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If E(1) denots the envent sof coming sum...

If `E_(1)` denots the envent sof coming sum 6 in throwing two dice and `E_(2)` be the event of coming 2 in any one of two, then `P(E_(2)//E_(1))` is

A

`1//5`

B

`4//5`

C

`3//5`

D

`2//5`

Text Solution

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The correct Answer is:
D

`E_(1)` can occur as `{(1,5),(5,1),(2,4),(4,2),(3,3)}`
`E_(2)` as `{(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(6,5),(5,2),(4,2),(3,2),(1,2)}`
`thereforeP(E_(2)//E_(1))` = Probability of `E_(2)` when `E_(1)` has occurred `=2//5`
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