In a family with 4 children, the probability | vedclass

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(D) Total number of outcomes $= 2^4 = 16$. Let $X$ be the number of girls. We need to find the probability of at least two girls, i.e.,$P(X \ge 2)$. T

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$\frac{11}{16}$

Vedclass · Answer

(D) Total number of outcomes $= 2^4 = 16$. Let $X$ be the number of girls. We need to find the probability of at least two girls, i.e.,$P(X \ge 2)$. This can be calculated as $P(X \ge 2) = 1 - \{P(X=0) + P(X=1)\}$. For $P(X=0)$ (no girls, all boys): The only outcome is $\{BBBB\}$, so $P(X=0) = \frac{1}{16}$. For $P(X=1)$ (exactly one girl): The outcomes are $\{GBBB, BGBB, BBGB, BBBG\}$, so $P(X=1) = \frac{4}{16}$. Therefore, $P(X \ge 2) = 1 - \left(\frac{1}{16} + \frac{4}{16}\right) = 1 - \frac{5}{16} = \frac{11}{16}$.

In a family with $4$ children, the probability that there are at least two girls is:

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