Question 250486: A couple decided to have 4 children.
What is the probability that they will have at least two boys?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the possible combinations of boys and girls are:
bbbb
bbbg
bbgg
bggg
gggg
all boys can occur in 1 way so p(4b) = 1 * .5^4 = .0625
3 boys 1 girl can occur in 4 ways so p(3b1g) = 4 * .5^4 = .25
2 boys 2 girls can occur in 6 ways so p(2b2g) = 6 * .5^4 = .375
1 boy 2 girls can occur in 4 ways so p(1b3g) = 4 * .5^4 = .25
all girls can occur in 1 way so p(4g) = 1 * .5^4 = .0625
sum of all probabilities should be equal to 1.
2 * .0625 + 2 * .25 + .375 = 1 so we're good there.
probability of at least 2 boys equals p(2b2g) + p(3b1g) + p(4b) = .375 + .25 + .0625 = .6875 = 68.75%
the number of ways you can get all boys is 1 as shown below:
bbbb
the number of ways you can get 3 boys and 1 girls is 4 as shown below:
bbbg
bbgb
bgbb
gbbb
the number of ways you can get 2 boys and 1 girl is 6 as shown below:
bbgg
bggb
ggbb
gbbg
bgbg
gbgb
since the order in which the children arrive has something to do with the probability of them arriving, these would be permutations.
since the probability of having a boy or a girl was the same (.5 in each case), the probability part of the equations was always the same (.5^4) regardless of whether it was 1 boy, 2 boys, 3 boys, or 4 boys.
for example:
p(bbbb) = .5*.5*.5*.5 = .5^4
p(gggg) = .5*.5*.5*.5 = .5^4
p(bbbg) = .5*.5*.5*.5 = .5^4
p(bbgg) = .5*.5*.5*.5 = .5^4
if the probability was different, i.e. p(b) = .75, p(g) = .25, then the probability for each of the permutations would have been different.
for example:
p(bbbb) = .25*.25*.25*.25 = .25^4
p(gggg) = .75*.75*.75*.75 = .75^4
p(bbbg) = .25*.25*.25*.75 = .25^3*.75
p(bbgg) = .25*.25*.75*.75 = .25^2*.75^2
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