Search any question & find its solution
Question:
Answered & Verified by Expert
In a family with 4 children, the probability that there are atleast two girls, is
Options:
Solution:
2802 Upvotes
Verified Answer
The correct answer is:
$\frac{11}{16}$
Total number of outcomes $=2^4=16$
Let $X$ be the number of girls.
Now, probability (atleast two girls)
$$
=1-\{P(X=0)+P(X=1)\}
$$
$P(X=0)$, i.e. no girl or all boys.
Favourable outcome $=\{B B B B\}=1$
$\begin{array}{ll}\therefore & P(X=0)=\frac{1}{16} \\ \text { and } & P(X=1), \text { i.e. one girl. }\end{array}$
Favourable outcomes
$$
\begin{aligned}
& =\{G B B B, B G B B, B B G B, B B B G\}=4 \\
& \therefore \quad P(X=1)=\frac{4}{16} \\
&
\end{aligned}
$$
Now, required probability
$$
\begin{aligned}
& =1-\{P(X=0)+P(X=1)\} \\
& =1-\left\{\frac{1}{16}+\frac{4}{16}\right\}=1-\frac{5}{16}=\frac{11}{16}
\end{aligned}
$$
Let $X$ be the number of girls.
Now, probability (atleast two girls)
$$
=1-\{P(X=0)+P(X=1)\}
$$
$P(X=0)$, i.e. no girl or all boys.
Favourable outcome $=\{B B B B\}=1$
$\begin{array}{ll}\therefore & P(X=0)=\frac{1}{16} \\ \text { and } & P(X=1), \text { i.e. one girl. }\end{array}$
Favourable outcomes
$$
\begin{aligned}
& =\{G B B B, B G B B, B B G B, B B B G\}=4 \\
& \therefore \quad P(X=1)=\frac{4}{16} \\
&
\end{aligned}
$$
Now, required probability
$$
\begin{aligned}
& =1-\{P(X=0)+P(X=1)\} \\
& =1-\left\{\frac{1}{16}+\frac{4}{16}\right\}=1-\frac{5}{16}=\frac{11}{16}
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.