Question

each of 14 mothers - to - be received 3d ultrasound scans, which showed that 5 of them will give birth to girls. assuming that simultaneous births do not happen (i.e., babies are born one after another), what is the probability that the first two babies that are born happen to be boys?
note: for any final answer that has up to four decimal places, enter your answer without rounding the number. for any answers with more than four decimal values, round your final answer to four decimal places.

Explanation:

Step1: Calculate the number of boys

Total mothers - girls = boys. So \(14 - 5=9\) boys.

Step2: Calculate the probability of the first - born being a boy

The probability \(P_1\) that the first baby is a boy is \(\frac{9}{14}\) (number of boys / total number of babies).

Step3: Calculate the probability of the second - born being a boy (after the first is a boy)

After one boy is born, there are \(9 - 1=8\) boys left and \(14 - 1 = 13\) babies left. So the probability \(P_2=\frac{8}{13}\).

Step4: Calculate the probability of both events (first and second are boys)

Since these are independent events (the birth of the first baby affects the sample space for the second), we use the multiplication rule for probabilities. \(P = P_1\times P_2=\frac{9}{14}\times\frac{8}{13}\).

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Answer:

\(0.3956\)