Assume that male and female births are equally likely and that the birth of any child does not affect the probability of the gender of any other children. Find the probability of at most three boys in ten births.

Please help me I don't get it.

do a simpler ex first.

what's the chance of flipping heads on a quarter? 1 in 2.

what's the chance of doing it again? 1 in 2.

again? 1 in 2.

each event is independent of the others.


but when you consider the odds of doing it three times in a row, its 1/2 * 1/2 * 1/2 = 1/8

if you talk about 2 births only, you'd say there are 2 outcomes possible for each birth. 2*2 means four overall possible outcomes. Boy/boy... boy/girl... girl/boy... girl/girl

then here, what is the chance of having one boy? 2 in 4.

expand this to three births. 2*2*2 = 8
b/b/b
b/b/g
b/g/b
g/b/b
b/g/g
g/b/g
g/g/b
g/g/g

chances here of 2 boys? 3 of 8.

the only issue I have here is for your scenario there is an obnoxious amount of possible combos. 2*2*2*2*2*2*2*2*2*2 = 1024.

but you also only need to figure out how many times you can get three bs in 10.

like

b/b/b/g/g/g/g/g/g/g
b/b/g/b/g/g/g/g/g/g
b/b/g/g/b/g/g/g/g/g
b/b/g/g/g/b/g/g/g/g
b/b/g/g/g/g/b/g/g/g
b/b/g/g/g/g/g/b/g/g
b/b/g/g/g/g/g/g/b/g
b/b/g/g/g/g/g/g/g/b
b/g/b/b/g/g/g/g/g/g
etc...

its work but you can do it by hand.

here's the catch. There might be a formula or an easier way to do it than by hand. Buts as its been too long, and I'm not sure where my stats book is, that's all I can do to help. Hopefully someone else will come in with a shortcut. In the meantime, do you see how I was approaching it, with sytematically varying the orders?

Try this approach:

C(10 , 0) = 1 -> 0 boys
C(10 , 1) = 10 -> 1 boy
C(10 , 2) = 45 -> 2 boys
C(10 , 3) = 120 -> 3 boys

[The above values can be calculated by formula, scientific calculator or using Pascal's Triangle.]

P(0) = (1/2)^10
P(1) = 10(1/2)^10
P(2) = 45(1/2)^10
P(3) = 120(1/2)^10

P(at most 3 boys) = (1/2)^10 + 10(1/2)^10 + 45(1/2)^10 + 120(1/2)^10
= 176(1/2)^10
= 176(1/1024)
= 11/64