Question

The probability of a bomb hitting a bridge is $\frac { 1 } { 2 }$and two direct hits are needed to destroy it. The least number of bombs required so that the probability of the bridge beeing destroyed is greater then 0.9, is

• A. 8
• B. 7
• C. 6
• D. 9

Answer 1

Answer: (A)

8

Answer 2

Let $n$ be the least number of bombs required and $X$ the number of bombs that hit the bridge. Then $X$ follows a binomial distribution with parameter $n$ and $p = \frac { 1 } { 2 }$

Now $P ( X \geq 2 ) > 0.9 \Rightarrow 1 - P ( X < 2 ) > 0.9$

$\Rightarrow P ( X = 0 ) + P ( X = 1 ) < 0.1$

$\Rightarrow ^ { n } C _ { 0 } \left( \frac { 1 } { 2 } \right) ^ { n } + { } ^ { n } C _ { 1 } \left( \frac { 1 } { 2 } \right) ^ { n - 1 } \left( \frac { 1 } { 2 } \right) < 0.1 \Rightarrow 10 ( n + 1 ) < 2 ^ { n }$

This gives $n \geq 8$