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

 

 

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