The probability of simultaneous occurrence of at least one of two events A and B is p. If the probability that exactly one of A, B occurs is q, then P (A′) + P (B′) = ?
The probability of simultaneous occurrence of at least one of two events A and B is p. If the probability that exactly one of A, B occurs is q, then P (A′) + P (B′) = ? Correct Answer <span class="fontstyle0">2 - 2</span><span class="fontstyle2">p </span><span class="fontstyle0">+ </span><span class="fontstyle2">q</span><span class="fontstyle0">.</span>
Concept:
- P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
- P(A') = 1 - P(A)
Calculation:
Given: Probability of simultaneous occurrence of at least one of two events A and B is p
So, P(A ∪ B) = p.
Since probability that exactly one of A, B occurs = q (given), we get:
P(A ∪ B) - P(A ∩ B) = q
⇒ p - P(A ∩ B) = q
⇒ P(A ∩ B) = p - q
As we know, P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
⇒ p = P(A) + P(B) - (p - q)
⇒ P(A) + P(B) = p + (p - q)
⇒ P(A) + P(B) = 2p - q
Now, P (A′) + P (B′) = 1 - P(A) + 1 - P(B)
= 2 -
= 2 - (2p - q)
= 2 - 2p + q.