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There is a random selection of six baby births.

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

The condition for 鈥渁t least one鈥� 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.

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 $P\left(\stackrel{炉}{A}\right)$, 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, $P\left(\stackrel{炉}{A}\right)$is equal to 0.455

Six births are selected at random.

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

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

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.

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.