simply "one of the two remaining cards has a black other side" because the cards are not equally likely once we condition on the observation.

   ---

   Concept and Intuition

   The pitfall is to think: "We see red, so the card is either the all-red or the mixed card. That's two possibilities, so the chance is 1/2." But this ignores that the all-red card has two red sides, while the mixed card has only one red side. When we pick a card at random and then look at a random side, each of the six sides is equally likely to be the one we see. Seeing red eliminates the three black sides, leaving only the three red sides. Among those three red sides, two belong to the all-red card and only one belongs to the mixed card. So the probability that the other side is black is 1/3.

   ---

   Step-by-step reasoning

   1. Label the sides.

      Let the cards be:

      • Card A: both sides red (sides $R_1, R_2$)
      • Card B: both sides black (sides $B_1, B_2$)
      • Card C: one red, one black (sides $R_3, B_3$)

   2. Count equally likely outcomes.

      When we pick a card uniformly at random and then look at a random side, there are $3 \times 2 = 6$ equally likely side-views. Each of the six sides has probability $1/6$ of being the one we see.

   3. Condition on seeing a red side.

      The red sides are: $R_1, R_2$ (from card A) and $R_3$ (from card C). That's 3 red sides. So the conditional space has 3 equally likely possibilities.

   4. Identify which of these have the other side black.

      Only the red side $R_3$ (from card C) has a black other side. The other two red sides ($R_1, R_2$) have red on the other side.

   5. Compute the conditional probability.

   $P(\text{other side black} \mid \text{top side red}) = \frac{\text{number of favorable red sides}}{\text{total number of red sides}} = \frac{1}{3}.$

   Watch out

   The common mistake is to forget that the all-red card contributes two red sides to the sample space, making it twice as likely as the mixed card once we condition on seeing red.

   Tip

   A neat way to think: "Probability = (number of red sides with black opposite) / (total number of red sides)." This avoids the card-counting trap entirely.

   ---

   ✓Final answer

   The correct option is (D).

   ANSWER: D

9. COMEDK 2023Set 2023-E1 markMCQ
   Q.

   [!FORMULA] $\text { If } P(B)=\frac{3}{5} \quad P(A / B)=\frac{1}{2} \text { and } P(A \cup B)=\frac{4}{5} \text { then } P(A \cup B)^{\prime}+P\left(A^{\prime} \cup B\right)=$

   (A) $\frac{4}{5}$ (B) $\frac{1}{2}$ (C) 1 (D) $\frac{1}{5}$

   ›Reveal solutionSolution

   Step 5: sum = 1/5 + 4/5 = 1.

   Concept: conditional probability, addition theorem, complements, De Morgan.

   Given P(B) = 3/5, P(A|B) = 1/2, P(A U B) = 4/5.

   Step 1: P(A ^ B) = P(A|B) P(B) = (1/2)(3/5) = 3/10.

   Step 2: from P(A U B) = P(A) + P(B) - P(A ^ B),

   P(A) = 4/5 - 3/5 + 3/10 = 1/5 + 3/10 = 1/2.

   Step 3: P((A U B)') = 1 - P(A U B) = 1 - 4/5 = 1/5.

   Step 4: P(A' U B) = 1 - P((A' U B)') = 1 - P(A ^ B') [De Morgan]

   P(A ^ B') = P(A) - P(A ^ B) = 1/2 - 3/10 = 1/5.

   So P(A' U B) = 1 - 1/5 = 4/5.

   Step 5: sum = 1/5 + 4/5 = 1.

   ✓Final answer

   The correct option is (C) — 1

   ANSWER: C

10. 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).

11. COMEDK 2025Set 2025-A1 markMCQ
    Q.Three bags contain a number of red and white balls are as follows. Bag I: 3 red balls Bag II: 2 red balls and 1 white ball Bag III: 3 White balls The probability that bag $i$ will be chosen and a ball is selected from it is $\frac{i}{6}, i=1,2,3$. If a white ball is selected, what is the probablity that it came from Bag III (A) $\frac{9}{11}$ (B) $\frac{2}{11}$ (C) $0$ (D) $\frac{1}{11}$

    ›Reveal solutionSolution

    This is a classic Bayes’ theorem problem: we are given prior probabilities for choosing each bag and the conditional probabilities of drawing a white ball from each bag; after observing a white ball, we want the posterior probability that it came from Bag III. The answer is $\frac{9}{11}$.

    We start by noting that the problem gives us three bags with different compositions, and a rule for choosing a bag: the probability of picking bag $i$ is $\frac{i}{6}$. Then, from that bag, we pick a ball uniformly at random. We are told that a white ball was selected, and we need the probability that it came from Bag III. This is a textbook case for Bayes’ theorem, which lets us “reverse” conditional probabilities: we know $P(\text{white} \mid \text{bag } i)$, and we want $P(\text{bag } i \mid \text{white})$.

    Let’s work through it step by step.

    1. Define events and write down given probabilities. Let $B_i$ be the event that bag $i$ is chosen ($i=1,2,3$). We are told:

    $P(B_1) = \frac{1}{6},\quad P(B_2) = \frac{2}{6},\quad P(B_3) = \frac{3}{6}.$

    Let $W$ be the event that a white ball is drawn.

    From the bag compositions:

    • Bag I: 3 red, 0 white → $P(W \mid B_1) = 0$.
    • Bag II: 2 red, 1 white → $P(W \mid B_2) = \frac{1}{3}$.
    • Bag III: 0 red, 3 white → $P(W \mid B_3) = 1$.

    2. Compute the total probability of drawing a white ball. By the law of total probability:

    $P(W) = P(B_1)P(W\mid B_1) + P(B_2)P(W\mid B_2) + P(B_3)P(W\mid B_3).$

    Substitute:

    $P(W) = \frac{1}{6}\cdot 0 + \frac{2}{6}\cdot\frac{1}{3} + \frac{3}{6}\cdot 1 = 0 + \frac{2}{18} + \frac{3}{6} = \frac{1}{9} + \frac{1}{2}.$

    Get a common denominator (18):

    $P(W) = \frac{2}{18} + \frac{9}{18} = \frac{11}{18}.$

    3. Apply Bayes’ theorem to find $P(B_3 \mid W)$. Bayes’ theorem says:

    $P(B_3 \mid W) = \frac{P(B_3)\,P(W\mid B_3)}{P(W)}.$

    Plug in the numbers:

    $P(B_3 \mid W) = \frac{\frac{3}{6} \cdot 1}{\frac{11}{18}} = \frac{\frac{1}{2}}{\frac{11}{18}} = \frac{1}{2} \cdot \frac{18}{11} = \frac{9}{11}.$

    Tip

    Notice that Bag I contributes zero probability of a white ball, so it can be ignored entirely in the numerator — it only affects the denominator by adding zero. This often simplifies Bayes’ calculations: only bags that can produce the observed outcome matter.

    Watch out

    A common mistake is to forget that the prior probabilities $\frac{i}{6}$ are not equal — they favor higher-numbered bags. If you mistakenly treated all bags as equally likely, you would get $\frac{1}{2}$ instead of $\frac{9}{11}$.

    Thus, given that a white ball was drawn, the probability it came from Bag III is $\frac{9}{11}$.

    ✓Final answer

    The correct option is (A).

    ANSWER: A

12. COMEDK 2025Set 2025-E1 markMCQ
    Q.Two numbers are selected at random from integers 1 to 9 . If their sum is even, what is the probability that both the numbers are odd? (A) $\frac{4}{9}$ (B) $\frac{5}{8}$ (C) $\frac{1}{6}$ (D) $\frac{2}{3}$

    ›Reveal solutionSolution

    This is a conditional probability problem: given that the sum of two numbers from 1–9 is even, we want the probability both are odd. The answer is 5/8, option (B).

    We are selecting two numbers from 1 to 9 without replacement (since "selected at random" from distinct integers usually implies no repetition). The sum is even only if both numbers are odd or both are even. So the condition restricts us to those pairs. The question asks: among those pairs with an even sum, what fraction consists of two odd numbers?

    ---

    1. Count total possible pairs (without replacement)

    From 1 to 9, there are 9 numbers. The number of ways to choose any two distinct numbers is

    $\binom{9}{2} = 36.$

    2. Count pairs with an even sum

    A sum is even when both numbers have the same parity.

    • Odd numbers from 1 to 9: 1, 3, 5, 7, 9 → 5 odds.

      Number of odd–odd pairs: $\binom{5}{2} = 10$.

    • Even numbers from 1 to 9: 2, 4, 6, 8 → 4 evens.

      Number of even–even pairs: $\binom{4}{2} = 6$.

    So total pairs with an even sum:

    $10 + 6 = 16.$

    3. Apply conditional probability

    We want

    $P(\text{both odd} \mid \text{sum even}) = \frac{\text{Number of odd–odd pairs}}{\text{Number of even-sum pairs}} = \frac{10}{16} = \frac{5}{8}.$

    Tip

    A common mistake is to treat this as an unconditional probability and compute $\frac{10}{36}$, which gives $\frac{5}{18}$ — not even among the options. The condition "given sum is even" changes the denominator from 36 to 16.

    Watch out

    Another pitfall: forgetting that selection is without replacement. If replacement were allowed, the counts would differ, but here the problem implies distinct integers.

    ---

    ✓Final answer

    The correct option is (B).

    ANSWER: B