Question

The probability that a randomly selected calculator from a store is of brand ' $r$ ' is proportional to $r(r=1,2,3\dots 6)$. Further, the probability of a calculator of brand ' $r$ ' being defective is $\frac{7-r}{21}$, $r=1,2,\dots 6$. If probability that a calculator randomly selected from the store being defective is $\frac{p}{q}$, where $p$ and $q$ are co-prime then find the value of $(p+q)$.

Answer 1

Answer: 0071

Probability that calculator of brand $r$ is selected and is defective
$=\sum _1^6(kr)\left(\frac{7-r}{21}\right)=\frac{k}{21}\sum _{r=1}^6\left(7r-r^2\right)=\frac{8k}{3}\dots$
$\Rightarrow$ Let $E_r$ denote the event that calculator of brand $r$ is selected.
$P\left(E_r\right)=kr$
Since $E_r(r=1,2,\dots ,6)$ are mutually exclusive and exhaustive events we must have
$\sum _{r=1}^{6}P\left(E_r\right)=1\Rightarrow \sum _{r=1}^{6}kr=1\Rightarrow k=\frac{1}{21}$
$\therefore$ Required probability $=\frac{8k}{3}=\frac{8}{63}=\frac{p}{q}$
So, $p=8$ and $q=63$.
Hence, $(p+q)=71$