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Six baby births are randomly selected.

The probability of a baby being a boy is equal to 0.512.

The probability of 鈥渁t least one鈥� occurrence of a given event is the probability that the event has occurred once or more than once.

Moreover, the occurrence of an event at least once complements the non-occurrence of the event. Mathematically,

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

Let A be the event of having a baby boy.

Then, is equal to 0.512.

Six random births are considered.

The probability of having no girl is the same as the event that all six are boys. This is computed as follows:

$$\begin{gathered} P\left(\text{no}\text{girl}\right)=\left(0.512\right)脳\left(0.512\right)脳\left(0.512\right)脳\left(0.512\right)脳\left(0.512\right)脳\left(0.512\right) \\ =0.0180 \end{gathered}$$

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

$$\begin{gathered} P\left(\text{at}\text{least}\text{1}\text{girl}\right)=1-P\left(\text{no}\text{girl}\right) \\ =1-0.01801 \\ =0.982 \end{gathered}$$

Therefore, the probability of having at least one girl is equal to 0.982.