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If barE and barF are the complementary e...

If `barE` and `barF` are the complementary events of E and F respectively and if `0 < P(F)<1`, then

A

`P(E//F)+P(overset(" "-)(E )//F)=1`

B

`P(E//F)+P(E //overset(" "-)(F))=1`

C

`P(overset(" "-)(E)//F)+P(E//overset(" "-)(F))=1`

D

`P(E//overset(" "-)(F))+P(overset(" "-)(E)//overset(" "-)(F))=1`

Text Solution

Verified by Experts

The correct Answer is:
A, B
`(a) P(E//F) + P(bar(E)//F) = (P (E nn F))/(P(F)) + (P(bar(E) nn F))/(P(F))`
` = (P(E nn F) + P( bar(E) nn F))/(P(F))`
` = (P(F))/(P(F)) = 1`
Therefore, option (a) is correct.
`(b) P(E//F) + P(E//bar(F)) = (P(E nn F))/(P(F)) + (P(E nn bar(F)))/(P(bar(F)))`
` = (P(E nn F))/(P(F)) + (P(E nn bar(F))/(1-P(F)) ne 1`
Therefore, option (b) is not correct.
`(c) P(bar(E)//F) + P(E// bar(F)) = (P(bar(E) nn F))/(P(F)) + (P(E nn bar(F)))/(P(bar(F)))`
`= (P (bar(E) nn F))/(P(F)) + (P(E nn bar(F)))/(1-P(F)) ne 1`
Therefore, option (c) is not correct.
`(d) P(E//bar(F)) + P(bar(E)//bar(F)) = (P(E nn bar(F)))/(P(bar(F))) + (P(bar(E) nn bar(F)))/(P(bar(F)))`
` = (P(bar(F)))/(P(bar(F))) = 1`
Therefore, option (d) is correct.
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