Assume that the probability of a boy being born is the same as the probability of a girl being born. Find the probability that a family with four children will have the given composition. (Enter your answer to four decimal places.)

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If the probability of a boy to be born is the same as of a girl, and there are no other possibilities, then both probabilities for each birth are equal to 1/2. Suppose also that each next birth outcome is independent of the previous outcomes.

It is not clear whether...

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If the probability of a boy to be born is the same as of a girl, and there are no other possibilities, then both probabilities for each birth are equal to 1/2. Suppose also that each next birth outcome is independent of the previous outcomes.

It is not clear whether a "composition" includes the order of birth. If yes, then any sequence, say boy-girl-boy-girl, girl-girl-girl-girl, or boy-boy-girl-boy has the same probability `1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 = 1 / 16 ` (it is not a surprise that there are `2^4 = 16 ` such sequences).

But if a composition includes only the total quantities, then the probability differs for different compositions. The probability of "4 boys, 0 girls" and of "4 girls, 0 boys" are still 1/16 each because only one ordered sequence gives each result.

The probability of "3 boys, 1 girl" is 4 times greater because this combination can be achieved `( ( 4 ) , ( 3 ) ) = ( ( 4 ) , ( 1 ) ) = 4 ` ways, i.e. it is 1/4. The same for "3 girls, 1 boy".

Finally, the probability of "2 boys, 2 girls" is `( ( 4 ) , ( 2 ) ) = 6 ` times greater than 1/16, i.e. it is equal to 3/8.

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