The probability that a randomly selected calculator from a store is of brand ' r ' is proportional to r(r=1,2,.....,6). Further, the probability of a calculator of brand ‘r’ being defective is 7−r21,r=1,2,3,4,5,6. If probability that a calculator randomly selected from the store being defective is pq, when p and q are co-prime then the value of (p+q) is

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The probability that a randomly selected calculator from a store is of brand ' $r$ ' is proportional to $r (r = 1, 2, ....., 6) .$ Further, the probability of a calculator of brand ‘r’ being defective is $\frac{7 - r}{21}, r = 1, 2, 3, 4, 5, 6.$ If probability that a calculator randomly selected from the store being defective is $\frac{p}{q},$ when $p$ and $q$ are co-prime then the value of $(p + q)$ is

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answer is 71.

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Detailed Solution

Probability that calculator of brand r is selected and is defective $= \sum_{1}^{6} (k r) (\frac{7 - r}{21}) = \frac{k}{21} \sum_{r = 1}^{6} (7 r - r^{2}) = \frac{8 k}{3}_____(1)$

$\Rightarrow$ Let $E_{r}$ denote the event that calculator of brand r is selected $P (E_{r}) = k r$

Since $E_{r} = (r = 1, 2, ...., 6)$ are mutually exclusive and exhaustive events we must have $\sum_{r = 1}^{6} P (E_{r}) = 1 \Rightarrow \sum_{r = 1}^{6} k r = 1 \Rightarrow k = \frac{1}{21}$

$∴ required probability = \frac{8 k}{3} = \frac{8}{63} = \frac{p}{q}$

$∴ (p + q) = 71$

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