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
$\therefore P$ (Having at least one boy) $=P(E)=\frac{\text { Number of outcomes favourable to } E}{\text { Number of all possible outcomes }}$
$=\frac{7}{8}$
Thus, the probability of having at least one boy is $\frac{7}{8}$.
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