Probability 37651

What is the probability that in a family with four children, there are:
a) at least three girls
b) at least one boy,

If the probability of a boy is 0.51?

Correct answer:

p₁ =  0.2977
p₂ =  0.9424

Step-by-step explanation:

$$\begin{gathered} p=0.51 \\ {C}_3(4)=\left(\genfrac{}{}{0ex}{}{4}{3}\right)=\frac{4!}{3!(4-3)!}=\frac{4}{1}=4 \\ {d}_3=\left(\genfrac{}{}{0ex}{}{4}{3}\right)\cdot (1-p{)}^3\cdot p^1=4\cdot (1-0.51{)}^3\cdot 0.51^1\doteq 0.24 \\ {C}_4(4)=\left(\genfrac{}{}{0ex}{}{4}{4}\right)=\frac{4!}{4!(4-4)!}=\frac{1}{1}=1 \\ {d}_4=\left(\genfrac{}{}{0ex}{}{4}{4}\right)\cdot (1-p{)}^4\cdot p^0=1\cdot (1-0.51{)}^4\cdot 0.51^0\doteq 0.0576 \\ {p}_1=d_3+d_4=0.24+0.0576=0.2977 \end{gathered}$$

$p_2=1-d_4=1-0.0576=0.9424$

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