SOLUTION: a company finds that 20% of thier employees resign within one year. find the probability that among 30 employees, exactly 5 will resign I got the answer of .421 but not sur

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Question 367089: a company finds that 20% of thier employees resign within one year.
find the probability that among 30 employees, exactly 5 will resign
I got the answer of .421 but not sure, how to check it
thank you

Found 2 solutions by stanbon, ewatrrr:
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
a company finds that 20% of thier employees resign within one year.
find the probability that among 30 employees, exactly 5 will resign
I got the answer of .421 but not sure, how to check it
----------------
Binomial Problem with n=30 and p = 0.2
---
P(x= 5) = 30C5*(0.2)^5*(0.8)^25 = 142506*0.00032*0.003778 = 0.172278..
==============
Cheers,
Stan H.

Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!

Hi,
Note: The probability of x successes in n trials is:
P = nCx* where p and q are the probabilities of success and failure respectively
In this case p =.20 and q = .80
nCx =
P(5 out of 30 will resign) = 30C5 *.2^5 * .8^25 = 142506 *.2^5 * .8^25 =.1723
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