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

blue marble is 15 times out of 50 = 15/50 reduces to 3/10

landing on tails is 24 out of 50 = 24/50 reduces to 12/25

to find the probability of both multiply together:

3/10 x 12/25 = 3 x 12 / 10 x 25 = 36/250

if you multiplied the numerator and denominator by 10 you get 360/2500.

the answer is 360/2500

c) the median for town a 30" is greater than the median for town b 25"

step-by-step explanation:

town a, median = q2 = 30

town b, median - q2 = 25

so the median for town a 30" is greater than the median for town b 25"