There are only 23=823=8 possible ways to arrange the genders boy/girl, with repetition. Since the probability of boy and girl are equal, the probability of each of these arrangements are also equal. We can therefore count the number of possible "good" answers, and divide by 8.
There are 3 possible ways to have 1 boy, that being "first/second/third child is boy", so the probability of exactly 1 boy is indeed 3838.
Having at most 2 girls is the opposite of having three girls, so since there are only one way of having three girls, there must be 8−1=78−1=7 ways of having at most two girls, so the probability of at most two girls is indeed 78
The probability of the event that if a family has four children (no twins), none of them are boys is 1/8.
Given
Assume the probability is one-half that a child born at any time is a boy and births are independent.
If there is no effect on the probability, then the events are independent.
The probability of a child being born is 1/2.
Therefore,
The probability of the event that if a family has four children (no twins), none of them are boys is;
[tex]\rm Probability = \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} \\\\Probability = \dfrac{1}{8}[/tex]
Hence, the probability of the event that if a family has four children (no twins), none of them are boys is 1/8.
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