Q.56 If A and B are events such that 𝑃𝑃(𝐴𝐴) = 0.3, 𝑃𝑃(𝐵𝐵) = 0.2 and 𝑃𝑃(𝐴𝐴 ∪ 𝐵𝐵) = 0.45, the value
of 𝑃𝑃(𝐴𝐴 ∩ 𝐵𝐵�) is ___________.
The value of P(A ∩ B̄) is 0.05.
Question Statement
If A and B are events such that:
- P(A) = 0.3
- P(B) = 0.2
- P(A ∪ B) = 0.45
Find P(A ∩ B̄).
Step 1: Use Union Formula to Find P(A ∩ B)
For any two events A and B:
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Rearrange to get the intersection:
P(A ∩ B) = P(A) + P(B) − P(A ∪ B)
Substitute the given values:
P(A ∩ B) = 0.3 + 0.2 − 0.45 = 0.5 − 0.45 = 0.05
So, P(A ∩ B) = 0.05.
Step 2: Express P(A) in Terms of Disjoint Parts
Event A can occur in two mutually exclusive ways:
- A occurs together with B: A ∩ B
- A occurs without B: A ∩ B̄
Hence,
P(A) = P(A ∩ B) + P(A ∩ B̄)
We know P(A) = 0.3 and P(A ∩ B) = 0.05.
Therefore:
P(A ∩ B̄) = P(A) − P(A ∩ B) = 0.3 − 0.25 = 0.05
Thus, P(A ∩ B̄) = 0.05.
Why This Method Works
- The union formula P(A ∪ B) = P(A) + P(B) − P(A ∩ B) adds the probabilities of A and B and subtracts the overlap so the common part is not counted twice.
- The event A is partitioned into “A with B” and “A without B”, which are disjoint, so their probabilities add to P(A).
- The probability of A intersection not B is the part of A that remains after subtracting the overlap with B.