Question

In a family of 11, what is the probability that there will be more boys than girls

Answer 1

P(all female)= 1/2 x 1/2 x 1/2 = 1/8

P(all male ) = 1/2 x 1/2 x 1/2 = 1/8

P(all one gender) = P(all female) + P(all male) = 1/8 + 1/8 = 1/4

What is the probability that a three-child family is two girls and one boy?

Each possible birth order has P=1/8. That is, P(G,G,B)=P(G,B,G)=P(B,G,G)=1/8.

So, P(2G,1B)= 3/8 and P(1G,2B)= 3/8.

This allows us to write the overall gender probability distribution for families of three children as follows:

1/8 will be three girls

3/8 will be two girls and one boy

3/8 will be one girl and two boys

1/8 will be three boys

Adding it all up, we have 1/8 + 3/8 + 3/8 + 1/8 = 1 (100%)

Explanation:

Answer 2

The probability of there being more boys than girls in a family of 11 is 50%.

To calculate the probability of there being more boys than girls in a family of 11, we need to consider all the possible combinations of boys and girls in the family.

Let's start with the simplest case where there are 0 girls and 11 boys. There is only 1 way for this to happen.

Next, we move on to the case where there is 1 girl and 10 boys. There are 11 ways for this to happen (since the girl can be in any of the 11 positions).

Similarly, for 2 girls and 9 boys, there are 55 ways; for 3 girls and 8 boys, there are 165 ways; and so on. We keep adding up these possibilities until we reach the case where there are 5 girls and 6 boys, where there are 462 ways.

To find the total number of possibilities, we add up all these values: 1 + 11 + 55 + 165 + 330 + 462 = 1,024.

Therefore, the probability of there being more boys than girls is 1,024 divided by the total number of possible combinations (2^11 = 2,048), which simplifies to 0.5, or 50%.