SOLUTION: I am confused on how to solve binomial probability problems. Question: Accordng to a U.S. Govt study 68% of children live in home with 2 married parents. Assuming that this resu

Question 458166: I am confused on how to solve binomial probability problems.
Question: Accordng to a U.S. Govt study 68% of children live in home with 2 married parents. Assuming that this result holds true today, find the probability that in a random sample of 12 children, the number who live in homes with 2 married parents is: a) exactly 6; b) none; C) exactly 9
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Question: Accordng to a U.S. Govt study 68% of children live in home with 2 married parents. Assuming that this result holds true today, find the probability that in a random sample of 12 children, the number who live in homes with 2 married parents is: a) exactly 6; b) none; C) exactly 9
-----------------
Binomial Problem with n = 12 and p = 0.68
---
a) P(x = 6)
= 12C6(0.68)^6(0.32)^6
= binompdf(12,0.68,6)
= 0.0981
--------------------
b) P(x=0)
= 12C0(0.68)^0*(0.32)^12
---
= 0.32^12
= 0.0000011529
---
c) P(x = 9)
= 12C9(0.68)^9(0.32)^3
---
= 0.2241
===============
Cheers,
Stan H.
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