Question 235235: How many different ways can a couple have 3 daughters and 2 sons in five distinct pregnancies (no multiple births). list all gender string (GGGBB etc) and compute the probability this happens.
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Here are all of the possible combinations of 5 children:
G,G,G,G,G
G,G,G,G,B
G,G,G,B,G
G,G,G,B,B
G,G,B,G,G
G,G,B,G,B
G,G,B,B,G
G,G,B,B,B
G,B,G,G,G
G,B,G,G,B
G,B,G,B,G
G,B,G,B,B
G,B,B,G,G
G,B,B,G,B
G,B,B,B,G
G,B,B,B,B
B,G,G,G,G
B,G,G,G,B
B,G,G,B,G
B,G,G,B,B
B,G,B,G,G
B,G,B,G,B
B,G,B,B,G
B,G,B,B,B
B,B,G,G,G
B,B,G,G,B
B,B,G,B,G
B,B,G,B,B
B,B,B,G,G
B,B,B,G,B
B,B,B,B,G
B,B,B,B,B
There are a total of 32 possible combinations.
From this list, we have the following combinations for 3 girls and 2 boys:
G,G,G,B,B
G,G,B,G,B
G,B,G,G,B
B,G,G,G,B
B,G,G,B,G
B,G,B,G,G
B,B,G,G,G
G,B,B,G,G
G,B,G,B,G
G,G,B,B,G
In this list, there are 10 different possible combinations.
So the probability of having 3 girls and 2 boys is:
P(3 girls & 2 boys) = 10/32 = 5/16 = 0.3125
which is a 31.25% chance
|
|
|
