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In a bombing attack there is 50% chance that a bomb will hit the target Atleast two independent hits are required to destroy the target completely. Then the minimum no of bombs that must be dropped to ensure that there is atleast 99% chance of completely destroying the target is ..
Option: 1 9
Option: 2 10
Option: 3 11
Option: 4 12

Answers (1)

best_answer

The probability of success in one strike is p = 1/2

probability of failure = q = 1/2

Now, the probability of r success 

\mathrm{P}(\mathrm{x}=\mathrm{r})={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \cdot\left(\frac{1}{2}\right)^{\mathrm{r}} \cdot\left(\frac{1}{2}\right)^{\mathrm{n}-\mathrm{r}}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\left(\frac{1}{2}\right)^{\mathrm{n}}

From the given question \mathrm{P}(\mathrm{x} \geq 2) \geq 0.99

\begin{array}{l} \Rightarrow 1-\mathrm{P}(\mathrm{x}<2) \geq 0.99 \\ \Rightarrow 1-\mathrm{P}(\mathrm{x}=0)-\mathrm{P}(\mathrm{x}=1) \geq 0.99 \\ \Rightarrow 1-0.99 \geq \frac{1+\mathrm{n}}{2^{\mathrm{n}}} \\ \Rightarrow 2^{\mathrm{n}} \geq 100+100 \mathrm{n} \\ \text { For } \mathrm{n}=10,2^{\mathrm{n}}<100+100 \mathrm{n} \\ \text { For } \mathrm{n}=11,12,13 \\ 2^{\mathrm{n}}>100+100 \mathrm{n} \end{array}

n = 11 is the least number.

Posted by

himanshu.meshram

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