The probability that randomly selected calculator from a store is of brand r is proportional to r, r=1,2,..,6. Further, the probability of a calucltor of brand r being defective is `(7-r )/(21), r=1,2,..,6`. Then the probability that a calculator randomly selected from the store being defective is

Options

A. `(8)/(63)`
B. `(13)/(63)`
C. `(55)/(63)`
D. `(50)/(63)`

Answer

A

Explanation

Let `C_(r )` denote the event that a calculator of brand r is selected and `D_(r )` denote the event that a calculate of brand r is defective. Then,
`P(C_(r ))=lamda r " and " P(D_(r )//C_(r ))=(7-r)/(21), r=1, 2,..,6`.
Now, `underset(r=1)overset(6)(sum)P(C_(r )=1`
`implies underset(r=1)overset(6)(sum) lamda r=1 implies lamda(underset(r=1)overset(6)(sum) r)=1 implies 21 lamda =1 implies lamda =(1)/(21)`
Required probability `= underset(r=1)overset(6)(sum) P(C_(r ) cap D_(r ))`
`= underset(r=1)overset(6)(sum) P(C_(r ))P(D_(r )//C_(r ))`
`= underset(r=1)overset(6)(sum) lamda r xx(7-r)/(21)`
`=(lamda)/(21) underset(r=1)overset(6)(sum) (7r - r^(2))`
`=(1)/(21^(2)){ underset( r=1)overset(6)(sum) 7r- underset(r=1)overset(6)(sum) r^(2)}`
`=(1)/(21^(2)){7xx21-(6xx7xx13)/(6)}=(8)/(63)`