In a family with two children, what are the probabilities of the following outcomes, assuming that the birth of boys and girls is equally likely?

a. Both are boys.
b. The first is a girl and the second a boy.
c. Neither is a girl.
d. At least one is a girl.

I was wondering about the formula....

is a. P(probability) = 1/2(boy) + 1/2(boy)
P = 1/2 + 1/2
so P = 2/2
to me this doesn't seem right

A) PR(B,B)=Pr(b)*Pr(b)=1/2*1/2

B) Pr(g)Pr(b)=1/2*1/2
C) you can get at least one girl these ways:
gb
bg
gg

Pr(at least one g)= 1/2*1/2*3

Thank you for answering my question.

Why do you multiply the probability of both boys, girl and boy, etc?

To calculate the probabilities of each outcome, we need to analyze the possible combinations of genders for the two children. Let's break it down step by step:

a. Both are boys:
There is only one combination where both children are boys: BB. Since the birth of boys and girls is equally likely, the probability of this outcome is 1/4.

b. The first is a girl and the second a boy:
There are two possible combinations where the first child is a girl and the second child is a boy: GB or BG. Again, since the birth of boys and girls is equally likely, the probability of each combination is 1/4. Therefore, the probability of this outcome is 1/4 + 1/4 = 1/2.

c. Neither is a girl:
In this case, there are two outcomes: BB (both boys) which we already calculated in (a), and GG (both girls). Therefore, the probability of this outcome is 1/4 + 1/4 = 1/2.

d. At least one is a girl:
To calculate the probability of at least one girl, we need to consider all the outcomes where at least one girl is present. The outcomes are: GB (girl-boy), BG (boy-girl), and GG (both girls). Therefore, the probability of this outcome is 1/4 + 1/4 + 1/4 = 3/4.

In summary:
a. Probability of both being boys: 1/4
b. Probability of the first being a girl and the second being a boy: 1/2
c. Probability of neither being a girl: 1/2
d. Probability of at least one being a girl: 3/4