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  • 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. 1/2 B. 1/3 C. 2/3 D. 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 chosen at random. The probability that the eldest child is a girl given that the family has at least one girl is
A. 1/2
B. 1/3
C. 2/3
D. 4/7

Answers (1)

The statement can be arranged in a set as S={(B,B,B),(G,G,G),(B,G,G),(G,B,G),(G,G,B),(G,B,B),(B,G,B),(B,B,G)}

Let A be Event that a family has at least one girl, therefore,
A={(G,B,B),(B,G,B),(B,B,G),(G,G,B),(B,G,G)(G,B,G),(G,G,G)
Let B be Event that eldest child is girl then, therefore,
B={(G,B,B)(G,G,B),(G,B,G),(G,G,G)

(A ∩ B)={(G,B,B),(G,G,B),(G,B,G,)(G,G,G)

since, \mathrm{P}(\mathrm{A} \mid \mathrm{B})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}$
P(A \mid B)=\frac{\frac{4}{8}}{\frac{7}{8}}=\frac{4}{7}$
Hence, \mathrm{P}\left(\mathrm{E}_{2} \mid \mathrm{E}_{1}\right)=\frac{4}{7}$

Option D is correct.

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infoexpert22

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