Assume the probability is 1/2 that a child born is a girl. If a family has three children, what is the probability that they have a) exactly one girl b) at most two boys?
a) So it could be
GBB or BGB or BBG
prob of that is 3(1/2)(1/2)(1/2) = 3/8
b) at most 2 boys ---> cannot have BBB
which has a prob of 1/8
so your case has prob of 1 - 1/8 = 7/8
sdf
To find the probability in these scenarios, we can use the concept of binomial probability, which calculates the probability of a certain number of successes in a fixed number of independent trials.
a) To find the probability of exactly one girl in a family with three children, we need to determine the probability of having one girl and two boys in any order. The probability of a single girl in each birth is 1/2, and the probability of two boys in each birth is also 1/2.
There are three possible combinations that result in exactly one girl: GBB, BGB, and BBG.
Let's calculate the probability of each combination and sum them up:
P(One girl) = P(GBB) + P(BGB) + P(BBG)
P(G) represents the probability of having a girl, which is 1/2, and P(B) represents the probability of having a boy, also 1/2.
P(One girl) = (1/2) * (1/2) * (1/2) (GBB)
+ (1/2) * (1/2) * (1/2) (BGB)
+ (1/2) * (1/2) * (1/2) (BBG)
P(One girl) = 1/8 + 1/8 + 1/8 = 3/8
Therefore, the probability that a family with three children has exactly one girl is 3/8.
b) To find the probability of at most two boys, we need to calculate the probability of having zero boys, one boy, or two boys.
P(At most two boys) = P(No boys) + P(One boy) + P(Two boys)
P(No boys) represents the probability of having all girls, which is (1/2) * (1/2) * (1/2) = 1/8.
P(One boy) was calculated in part a, and we found it to be 3/8.
P(Two boys) represents the probability of having exactly two boys and one girl, which is also 3/8.
P(At most two boys) = 1/8 + 3/8 + 3/8 = 7/8
Therefore, the probability that a family with three children has at most two boys is 7/8.