A sample of four families with three children each was taken.find the probability that one family has two girls,two families have three girls and one family has three boys.
Pr(three boys,1g)=1/16
pr(2g,2b)=1/16
pr(3g,1b)=1/16
Pr(above)=4*pr(3b)*3pr(3g)*2pr(3g)*pr(2g)=4*1/16*3*1/16*2*1/16*1/16
=4! (1/16)^4
check that.
I disagree
The probs that Bob found would each have 4 children, not three
prob(of 3 children, two are girls)
= C(3,2)(1/2)^2 (1/2) = 3/8
illustration:
BBB
BBG
BGB
GBB
GGG
GGB -- 2 GIRLS
GBG -- 2 GIRLS
BGG -- 2 GIRLS
So there are 3 cases of 2 girls
prob(2 girls) = 3/8
prob(3 girls)
= 1/8
prob(3 boys)
= 1/8
so prob(as stated in problem)
= (3/8)(1/8)(1/8)(1/8)
= 3/4096
To find the probability, we need to determine the total number of possible outcomes and the number of favorable outcomes.
First, let's calculate the total number of possible outcomes. We have four families, and each family can have either three boys or three girls. So, for each family, there are 2 possibilities (2 outcomes) - either three boys or three girls. Since we have four families, the total number of possible outcomes is 2^4 = 16.
Next, let's determine the number of favorable outcomes. We want one family to have two girls, two families to have three girls, and one family to have three boys.
To calculate the number of ways we can have one family with two girls out of four families, we use combination notation. We have 4 families, and we need to choose 1 to have two girls. Therefore, the number of ways to choose 1 family out of 4 is C(4, 1) = 4.
Now, for the chosen family with two girls, we need to determine the number of ways to distribute the two girls across the three children. Since each child can either be a boy or a girl, there are 2^3 = 8 possible outcomes. However, we don't need to consider the order in which the girls appear, so we divide by 2! (the number of ways to arrange two items) to get the number of combinations. Therefore, the number of ways to have two girls in one family is 8/2! = 4.
Similarly, for the two families with three girls, we need to choose 2 families out of 3 (excluding the previously chosen family). Using combination notation, we get C(3, 2) = 3.
For each of these two families, we again need to consider the number of ways to distribute the three girls across the three children. This is the same calculation as before, so we have 8/2! = 4 combinations for each family. Since there are two families with three girls, the total number of ways to have three girls in two families is 4 * 4 = 16.
Lastly, for the remaining family, we need three boys. Similarly, we only have one family to consider in this case, so the number of ways is 1.
Now, let's calculate the total number of favorable outcomes. We multiply the number of ways for each event together: 4 * 16 * 1 = 64.
Therefore, the probability of having one family with two girls, two families with three girls, and one family with three boys is 64/16 = 4.
Hence, the probability is 4/16 = 1/4.