Question

At Least One. In Exercises 5–12, find the probability.

Births in China In China, where many couples were allowed to have only one child, the probability of a baby being a boy was 0.545. Among six randomly selected births in China, what is the probability that at least one of them is a girl? Could this system continue to work indefinitely? (Phasing out of this policy was begun in 2015.)

Short answer

The probability that at least one of them is a girl is equal to 0.974.

Yes, this system can continue to work indefinitely.

Step-by-step solution

Given information

There is a random selection of six baby births.

The probability of a baby being a boy in China is 0.545.

Probability of “at least” one

The condition for “at least one” implies that one or more such events appear while experimenting. It can also be computed as follows:

$P\left(\text{occurrence}\text{of}\text{A}\text{at}\text{least}\text{once}\right)=1-P\left(\text{non - occurrence}\text{of}\text{A}\right)$

It explains that the occurrence of an event at least once and no occurrence of the event arecomplementary.

Compute the probability of a girl being born

Let A be the event of having a baby boy in China.

Then, $P\left(A\right)$is equal to 0.545.

The complement of the event is denoted as $P\left(\overline{A}\right)$, which is defined as the probability of a girl being born in China. It is computed as follows:

$$\begin{gathered} 1-P\left(A\right)=1-0.545 \\ =0.455 \end{gathered}$$

Thus, $P\left(\overline{A}\right)$is equal to 0.455

Compute the probability that no girl is born in six births

Six births are selected at random.

The probability of having no girl or all six boys is given by:

$$\begin{gathered} P\left(\text{no}\text{girl}\right)=\left(0.545\right)\times \left(0.545\right)\times \left(0.545\right)\times \left(0.545\right)\times \left(0.545\right)\times \left(0.545\right) \\ =0.0262 \end{gathered}$$

The probability of having at least one girl is equal to one minus the probability of having no girl out of six births.

$$\begin{gathered} P\left(\text{at}\text{least}\text{1}\text{girl}\right)=1-P\left(\text{no}\text{girl}\right) \\ =1-0.2621 \\ =0.974 \end{gathered}$$

Therefore, the probability of having at least one girl is equal to 0.974.

Extension to an indefinite sample size

The process of computing the probability of at least one girl in indefinite sample births would remain the same.

In other words, if the sample of births selected is extended to a very large number, the above-mentioned method of computing the probability of at least one girl will still be valid.