Question

Shots are being fired independently from a gun at a target with a maximum score of 10 points per shot. Given that the probability of scoring 30 points in 3 shots is $\frac{1}{64}$, probability of scoring eight points in one shot is $\frac{1}{5}$ and less than eight points in one shot is $\frac{2}{5}$. If the probability of scoring at least 28 points in 3 shots is $\frac{ p }{1600}$, then find the value of $p$.

Answer 1

$P($ atleast 28$)= P ((10,10,10)$ or $(10,10,9)$ or $(10,10,8)$ or $(10,9,9))$
Now $P(30)=\frac{1}{64}$ (Given) i.e. $P(10$ and 10 and 10$)=\frac{1}{64}$. Hence $(P(10))^3=\frac{1}{64} \Rightarrow P(10)=\frac{1}{4}$
$\therefore P (10$ in one shot $)=\frac{1}{4}$
$P (8$ points in one shot $)=\frac{1}{5}$ and $P ($ less than 8 points $)=\frac{2}{5}$
$\therefore P (\geq 8$ points $)=1-\frac{2}{5}=\frac{3}{5}$ or $P (8$ or 9 or 10$)=\frac{3}{5}$
Hence $P (8)+ P (9)+ P (10)=\frac{3}{5}$
$\frac{1}{5}+ P (9)+\frac{1}{4}=\frac{3}{5} \Rightarrow P (9)=\frac{3}{20}$
$P ($ atleast 28 in 3 shots $)= P (10,10,10)+3 P (10,10,9)+3 P (10,10,8)+3 P (9,9,10)$
$=\frac{1}{64}+3 \cdot \frac{1}{4} \cdot \frac{1}{4} \cdot \frac{3}{20}+3 \cdot \frac{1}{4} \cdot \frac{1}{4} \cdot \frac{1}{5}+3 \cdot \frac{3}{20} \cdot \frac{3}{20} \cdot \frac{1}{4}=\frac{1}{64}+\frac{9}{320}+\frac{3}{80}+\frac{27}{1600}$
$=\frac{25}{1600}+\frac{45}{1600}+\frac{60}{1600}+\frac{27}{1600}=\frac{157}{1600}$
Hence $p=157$