Question 1089547: One day, eleven babies are born at a hospital. Assuming each baby has an equal chance of being a boy or a girl, what is the probability that at most nine of the eleven babies are girls?
Answer by mathmate(429)   (Show Source):
You can put this solution on YOUR website! Question:
One day, eleven babies are born at a hospital. Assuming each baby has an equal chance of being a boy or a girl, what is the probability that at most nine of the eleven babies are girls?
Solution:
We will first check if this problem can be modelled using the binomial distribution.
1. Bernoulli trials, i.e. exactly two possible outcomes (girl or boy)
2. Number of trials (n) is known before and constant throughout the experiment, i.e. independent of outcomes. (n=11)
3. All trials are independent of each other. (assumed from context).
4. Probability (p) of success is known, and remain constant throughout trials. (p=0.5)
Since all criteria are satisfied, we can model this situation with binomial distribution, where the probability of x successes out of N trials each with probability of success p is given by
P(x)=C(N,x)(p^x)(1-p)^(N-x)
and,
C(N,x) is number of combinations of selecting x objects out of N.
n=11
p=0.5
x<=9
P(x<=9)=1-(P(x=10)+P(x=11)
=1-(C(11,10)0.5^10*0.5^1+C(11,11)0.5^11*0.5^0)
=1-(0.005371+0.000488)
=0.994141
The probability that at most 9 out of 11 newborns are girls is 0.994141, assuming boys and girls are equally probable.
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