Assume that in a family, each child is equally likely to be a boy or a girl. A family with three children is chosen at random. The probability that the eldest child is a girl given that the family has at least one girl is (A) $\dfrac{1}{2}$ (B) $\dfrac{1}{3}$ (C) $\dfrac{2}{3}$ (D) $\dfrac{4}{7}$

Assume that in a family, each child is equally likely to be a boy or a girl. A family with three children is c

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\dfrac{1}{2}

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We use conditional probability: reduce the sample space to families with at least one girl, then count how many of those have the eldest child a girl. The answer is $\frac{4}{7}$ , option (D).

Why conditional probability?

The question asks: Given that the family has at least one girl, what is the probability the eldest is a girl? This is a classic "restricted sample space" problem. Instead of all 8 equally likely outcomes for three children, we only consider those outcomes that satisfy the condition (at least one girl). Within that smaller set, we count how many also have the eldest child a girl.

The formula is:

P(\text{eldest is girl} \mid \text{at least one girl}) = \frac{P(\text{eldest is girl AND at least one girl})}{P(\text{at least one girl})}

But since "eldest is girl" already implies at least one girl, the numerator is just $P(\text{eldest is girl})$ .

Step-by-step

1. List all possible outcomes for three children (B = boy, G = girl).

Each child is independent with probability $\frac12$ for each gender. The 8 equally likely outcomes are:

#	Outcome
1	BBB
2	BBG
3	BGB
4	BGG
5	GBB
6	GBG
7	GGB
8	GGG

2. Identify the condition: at least one girl.

Outcomes with no girls: only BBB. So the condition "at least one girl" includes outcomes 2 through 8 — that's 7 outcomes.

3. Identify the event: eldest child is a girl.

The eldest is the first child listed. Outcomes where the first child is G: GBB, GBG, GGB, GGG — that's 4 outcomes.

4. Apply conditional probability.

Since all outcomes are equally likely:

$P(\text{eldest is girl} \mid \text{at least one girl}) = \frac{\text{Number of outcomes with eldest girl AND at least one girl}}{\text{Number of outcomes with at least one girl}}$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CA Foundation 2026Set jan-20261 markMCQ
$\frac{3}{7}$
›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

$P(S \mid M) = \frac{P(M \cap S)}{P(M)} = \frac{0.5}{0.7} = \frac{5}{7}.$ …
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CA Foundation 2026Set jan-20261 markMCQ
$\frac{1}{2}$
›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.

$P = \frac{3}{6} = \frac{1}{2}.$ …
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CA Foundation 2023Set jun-20231 markMCQ
$P(A)=\dfrac{1}{3}, P(B)=\dfrac{1}{4}, P(A/B)=\dfrac{1}{6}$
›Reveal solutionSolution

P(B/A) = P(A∩B)/P(A) = (1/24)/(1/3) = 1/8.

Step 1 — Find the joint probability
$P(A\cap B) = P(A/B)\,P(B) = \frac{1}{6}\times\frac{1}{4} = \frac{1}{24}$
Step 2 — Apply the definition of conditional probability
$P(B/A) = \frac{P(A\cap B)}{P(A)} = \frac{1/24}{1/3} = \frac{3}{24} = \frac{1}{8}$
Watch out

P(A/B) and P(B/A) are not equal — you must recompute the joint probability first, then divide by P(A), not P(B). …

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CA Foundation 2022Set dec-20221 markMCQ
$P(A)=\dfrac{1}{3}$
›Reveal solutionSolution

P(A∩B)=1/6, so P(B|A)=(1/6)/(1/3)=1/2.

Step 1 — Intersection via the addition rule
$P(A\cap B)=P(A)+P(B)-P(A\cup B)=\frac13+\frac34-\frac{11}{12}=\frac{4+9-11}{12}=\frac{2}{12}=\frac16$
Step 2 — Apply the conditional-probability formula
$P\!\left(\frac{B}{A}\right)=\frac{P(A\cap B)}{P(A)}=\frac{1/6}{1/3}=\frac12$
Watch out

Option (A) 1/6 is just P(A∩B) — you must still divide by P(A) to get the conditional probability. …

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CA Foundation 2021Set dec-20211 markMCQ
$A$
›Reveal solutionSolution

Divide the joint probability by the conditioning event's probability: $P(A\mid B)=5/9$ , $P(B\mid A)=6/11$ .

Step 1 — Apply the conditional probability formula for $P(A\mid B)$
$P(A\mid B) = \frac{P(A\cap B)}{P(B)} = \frac{10/33}{6/11} = \frac{10}{33}\times\frac{11}{6} = \frac{110}{198} = \frac{5}{9}$
Step 2 — Apply it for $P(B\mid A)$
$P(B\mid A) = \frac{P(A\cap B)}{P(A)} = \frac{10/33}{5/9} = \frac{10}{33}\times\frac{9}{5} = \frac{90}{165} = \frac{6}{11}$
Step 3 — Sanity check

Since $P(A)P(B) = (5/9)(6/11) = 10/33 = P(A\cap B)$ , the conditionals collapse to the marginals — consistent with the computed values. …
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CA Foundation 2021Set dec-20211 markMCQ
$Q. In a group of 20 males and 15 females, 12 males and 8 females are service holders. What is the probability that a person selected at random from the group is a service holder given that the selected person is a male? (A) 0.40 (B) 0.60 (C) 0.45 (D) 0.55$
›Reveal solutionSolution

Condition on males only: 12 service holders out of 20 males = 0.60.

Step 1 — Identify the reduced sample space

Given the person is male, only the 20 males matter.

Step 2 — Apply the conditional formula
$P(\text{service}\mid \text{male}) = \frac{\text{male service holders}}{\text{total males}} = \frac{12}{20} = 0.60$
Watch out

Do not divide by the full group of 35 — the condition 'given male' shrinks the denominator to 20. …

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Q.Assume that in a family, each child is equally likely to be a boy or a girl. A family with three children is chosen at random. The probability that the eldest child is a girl given that the family has at least one girl is
(A) $\dfrac{1}{2}$
(B) $\dfrac{1}{3}$
(C) $\dfrac{2}{3}$
(D) $\dfrac{4}{7}$

Concept understanding — Conditional Probability

Conditional Probability

The Idea: Shrink the Sample Space

The Definition

Two Cautions

Why conditional probability?

Step-by-step

Method: Conditional probability on a finite equally-likely sample space

Steps

Common Mistakes

Step 1 — identify the probabilities

Step 2 — apply the conditional-probability formula

Step 1 — list the outcomes with sum 7.

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

Step 3 — conditional probability.

Step 1 — Find the joint probability

Step 2 — Apply the definition of conditional probability

Step 1 — Intersection via the addition rule

Step 2 — Apply the conditional-probability formula

Step 1 — Apply the conditional probability formula for $P(A\mid B)$

Step 2 — Apply it for $P(B\mid A)$

Step 3 — Sanity check

Step 1 — Identify the reduced sample space

Step 2 — Apply the conditional formula

Why conditional probability?

Step-by-step

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