##### Bionomial Distribution Example(2)

The probability of a man hitting a target is 1/3.

How many times he must hit so that the probability of hitting a target at least once is more than 90 %?

**Answer**

**Solution:**

From the question, we can conclude that p = 1/3 and 1-p =2/3

and they are saying that "probability of hitting a target at least once", It means **p(i>=1) **

**p(i>=1) = 1- p(i=0)**

0.9 >= 1- { (n c 0) (1/3)^{0} (2/3)^{n} }

1- 0.9 >= (2/3)^{n}

0.1 > = (2/3)^{n}

on solving you will get n=6