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Worked Examples · Example 3

Q.A family has two children. What is the probability that both the children are boys given that at least one of them is a boy?

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Using conditional probability on the 4-outcome family sample space, P(both boys∣at least one boy)=13P(\text{both boys}\mid\text{at least one boy})=\dfrac13.

Conditional probability of event AA given event BB has occurred:

P(A∣B)=P(A∩B)P(B),P(B)≠0P(A\mid B)=\frac{P(A\cap B)}{P(B)},\qquad P(B)\neq 0

  1. List the sample space. A family with two children; recording (elder child, younger child) with B=B=boy, G=G=girl, each child equally likely to be a boy or a girl:

S={BB, BG, GB, GG},P(BB)=P(BG)=P(GB)=P(GG)=14S=\{BB,\ BG,\ GB,\ GG\},\qquad P(BB)=P(BG)=P(GB)=P(GG)=\frac14

  1. Define the events.

A="both children are boys"={BB}A=\text{"both children are boys"}=\{BB\}

B="at least one boy"={BB,BG,GB}B=\text{"at least one boy"}=\{BB,BG,GB\}

  1. Compute P(B)P(B).

P(B)=P(BB)+P(BG)+P(GB)=14+14+14=34P(B)=P(BB)+P(BG)+P(GB)=\frac14+\frac14+\frac14=\frac34

  1. Compute P(A∩B)P(A\cap B). Since A⊂BA\subset B (if both are boys, at least one certainly is), A∩B=A={BB}A\cap B=A=\{BB\}: P(A∩B)=P(BB)=14P(A\cap B)=P(BB)=\frac14 …

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