Question 1028853: In a certain country, the true probability of a baby being a girl
is 0.485.
Among the next nine randomly selected births in the country, what is the probability that at least one of them is a boy?
Answer by mathmate(429)   (Show Source):
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Question:
In a certain country, the true probability of a baby being a girl is 0.485.
Among the next nine randomly selected births in the country, what is the probability that at least one of them is a boy ?
Solution:
Since the probability remains constant, and we know the number of steps (9) of the random experiments. Each step's outcome is either boy or girl (Bernoulli trial), we can use the binomial distribution, given by:
P(x,n,p)=C(n,x)*p^x*(1-p)^(n-x)
x=number of successful outcomes
n=number of trials
p=probability of success in each trial
C(n,x)=binomial coefficient equal to n!/(x!(n-x)!).
Here, the probability of success (boy) is 1-0.485=0.515
number of trials = 9
we need at least one boy out of 9, means that we cannot have zero boys.
So we use the complement of P(1≤x&;9)=1-P(0).
So the probability is
P(1≤x≤9, 9, 0.515)
=1-P(0,9,0.515)
=1-C(9,0)*0.515^0*0.485^9
=1-1*1*0.485^9
=1-0.001484826
=0.9985152
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