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Question
In a family of 3 children, find the probability of having at least one boy.
Solution
All possible outcomes are BBB, BBG, BGB, GBB, BGG, GBG, GGB and GGG.
Number of all possible outcomes = 8
Let E be the event of having at least one boy.
Then the outcomes are BBB, BBG, BGB, GBB, BGG, GBG and GGB.
Number of possible outcomes = 7
∴ P(Having at least one boy) = P(E) =`"Number of favourable outcomes to E"/"Number of all possible outcome"=7/8`
Thus, the probability of having at least one boy is `7/8`
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