# dbinom for Bernoulli trials

I have this question:

"There are a 100 families each with 5 children. Given that the null probability of having a boy is $p=0.5$, what is the probability of a family having 0,1,2,3,4,5 boys"

We have been asked specifically to use dbinom from R.

My solution:

dbinom(0:5,5,0.5)

0.03125

0.15625

0.31250

0.31250

0.15625

0.03125


My question:

Why are the numbers similar, as in why is the probability of having 0 boys the same as 5 boys, or why the probability of having 2 boys the same as having 3 boys?

• Try writing out the math involved, i.e., the formula for the binomial distribution with $p = 0.5$ and $n = 5$, and it will become clear. Jun 4 '18 at 4:35
• Hint: when there are two boys how many girls are there? Jun 4 '18 at 12:25
• Shouldn't the probability of 3 girls and 2 boys be the same as having 3 boys and 2 girls since p=0.50 Jun 4 '18 at 14:12
• The results show a symmetry which should be the case when p=0.50. Jun 4 '18 at 14:13 prop.table(c(1,5,10,10,5,1))