A couple plans to have three children. What is the probability that
a) they have all boys?
b) they have at least one girl?
prob(3 boys) = (1/2)^3 = 1/8
prob(at least 1 girl) = 1 - prob(all boys)
= 1 - 1/8 = 7/8
I apologize for the error in my previous answer. You are correct that the probability of having all boys is 1/8 as there is only one way to have all boys out of the eight possible outcomes (BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG). The probability of having at least one girl is indeed 7/8 as there are seven outcomes that have at least one girl. Thank you for correcting me.
From a group of 7 people, you randomly select 5 of them.
What is the probability that they are the 5 oldest people in the group?
Contestants A, B, C, D, E, F, G, and I are in a 10K race.
How many different ways can a race with 8 runners be completed? (Assume there is no tie.)
Your answer would be 1/8 for all boys and 7/8 for at least 1 girl. There are 8 total things that could happen, and ONE of those possibilities is 3 boys. All the rest of the options have at least one girl in it. So, 7/8 is the probability of at least one girl, and the probability for all boys is ONE(1) of the options, so 1/8.
Boys and Girls: Suppose a couple plans to have two children and the probability of having a girl is 0.50.
(a) What is the sample space for the gender outcomes?
{bb, bg, gg}
{bb, bg, gb, gg}
{b,g}
{bb, gg}