Talamancoeb

2021-12-16

A family has five kids. Assume that each child has an equal chance of being a male or a girl. Find the probability that the family has 5 girls if it is known the family has at least one girl.

ol3i4c5s4hr

Expert

Step 1
Given that the family has 5 children.
Assume that each child is as likely to be a boy as it is to be a girl.
That is, $P\left(girl\right)=P\left(boy\right)=\frac{1}{2}$.
Consider,
$P\left(5\text{girls|at least one girl}\right)=P\frac{5\text{girls and at least one girl}}{P}\left(\text{at least one girl}\right)$
$=P\frac{5\text{girls}}{P}\left(\text{at least one girl}\right)$ (1)
Step 2
Let us define the random variable X as the number of girls follows Binomial distribution with $n=5$ and probability of girl child is $p=\frac{1}{2}$.
$P\left(X=x\right)=\left(5Cx\right){\left(\frac{1}{2}\right)}^{x}{\left(1-\frac{1}{2}\right)}^{5-x}=\left(5Cx\right){\left(\frac{1}{2}\right)}^{x}{\left(\frac{1}{2}\right)}^{5-x}$.
$P\left(X=5\right)=\left(5C5\right){\left(\frac{1}{2}\right)}^{5}{\left(\frac{1}{2}\right)}^{5}-5=0.03125$ (2)
$P\left(X\ge 1\right)=1-P\left(X<1\right)=1-P\left(X=0\right)=1-\left(5C0\right){\left(\frac{1}{2}\right)}^{0}{\left(\frac{1}{2}\right)}^{5-0}=1-0.03125=0.96875$ (3)
Substitute (2) and (3) in equation (1)
$P\left(5\text{girls|at least one girl}\right)=P\frac{5\text{girls}}{P}\left(\text{at least one girl}\right)$
$=\frac{0.03125}{0.96875}$
$=0.032258$

raefx88y

Expert

Step 1
A family has five kids. Assume that each child has an equal chance of being a boy or a girl. If it is known that the family has at least one girl, calculate the likelihood that there are 5 girls.
So, there are
$2×2×2×2×2$ (2 raised to the 5th power)
$=32$ sequences of 5 children.
As we already know that only one of them is all female, the probability of five female births in a row when all births are independent and
$p\left(B\right)=p\left(G\right)=0.5$ is $p\left(G,G,G,G,G\right)=\frac{1}{32}=0.03125$
Note, that this comes out to be the same as simply multiplying the probability of a girl five times because there is only one of these sequences. if you wanted to know the probability of 2 girls and 3 boys, you have to know how many sequences have 2 girls and 3 boys and adjust accordingly.

nick1337

Expert

You have a $\frac{50}{50}$ chance of it becoming an a boy or a girl.
The chance the first is a girl is 50%. The chance the second is also 50%, and so on.
Therefore the chance that all five are girls is $\left(0.5{\right)}^{2}$ or 3.125%

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