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One ticket is selected at random from 100 tickets numbered `00,01,02, …, 99.` Suppose A and B are the sum and product of the digit found on the ticket, respectively. Then `P((A=7)//(B=0))` is given by

A

`2//13`

B

`2//19`

C

`1//50`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
B

The sum of the digits can be 7 in the following ways: `07,16,25,34,43,52,61,70.`
`therefore(A=7)={07,16,25,34,43,52,61,70}`
Similarly, `(B=0)={00,01,02...,10,20,30,...,90}`
Thus, `(A=7)nn(B=0)={09,70}`
`thereforeP((A=7)nn(B=0))=2/100,P((B=0))=19/100`
`P(A=7)|B=0=(P((A=7)nn(B=0)))/(P(B=0))`
`=(2/100)/(19/100)=2/19`
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