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Q12(ii):
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 (ii) at least one is a girl?

Solution :

Step 1: Define the Sample Space

Let $B$ denote a boy and $G$ denote a girl. For a family with two children, the sample space $S$ is the set of all possible ordered pairs $(c_1, c_2)$, where $c_1$ is the first child and $c_2$ is the second child.

$S = \{ (B, B), (B, G), (G, B), (G, G) \}$

The total number of equally likely outcomes is $n(S) = 4$.

Step 2: Identify Event Sets

Let $E$ be the event that both children are girls, and $F$ be the event that at least one child is a girl.

  • Event $E$ (Both are girls): $E = \{ (G, G) \}$
  • Event $F$ (At least one is a girl): $F = \{ (B, G), (G, B), (G, G) \}$

[By the definition of conditional probability, we require the intersection of these sets.]

Step 3: Intersection and Probabilities

Find the intersection $E \cap F$:

$E \cap F = \{ (G, G) \} \cap \{ (B, G), (G, B), (G, G) \} = \{ (G, G) \}$

The number of favorable outcomes are:

  • $n(E \cap F) = 1$
  • $n(F) = 3$

Since each outcome is equally likely with probability $P(\omega) = \frac{1}{4}$, the probabilities are:

$P(E \cap F) = \frac{n(E \cap F)}{n(S)} = \frac{1}{4}$

$P(F) = \frac{n(F)}{n(S)} = \frac{3}{4}$

Step 4: Applying the Conditional Probability Formula

The conditional probability $P(E|F)$ is defined as:

$P(E|F) = \frac{P(E \cap F)}{P(F)}$ [Per the definition of Conditional Probability]

Substituting the calculated values:

$P(E|F) = \frac{1/4}{3/4}$

$P(E|F) = \frac{1}{3}$

Visual Representation of Sample Space

(B,B) (B,G) (G,B) (G,G) Sample Space S Red/Blue = Event F; Blue = Event E

Final Solution: The conditional probability that both children are girls given that at least one is a girl is $\frac{1}{3}$.


More Questions from Class 12 Mathematics Probability EXERCISE 13.1


CBSE Solutions for Class 12 Mathematics Probability


Chapters in CBSE - Class 12 Mathematics


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