Question

The probability that a randomly selected calculator from a store is of brand ' $r$ ' is proportional to $r ( r =1,2,3 \ldots 6)$. Further, the probability of a calculator of brand ' $r$ ' being defective is $\frac{7- r }{21}$, $r=1,2, \ldots 6$. 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

Probability that calculator of brand $r$ is selected and is defective
$=\displaystyle\sum_1^6(k r)\left(\frac{7-r}{21}\right)=\frac{k}{21} \sum_{r=1}^6\left(7 r-r^2\right)=\frac{8 k}{3} \ldots$
$\Rightarrow $ Let $E _{ r }$ denote the event that calculator of brand $r$ is selected.
$P \left( E _{ r }\right)= kr$
Since $E _{ r }( r =1,2, \ldots, 6)$ are mutually exclusive and exhaustive events we must have
$\displaystyle\sum_{ r =1}^6 P \left( E _{ r }\right)=1 \Rightarrow \displaystyle\sum_{ r =1}^6 kr =1 \Rightarrow k =\frac{1}{21}$
$\therefore$ Required probability $=\frac{8 k }{3}=\frac{8}{63}=\frac{ p }{ q }$
So, $p=8$ and $q=63$.
Hence, $(p+q)=71$