# A family has 3 children. Find the probabilities. A) all boys B) All boys or all girls C) Exactly two boys or two girls D) At least one child of each gender What are the probabilities?

##### 2 Answers

#### Answer:

See below

#### Explanation:

We have 3 children, with each one possibly being a boy or a girl. And so for each child, the probability of being a boy is

**A**

Therefore, for there to be all boys, the probability is:

**B**

The probability of all three children being girls is the same as that as being all boys,

**C**

We know that there are 8 possible ways for the genders of the children to be. We know too that of those 8 ways, 2 are for where all three children are the same gender, leaving 6 ways.

Of those 6 ways, half will be for having 2 boys and 1 girl and the other half will be for having 2 girls and 1 boy.

And so we can say:

and

**D**

To have at least one child of each sex, we need to subtract out from the 8 ways the genders can be the 2 that are single sex (i.e. all boys and all girls). So we get:

#### Answer:

P(A)=P(all boys)

P(B)=P(all boys or all girls)

P(C)=P(2 boys or 2 girls)

P(D)=P(at least 1 of each gender)

#### Explanation:

It is customary to say that the probability of getting a boy is

In addition statistics show that the probability of the child having a specific sex is correlated with the sex of the previous children. If the family for instance already have 2 boys, the probability of the third child being a boy is a little more than 52%. To get a more precise figure we should really take these facts into account.

But as a rough guide a probability of

As the calculation is already shown in another answer, I won't repeat it here, only give the rough answers:

P(A)=P(all boys)

P(B)=P(all boys or all girls)

P(C)=P(2 boys or 2 girls)

P(D)=P(at least 1 of each gender)