Answer :
Answer:
P( at least 3 girls) = 0.3125
P( at most 2 girls) = 0.6875
Explanation:
To know the probability that the Wilson family had at least 3 girls, we will use the binomial distribution because we have n identical events with a probability p of success and probability (1 - p) of fail.
In this case, n is the number of children and we consider success if they have a girl, so p is 0.5 and (1-p) is also 0.5 because:
1 - p = 1 - 0.5 = 0.5
Then, the probability to have x number of girls can be calculated as:
[tex]P(x)=\text{nCx}\cdot p^x\cdot(1-p)^{n-x}[/tex]Where nCx is equal to:
[tex]\text{nCx}=\frac{n!}{x!(n-x)!}[/tex]Now, the probability that the Wilson family had at least 3 girls is the probability to have 3 girls added to the probability to have 4 girls, so:
[tex]P(x\ge3)=P(3)+P(4)[/tex]Where P(3) and P(4) are equal to:
[tex]\begin{gathered} P(3)=4C3\cdot0.5^3\cdot0.5^1=\frac{4!}{3!(4-3)!}\cdot0.5^3\cdot0.5^1=0.25 \\ P(4)=4C4\cdot0.5^4_{}\cdot0.5^0=\frac{4!}{4!(4-4)!}\cdot0.5^3\cdot0.5^1=0.0625 \end{gathered}[/tex]Therefore, the probability that the Wilson family had at least 3 girls is equal to:
[tex]P(x\ge3)=0.25+0.0625=0.3125[/tex]In the same way, the probability that the Wilson family had at most 2 girls can be calculated as:
[tex]P(x\leq2)=P(0)+P(1)+P(2)[/tex][tex]\begin{gathered} P(0)=4C0\cdot0.5^0\cdot0.5^4=0.0625 \\ P(1)=4C1\cdot0.5^1\cdot0.5^3=0.25 \\ P(2)=4C2\cdot0.5^2\cdot0.5^2=0.375 \end{gathered}[/tex][tex]P(x\leq2)=0.0625+0.25+0.375=0.6875[/tex]Therefore, the answers are:
P( at least 3 girls) = 0.3125
P( at most 2 girls) = 0.6875
