Question

If E and F are two events such that P(E) > 0 and P(F) ≠ 1, then `P(barE//barF)` is ______.

Options

• `(P(barE))/(P(barF))`

• `1 - P(barE//F)`

• 1 − P(E/F)

• `(1 - P(E ∪ F))/(P(barF)`

Answer 1

If E and F are two events such that P(E) > 0 and P(F) ≠ 1, then `P(barE//barF)` is `bbunderline((1 - P(E ∪ F))/(P(barF))`.

Explanation:

`P(barE//barF) = (P(barE ∩ barF))/(P(barF))     ...["Using De-morgan’s law" (A ∪ B)' = A' ∩ B']`

= `(P(E ∪ F))/(P(barF)`

= `(1 - P(E ∪ F))/(P(barF)`