the probability that in a family of 3 children, there will be at least one boy, is (a) 1/7 (b) 1/8, (c) 3/8 (d) 7/8?

P(boy) = P(girl) = 1/2

P(at least one boy in 3 children)
= 1 - P(all 3 children are boys)
= 1 - (1/2)³
= 1 - (1/8)
= 7/8

We assume that there are only two possible outcomes of each birth (boy or girl) and they have equal likelihood and are all independent.

There are 2 choices for the sex of each child. With 3 children, the total possible outcomes would be:
2 x 2 x 2 = 8 outcomes

All of these would involve at least one boy, except for the 1 case of all girls. So there are 7 outcomes with at least one boy.

At least one boy:
GGB
GBG
GBB
BGG
BGB
BBG
BBB

No boys:
GGG

There are 7 desired outcomes (at least one boy) out of 8 possible outcomes.

Answer:
(d) 7/8

　
1 − P(0 boys) = 1 − P(3 girls) = 1 − (1/2)³ = 1−1/8 = 7/8

there is 1/8 chance of all girls, so 7/8 at least one boy

d. 7/8