A family with three children is chose at random. The probability that the oldest and youngest children are…

A family with three children is chose at random. The probability that the oldest and youngest children are of the same gender is
  1. $\frac{3}{8}$
  2. $\frac{1}{2}$
  3. $\frac{1}{8}$
  4. $\frac{2}{8}$

Solution

$\mathrm{S}=\{\mathrm{BBB}, \mathrm{BBG}, \mathrm{BGB}, \mathrm{BGG}, \mathrm{GBB}, \mathrm{GBG}, \mathrm{GGB}, \mathrm{GGG}\}$ $\mathrm{E}=\{\mathrm{BBB}, \mathrm{BGB}, \mathrm{GBG}, \mathrm{GGG}\}$ $P(\mathrm{E})=\frac{n(\mathrm{E})}{n(\mathrm{~S})}=\frac{4}{8}=\frac{1}{2}$

Asked in: MHT CET 2022 (07 Aug Shift 2)

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