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The Jones family was one of the first to come to the U.S. They had 5 children. Assuming that the probability of a child being a...Asked by Kevin
The Jones family was one of the first to come to the U.S. They had 9 children. Assuming that the probability of a child being a girl is .5, find the probability that the Jones family had at least 5 girls.
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Answered by
MathMate
Use binomial distribution because:
1. Bernoulli trials (either girl or boy)
2. number of trials (n=9) known.
3. probability (p=0.5) of outcome known and remains constant throughout trials
4. trials are independent of each other.
n=9
p=0.5
P(X=k)=C(n,k)*p^k*(1-p)^(n-k)
where
C(n,k) is the number of combinations of k objects taken from n.
At least 5 girls means X=5,6,7,8,9
So
P(X>=5)
=P(X=5)+P(X=6)+P(X=7)+P(X=8)+P(X=9)
=0.24609+0.16406+0.07031+0.01758+0.00195
=0.5000
1. Bernoulli trials (either girl or boy)
2. number of trials (n=9) known.
3. probability (p=0.5) of outcome known and remains constant throughout trials
4. trials are independent of each other.
n=9
p=0.5
P(X=k)=C(n,k)*p^k*(1-p)^(n-k)
where
C(n,k) is the number of combinations of k objects taken from n.
At least 5 girls means X=5,6,7,8,9
So
P(X>=5)
=P(X=5)+P(X=6)+P(X=7)+P(X=8)+P(X=9)
=0.24609+0.16406+0.07031+0.01758+0.00195
=0.5000
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