Question

The Jones family was one of the first to come to the U.S. They had 7 children. Assuming that the probability of a child being a girl is 0.5, find the probability that the Jones family had at least 5 girls? Enter an integer or decimal number (more..)

Answer 1

Let's list down the given information in the question.

n = 7 children

probability of a girl (p) = 0.5

Hence, we can say that probability of a boy (q) = 0.5 too.

For this situation, we can use the formula for Binomial Distribution.

$P(x)=_nC_rp^xq^{n-x}$

To get the probability of at least 5 girls, we need to find the probability of having exactly 5, 6, and 7 girls then, add the results.

At x = 5 girls, plug in the value of n, p, q that was listed above under the given information.

$\begin{gathered} P(5)=_7C_5(0.5)^5(0.5)^2 \\ P(5)=21(0.03125)(0.25) \\ P(5)=0.1640625 \end{gathered}$

At x = 6 girls, we have:

$\begin{gathered} P(6)=_7C_6(0.5)^6(0.5)^1_{} \\ P(6)=7(0.015625)(0.5) \\ P(6)=0.0546875 \end{gathered}$

At x = 7 girls, we have:

$\begin{gathered} P(7)=_7C_7(0.5)^7(0.5)^0_{} \\ P(7)=1(0.0078125)(1) \\ P(7)=0.0078125 \end{gathered}$

Let's add each probability to get the probability of having at least 5 girls.

$\begin{gathered} P(x\ge5)=P(5)+P(6)+P(7) \\ P(x\ge5)=0.1640625+0.0546875+0.0078125 \\ P(x\ge5)=0.2265625 \end{gathered}$

The probability of the Jones family having at least 5 girls is 0.2266.

To express the probability in percent form, let's multiply the results by 100.

$0.2265625\times100=22.65625\approx22.66$

Therefore, the probability of the Jones family having at least 5 girls is 22.66%.