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Question
In a family of 3 children, the probability of having at least one boy is
Options
\[\frac{7}{8}\]
\[\frac{1}{8}\]
\[\frac{5}{8}\]
\[\frac{3}{4}\]
Solution
The possible outcomes are BBB, BBG, BGB, GBB, BGG, GBG, GGB, GGG.
∴ Total number of outcomes = 8
The favourable outcomes are BBB, BBG, BGB, GBB, BGG, GBG, GGB.
So, the favourable number of outcomes are 7.
∴ P(at least one boy) = \[\frac{\text{ Favourable number of outcomes }}{\text{ Total number of outcomes }} = \frac{7}{8}\]
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