A family plans to have 3 children. For each birth, assume that the probability of a boy is the same as the probability of a girl. What is the probability that they will have at least one boy and at least one girl?

a.0.5
b.0.125
c.0.75
d.0.875

(c), 0.75

Step-by-step explanation:

It is given that, the probability of boy is same as the probability of girl.

Number of children family plans to have = 3

Consider, B be the event that represents that child is a boy

G be the event that represents that child is a girl.

The simple events for the provided case could be written as:

S = {(GGG, BBB, GBB, BGB, BBG, BBG, GBG, GGB)}

From above simple event, it is clear that

P(3 boys) = 1/8 and P(3 Girls) = 1/8

Thus, the probability of having at least one boy and at least one girl can be calculated as:

Thus, the required probability is 0.75.

Hence, the correct option is (c), 0.75

1/800 is the right answer

i think there was a typo. you wrote pat brings 3 quarts of ice-cream to the party. each serving of ice-cream is 1 6 quart. how many total servings of ice-cream does pat bring to the party? i think the bold is supposed to be 1/6. so 1 quart of ice cream = 6 servings. 6 * 3 = 18.