Question

For three events A, B and C,  P(Exactly one of A or B occurs)  = P(Exactly one of B or C occurs) = P(Exactly one of C or A occurs) = `1/4` and P(All three events occur simultaneously) = `1/16`. Then the probability that at least one of the events occurs is ______.

Options

• `3/16`

• `7/32`

• `7/16`

• `7/64`

Answer 1

For three events A, B and C,  P(Exactly one of A or B occurs)  = P(Exactly one of B or C occurs) = P(Exactly one of C or A occurs) = `1/4` and P(All three events occur simultaneously) = `1/16`. Then the probability that at least one of the events occurs is `underlinebb(7/16)`

Explanation:

P(exactly one of A or B occurs)

= P(A) + P(B) – 2P(A ∩ B) = `1/4`  ...(i)

P(Exactly one of B or C occurs)

= P(B) + P(C) – 2P(B ∩ C) = `1/4`  ...(ii)

P(Exactly one of C or A occurs)

= P(C) + P(A) – 2P(C ∩ A) = `1/4`  ...(iii)

Adding (i), (ii) and (iii), we get

`2sumP(A) - 2sumP(A ∩ B)` = `3/4`

∴ `sumP(A) - sumP(A ∩ B)` = `3/8`

Now, P(A ∩ B ∩ C) = `1/16`

∴ P(A ∪ B ∪ C) = `sumP(A) - sumP(A ∩ B) + P(A ∩ B ∩ C)`

= `3/8 + 1/16`

= `7/16`