Question

A certain type of missile hits the target with a probability of $p=0.3$. What is the least number of missiles that should be fired so that there is at least $80\%$ probability that the target is hit?

• A. $5$
• B. $6$
• C. $7$
• D. None of these

Answer 1

The correct option is A

$5$

The explanation for the correct option:

Finding the least number of missiles :

Given, that a certain type of missile hits the target with a probability of $p=0.3=\frac{3}{10}$

then, the missile does not the target with a probability of $\overline{p}=1-\frac{3}{10}=\frac{7}{10}$

Let $n$be the number of times that the missile is fired.

Given condition is $n$the least number of missiles that should be fired so that there is at least an 80% probability that the target is hit.

$$\begin{gathered} \Rightarrow \frac{3}{10}+\left(\frac{7}{10}\right)\times \left(\frac{3}{10}\right)+\left(\frac{7}{10}\right)\times \left(\frac{7}{10}\right)\times \left(\frac{3}{10}\right)+.........+\left(\frac{3}{10}\right)\times {\left(\frac{7}{10}\right)}^{n-1}\ge \frac{80}{100} \\ \Rightarrow \left(\frac{3}{10}\right)\left[1+\left(\frac{7}{10}\right)+{\left(\frac{7}{10}\right)}^2+........+{\left(\frac{7}{10}\right)}^{n-1}\right]\ge 0.8 \\ \Rightarrow \left(\frac{3}{10}\right)\left[\frac{1-{\left(\frac{7}{10}\right)}^n}{1-\left(\frac{7}{10}\right)}\right]\ge 0.8\left[\because G.P\mathrm{sum}=\frac{a(r^n-1)}{r-1}\right] \\ \Rightarrow \left(\frac{3}{10}\right)\left[\frac{1-{\left(\frac{7}{10}\right)}^n}{{\displaystyle \frac{3}{10}}}\right]\ge 0.8 \\ \Rightarrow \left[1-{\left(\frac{7}{10}\right)}^n\right]\ge 0.8 \\ \Rightarrow 0.2\ge 0.7^n.........\left(i\right) \end{gathered}$$

Now, put $n=5$ in equation (i).

$$\begin{gathered} \Rightarrow 0.2\ge {(0.7)}^5 \\ \Rightarrow 0.2\ge 0.168 \end{gathered}$$

So, this inequality is hold for $n=5$.

Thus $n=5$ is the least number.

Hence the correct option is A.