The probability that a randomly selected calculator from a store is of brand r is proportional to r r - 1- 2- 3 - 6- Further- the probability of a calculator of brand r being defective is 7-r21- r-1-2-6- If probability that a calculator randomly selected from the store being defective is pq- where p and q are co-prime- then the value of p - q is

Probability that a calculator randomly selected from the store being defective is =∑_r=1^6(kr) ( 7 r21 ) =k21 ∑ _r=1^6(7r r^2) =8k3 Let E_r nbsp;denote the ...

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Standard XII

Mathematics

Bayes' Theorem

The probabili...

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

71.00

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Solution

Probability that a calculator randomly selected from the store being defective is
$= 6 \sum r = 1 (k r) (\frac{7 - r}{21}) = \frac{k}{21} 6 \sum r = 1 (7 r - r^{2}) = \frac{8 k}{3}$
Let $E_{r}$ denote the event that calculator of brand $r$ is selected.
$P (E_{r}) \propto r$
$\Rightarrow P (E_{r}) = k r$
Since $E_{r} (r = 1, 2, . . ., 6)$ are mutually exclusive and exhaustive events, so we have
$6 \sum r = 1 P (E_{r}) = 1 \Rightarrow 6 \sum r = 1 k r = 1 \Rightarrow k = \frac{1}{21}$
Required probability $= \frac{8 k}{3} = \frac{8}{63} = \frac{p}{q}$
So, $p = 8$ and $q = 63.$
Hence, $(p + q) = 71$

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