Question 456624: Find the probability of at least 2 girls in 7 births. assume that male and female births are equally likely and that the births are independent events.
Answer: using binomcdf function 1-binomcdf(7,1/2,1)= .09375
I get this what I am not understanding is Why? I understand why I would use cdf vs pdf, because we are not looking for an exact count and each birth has a 1 out of 2 chance at being a girl. What I am confused about is that if n=7 p=1/2, and x=1 which is less than n. why is it that this formula is the only one that gives me the right answer?
Signed,
Just trying to understand
Answer by sudhanshu_kmr(1152)   (Show Source):
You can put this solution on YOUR website!
Find the probability when no girl and when 1 girl , add them and subtract by 1.
probability when no girl = 7C0 * (1/2)^0 *(1/2)^7 = 0.0078
probability when one girl = 7C1 * (1/2)^1 *(1/2)^6 = 0.0546
probability when at least 2 girls = 1 - ( 0.0078 + 0.0546 )
= 0.93751 (approx)
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