Question

What is the probability of a bomb hitting a bridge is $\dfrac{1}{2}$ 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?
A. 8
B.7
C. 6
D. 9

Answer 1

Hint: Given that 2 direct hits are needed to destroy the bridge. So, we will apply the binomial distribution for $x>2$. Since the probability of the bridge being destroyed is greater than 0.9. Make an inequality such that $P(x>2) > 0.9$ and calculate the value of n.

Formula Used:
Binomial distribution: $P\left( x \right){ = ^n}{C_r}{\left( p \right)^r}{\left( q \right)^{n - r}}$

Complete step by step solution:
$\left[ {P[x = a] = \dfrac{{{^n}{C_a}}}{{{2^n}}}} \right]$and $P[X = 0] + P[X = 1] < \text{Probability}$.
Let $n$be the least number of bombs required to destroy the bridge,
$ \Rightarrow p = \dfrac{1}{2}$, which is the probability of hitting the bridge,
Then as by binomial distribution,
$ \Rightarrow q = 1 - p$, which the failure of the bombs not hitting the bridge,
$ \Rightarrow q = 1 - \dfrac{1}{2}$
Then as two direct hits are required to destroy the bridge the probability will be,
$ \Rightarrow P[x \ge 2] > 0.9$
Then as per binomial distribution for the bridge not to be destroyed,
$ \Rightarrow 1 - P[x \le 2] > 0.9$
$ \Rightarrow P[x < 2] < 1 + 0.9 = 0.1$
Now using binomial distribution formula $\left[ {P[x = a] = \dfrac{{{^n}{C_a}}}{{{2^n}}}} \right]$,
$P[X = 0] + P[X = 1] < \text{Probability}$
$ \Rightarrow P[X = 0] + P[X = 1] < 0.1$
$ \Rightarrow \dfrac{{^n{C_0}}}{{{2^n}}} + \dfrac{{^n{C_1}}}{{{2^n}}} < 0.1$
$ \Rightarrow \dfrac{1}{{{2^n}}} + \dfrac{n}{{{2^n}}} < 0.1$
$ \Rightarrow \dfrac{{1 + n}}{{{2^n}}} < \dfrac{1}{{10}}$
Send the denominator of LHS to RHS and the denominator of RHS to LHS,
$ \Rightarrow 10[1 + n] < {2^n}$
Thus, this binomial equation is what will give us our needed answer,
We now need to substitute $n$ till it conforms to the above equation, i.e., until RHS is greater than LHS,
From the equation itself we can figure out that for RHS>LHS, $n \ge 7$

Option ‘B’ is correct

Note: In the given question, there is one event. We can find the probability of success and failure. So that we can apply binomial distribution. First, find the probability of failure. Then we will make an inequality to calculate the value of n.