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 $\displaystyle \frac14.$How many times must he fire so that the probability of his hitting the target at least once is greater than$\displaystyle \frac23$ ?

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 $\displaystyle \frac23$

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 $\displaystyle (\frac{3}{4})^n<\frac{1}{3} \implies (\frac{4}{3})^n>3 \implies n\ \ln \frac{4}{3}>\ln 3 \implies$

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

http://www.sv-luka.org/ikone/ikone180a.jpg

Marry Christmas from Serbia

$\displaystyle \chi$ $\displaystyle \sigma$