# Binomial distribution

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• December 25th 2011, 03:25 AM
Vinod
Binomial distribution
This problem have been taken from the old paper of MASTER of COMMERCE Examination of well-known University in India.
The probability of a man hitting a target is $\frac14.$How many times must he fire so that the probability of his hitting the target at least once is greater than $\frac23$ ?

Answer: -
3.818 times i-e approximately 4 times
A man must fire 4 times so that the probability of his hitting the target at least once is greater than $\frac23$
Verify the answer, if found wrong reply me.
Answer is not available in the university paper set.So I don't know whether I solved the problem wrongly or correctly.
• December 25th 2011, 08:38 AM
chisigma
Re: Binomial distribution
Quote:

Originally Posted by Vinod
This problem have been taken from the old paper of MASTER of COMMERCE Examination of well-known University in India.
The probability of a man hitting a target is $\frac14.$How many times must he fire so that the probability of his hitting the target at least once is greater than $\frac23$ ?

Answer: -
3.818 times i-e approximately 4 times
A man must fire 4 times so that the probability of his hitting the target at least once is greater than $\frac23$
Verify the answer, if found wrong reply me.
Answer is not available in the university paper set.So I don't know whether I solved the problem wrongly or correctly.

By definition it must be $(\frac{3}{4})^n<\frac{1}{3} \implies (\frac{4}{3})^n>3 \implies n\ \ln \frac{4}{3}>\ln 3 \implies$

$\implies n> \frac{\ln 3}{\ln \frac{4}{3}} = 3.8188416793...$

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Marry Christmas from Serbia

$\chi$ $\sigma$