Q.Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that

Showing the 12 most recent of 45 on this concept.

1. COMEDK 2023Set 2023-E1 markMCQ
   Q.A die is thrown twice and the sum of numbers appearing is observed to be 8 . What is the conditional probability that the number 5 has appeared atleast once? (A) $\frac{5}{36}$ (B) $\frac{2}{5}$ (C) $\frac{1}{18}$ (D) $\frac{1}{3}$

   ›Reveal solutionSolution

   Given the sum is $8$, the sample space is the $5$ ordered pairs summing to $8$; two of them include a $5$, giving probability $\frac{2}{5}$.

   The ordered outcomes with sum $8$ are

   $(2,6),(3,5),(4,4),(5,3),(6,2)\quad(5\text{ outcomes}).$ …

   🔒 Start your 14-day free trial to see the full solution →
2. COMEDK 2024Set 2024-A1 markMCQ
   Q.P and Q are considering to apply for a job. The probability that P applies for the job is $\frac{1}{4}$. The probability that $\mathrm{P}$ applies for the job given that $\mathrm{Q}$ applies for the job is $\frac{1}{2}$, and the probability that Q applies for the job given that P applies for the job is $\frac{1}{3}$. Then the probability that $\mathrm{P}$ does not apply for the job given that $\mathrm{Q}$ does not apply for the job is (A) $\frac{4}{5}$ (B) $\frac{7}{8}$ (C) $\frac{5}{6}$ (D) $\frac{11}{12}$

   ›Reveal solutionSolution

   We are given conditional probabilities and need to find $P(\overline{P} \mid \overline{Q})$. Using the definitions of conditional probability and the law of total probability, we compute $P(Q)$ and $P(\overline{Q})$, then apply Bayes' theorem to get $\frac{4}{5}$, which corresponds to option (A).

   Concept and intuition:

   This is a classic problem of working backwards from conditional probabilities to find a joint probability table. We know $P(P)$, $P(P \mid Q)$, and $P(Q \mid P)$. From these, we can find $P(P \cap Q)$ in two ways, which lets us solve for $P(Q)$. Then we can compute the desired conditional probability $P(\overline{P} \mid \overline{Q})$ using the complement rule and the definition of conditional probability.

   Step-by-step solution:

   1. Write down what is given. Let $P$ = event that P applies, $Q$ = event that Q applies. We have:

   $P(P) = \frac{1}{4}, \quad P(P \mid Q) = \frac{1}{2}, \quad P(Q \mid P) = \frac{1}{3}.$

   2. Use the definition of conditional probability to express $P(P \cap Q)$ in two ways. From $P(P \mid Q) = \frac{P(P \cap Q)}{P(Q)}$, we get

   $P(P \cap Q) = P(P \mid Q) \cdot P(Q) = \frac{1}{2} P(Q).$

   From $P(Q \mid P) = \frac{P(P \cap Q)}{P(P)}$, we get

   $P(P \cap Q) = P(Q \mid P) \cdot P(P) = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12}.$

   3. Equate the two expressions for $P(P \cap Q)$ to find $P(Q)$.

   $\frac{1}{2} P(Q) = \frac{1}{12} \quad \Rightarrow \quad P(Q) = \frac{1}{6}.$

   4. Find $P(\overline{Q})$ and $P(P \cap \overline{Q})$.

   $P(\overline{Q}) = 1 - P(Q) = 1 - \frac{1}{6} = \frac{5}{6}.$

   Also, $P(P \cap \overline{Q}) = P(P) - P(P \cap Q) = \frac{1}{4} - \frac{1}{12} = \frac{3}{12} - \frac{1}{12} = \frac{2}{12} = \frac{1}{6}.$

   5. Find $P(\overline{P} \cap \overline{Q})$. Since $\overline{P} \cap \overline{Q}$ is the complement of $P \cup Q$, we can use:

   $P(\overline{P} \cap \overline{Q}) = 1 - P(P \cup Q).$

   First, $P(P \cup Q) = P(P) + P(Q) - P(P \cap Q) = \frac{1}{4} + \frac{1}{6} - \frac{1}{12}.$ …

   🔒 Start your 14-day free trial to see the full solution →
3. 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): …

   🔒 Start your 14-day free trial to see the full solution →
4. COMEDK 2024Set 2024-E1 markMCQ
   Q.A coin is tossed until a head appears or until the coin has been tossed three times. Given that 'head' does not appear on the first toss, what is the probability that the coin is tossed thrice? (A) $\frac{1}{2}$ (B) $\frac{3}{8}$ (C) $\frac{1}{8}$ (D) $\frac{1}{4}$

   ›Reveal solutionSolution

   Given the first toss is a tail, the coin is tossed a third time only if the second toss is also a tail — probability $\tfrac12$.

   The coin stops the moment a head appears, or after three tosses. We are told the first toss is a tail (no head on toss 1).

   Now the second toss decides:

   • Head on toss 2 $\Rightarrow$ stop after 2 tosses.
   • Tail on toss 2 $\Rightarrow$ a third toss is made. …

   🔒 Start your 14-day free trial to see the full solution →
5. 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 $7$). …

   🔒 Start your 14-day free trial to see the full solution →
6. COMEDK 2024Set 2024-M1 markMCQ
   Q.A and B each have a calculator which can generate a single digit random number from the set $\{1,2,3,4,5,6,7,8\}$. They can generate a random number on their calculator. Given that the sum of the two numbers is 12 , then the probability that the two numbers are equal is (A) $\frac{5}{64}$ (B) $\frac{1}{5}$ (C) $\frac{1}{16}$ (D) $\frac{1}{8}$

   ›Reveal solutionSolution

   We are asked for the conditional probability that two numbers are equal given their sum is 12.

   The only equal pair summing to 12 is (6,6), and there are 5 total pairs summing to 12.

   So the probability is $\frac{1}{5}$, which corresponds to option (B).

   ---

   Concept and intuition

   This is a classic conditional probability problem: we are not interested in all possible outcomes, only those where the sum is exactly 12. The phrase “given that the sum is 12” means we restrict our universe to those pairs. Then we count how many of those restricted outcomes have the two numbers equal. The trap is to forget to restrict the denominator — many students mistakenly use the total number of all possible pairs (64) instead of only the favorable-sum pairs.

   ---

   Step-by-step solution

   1. Identify the sample space

      Each of A and B picks a digit from $\{1,2,\dots,8\}$.

      Total possible ordered pairs $(a,b)$: $8 \times 8 = 64$.

   2. List all pairs with sum 12

      We need $a + b = 12$, with $1 \le a,b \le 8$.

      Possible values for $a$:

      • If $a = 4$, then $b = 8$
      • If $a = 5$, then $b = 7$
      • If $a = 6$, then $b = 6$
      • If $a = 7$, then $b = 5$
      • If $a = 8$, then $b = 4$

      So the pairs are:

   $(4,8),\ (5,7),\ (6,6),\ (7,5),\ (8,4)$

   That’s 5 ordered pairs.

   3. Count the favorable outcomes “The two numbers are equal” means $a = b$. …

   🔒 Start your 14-day free trial to see the full solution →
7. 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. …

   🔒 Start your 14-day free trial to see the full solution →
8. 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): …

   🔒 Start your 14-day free trial to see the full solution →
9. KCET 2020Set A-11 markMCQ
   Q.If A and B are two events such that $P(A) = \frac{1}{3}$, $P(B) = \frac{1}{2}$ and $P(A \cap B) = \frac{1}{6}$, then $P(A'/B)$ is (A) $\frac{2}{3}$ (B) $\frac{1}{3}$ (C) $\frac{1}{2}$ (D) $\frac{1}{12}$

   ›Reveal solutionSolution

   Use $P(A'|B) = 1 - P(A|B)$ (equivalently $\frac{P(B)-P(A\cap B)}{P(B)}$), which gives $\frac{2}{3}$.

   Step 1 — The definition of conditional probability.

   For $P(B) > 0$,

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

   Conditioning on $B$ means we restrict the sample space to $B$ and ask what fraction of $B$ lies outside $A$.

   Step 2 — Split $B$ into the part inside $A$ and the part outside $A$.

   The events $A \cap B$ and $A' \cap B$ are disjoint and their union is exactly $B$:

   $P(B) = P(A \cap B) + P(A' \cap B)$

   $\Rightarrow P(A' \cap B) = P(B) - P(A \cap B) = \frac{1}{2} - \frac{1}{6} = \frac{3-1}{6} = \frac{2}{6} = \frac{1}{3}$

   Step 3 — Divide by $P(B)$.

   $P(A'|B) = \frac{P(A' \cap B)}{P(B)} = \frac{\tfrac{1}{3}}{\tfrac{1}{2}} = \frac{1}{3} \times \frac{2}{1} = \frac{2}{3}$

   Step 4 — Cross-check with the complement rule. …

   🔒 Start your 14-day free trial to see the full solution →
10. 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

    $$$$

    🔒 Start your 14-day free trial to see the full solution →
11. KCET 2021Set A-11 markMCQ
    Q.Two dice are thrown. If it is known that the sum of numbers on the dice was less than 6 the probability of getting a sum as 3 is (A) $\dfrac{1}{18}$ (B) $\dfrac{5}{18}$ (C) $\dfrac{1}{5}$ (D) $\dfrac{2}{5}$

    ›Reveal solutionSolution

    We use conditional probability: reduce the sample space to only those outcomes where the sum is less than 6, then find the fraction of those that give a sum of exactly 3. The answer is $\frac{1}{5}$.

    The key idea here is conditional probability — the probability of one event given that another event has already occurred. When we say "if it is known that the sum was less than 6", we are no longer considering all 36 possible outcomes of throwing two dice. Instead, we restrict ourselves to only those outcomes that satisfy the condition, and then ask: within this smaller set, what fraction gives a sum of 3?

    Let’s work it out step by step.

    1. List all possible outcomes when two dice are thrown.

       Each die shows a number from 1 to 6. The total number of ordered pairs $(a,b)$ is $6 \times 6 = 36$.

    2. Identify the condition: sum less than 6.

       The possible sums are 2, 3, 4, and 5 (since sum = 1 is impossible with two dice). Let’s list all ordered pairs that give each sum:

       • Sum = 2: $(1,1)$ → 1 outcome
       • Sum = 3: $(1,2), (2,1)$ → 2 outcomes
       • Sum = 4: $(1,3), (2,2), (3,1)$ → 3 outcomes
       • Sum = 5: $(1,4), (2,3), (3,2), (4,1)$ → 4 outcomes

       Total outcomes with sum < 6: $1 + 2 + 3 + 4 = 10$.

       Watch out

       A common mistake is to forget that $(1,2)$ and $(2,1)$ are different outcomes. Dice are distinct, so order matters. Always count ordered pairs unless the problem explicitly says "identical dice" (which is rare in such problems).

    3. Identify the event of interest: sum = 3. …

    🔒 Start your 14-day free trial to see the full solution →
12. KCET 2020Set A-11 markMCQ
    Q.The probability of solving a problem by three persons A, B and C independently is $\frac{1}{2}$, $\frac{1}{4}$ and $\frac{1}{3}$ respectively. Then the probability of the problem is solved by any two of them is (A) $\frac{1}{12}$ (B) $\frac{1}{4}$ (C) $\frac{1}{24}$ (D) $\frac{1}{8}$

    ›Reveal solutionSolution

    The problem asks for the probability that exactly two of the three persons solve it. We compute the sum of probabilities for each pair solving while the third fails, giving $\frac{1}{8}$.

    The key idea: "solved by any two of them" means exactly two solve it, not at least two. The third person must fail. Since A, B, and C work independently, we multiply their individual probabilities for each specific outcome and then add the three possible cases.

    A common mistake is to include the case where all three solve it. That would be "at least two", not "any two". The phrase "any two" in probability problems almost always means exactly two.

    Let’s denote:

    • $P(A) = \frac{1}{2}$, so $P(\text{A fails}) = 1 - \frac{1}{2} = \frac{1}{2}$
    • $P(B) = \frac{1}{4}$, so $P(\text{B fails}) = 1 - \frac{1}{4} = \frac{3}{4}$
    • $P(C) = \frac{1}{3}$, so $P(\text{C fails}) = 1 - \frac{1}{3} = \frac{2}{3}$

    We want exactly two successes. There are three mutually exclusive ways this happens:

    1. A and B solve, C fails

       Probability = $P(A) \times P(B) \times P(\text{C fails})$

       $= \frac{1}{2} \times \frac{1}{4} \times \frac{2}{3} = \frac{2}{24} = \frac{1}{12}$

    2. A and C solve, B fails

       Probability = $P(A) \times P(C) \times P(\text{B fails})$

       $= \frac{1}{2} \times \frac{1}{3} \times \frac{3}{4} = \frac{3}{24} = \frac{1}{8}$

    3. B and C solve, A fails

       Probability = $P(B) \times P(C) \times P(\text{A fails})$ …

    🔒 Start your 14-day free trial to see the full solution →