Question

The probability of a bomb hitting a bridge is 1/2 and two direct hits are needed to destroy it. The least number of bombs required so that the probability of the bridge being destroyed is greater than 0.9 is
1) 8 2) 7
3) 6 4) 9
{ What is meant by 'Direct Hits' - two consecutive hits or any two hits }

Answer 1

Thank you for asking this question. Here is your answer:

The probability of the bridge being destroyed is greater than 0.9 that means  probability of the bridge being not destroyed is lesser than 0.1 .

We will let n bombs are needed

Now we will find all probabilities of the bridge being not destroyed by binomial distribution i.e.

[nC0+nC1]*(0.5)^n < 0.1

Now we will examine it for values n = 1,2,3,4,,....

You will see that n = 1 to 6

The above inequality doesn't satisfy

It first time satisfies for n = 7

So the final answer for this question is 7

If there is any confusion please leave a comment below.