Question

A boy has 20% chance of hitting at a target. Let p denote the probability of hitting the target for the first time at the nth trial. If $p$ satisfies the inequality $625p^2-175p+12<0$ then value of $n$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is C $3$

Let p(n)=Hitting for the 1st time at the n-th trial

$⟹p\left(n\right)={0.8}^{n-1}\times 0.2$

The boundary of the inequality $62p^2-175p+12=0$

is the solution of a quadratic equation in p :

or, ${175}^2-4\times 12\times 625=625={25}^2$

$⟹p=\frac{175\pm 25}{1250}=\frac{3}{25}$ or $\frac{4}{25}$

So p(n) is negative between those two values.

$p\left(n\right)=\frac{3}{25}={0.8}^{n-1}\times 0.2⟹\frac{3}{5}={0.8}^{n-1}⟹\log \left(\frac{3}{5}\right)=(n-1)\log \left(0.8\right)⟹n=1+\frac{\log \left(\frac{3}{5}\right)}{\log \left(0.8\right)}=3.289$

$p\left(n\right)=\frac{4}{25}={0.8}^{n-1}\times 0.2⟹\frac{4}{5}={0.8}^{n-1}⟹\log \left(\frac{4}{5}\right)=(n-1)\log \left(0.8\right)⟹n=1+\frac{\log \left(\frac{4}{5}\right)}{\log \left(0.8\right)}=2$

Hence, $2<n<3.289⟹n=3$