Talamancoeb

Answered

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.

Answer & Explanation

ol3i4c5s4hr

Expert

2021-12-17Added 48 answers

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 $p=\frac{1}{2}$ .

$P(X=x)=\left(5Cx\right){\left(\frac{1}{2}\right)}^{x}{(1-\frac{1}{2})}^{5-x}=\left(5Cx\right){\left(\frac{1}{2}\right)}^{x}{\left(\frac{1}{2}\right)}^{5-x}$ .

$P(X=5)=\left(5C5\right){\left(\frac{1}{2}\right)}^{5}{\left(\frac{1}{2}\right)}^{5}-5=0.03125$ (2)

$P(X\ge 1)=1-P\left(X<1\right)=1-P(X=0)=1-\left(5C0\right){\left(\frac{1}{2}\right)}^{0}{\left(\frac{1}{2}\right)}^{5-0}=1-0.03125=0.96875$ (3)

Substitute (2) and (3) in equation (1)

$P\left(5\text{girls|at least one girl}\right)=P\frac{5\text{girls}}{P}\left(\text{at least one girl}\right)$

$=\frac{0.03125}{0.96875}$

$=0.032258$

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,

Consider,

Step 2

Let us define the random variable X as the number of girls follows Binomial distribution with

Substitute (2) and (3) in equation (1)

raefx88y

Expert

2021-12-18Added 26 answers

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

$2\times 2\times 2\times 2\times 2$ (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 $p(G,G,G,G,G)=\frac{1}{32}=0.03125$

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

2021-12-28Added 573 answers

You have a

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

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