I do not understand how this works for mutually exclusive events for probabilty can you help explain it.. Example: P(A) = 0⋅3, P(B) = 0⋅9 and P(A′∩ B′) = 0⋅1.
Prove that A and B are mutually exclusive.
Solution: A′∩ B′ is shaded in the diagram
⇒ P(A′∩ B′) = 1 – P(A ∪ B)
⇒ P(A ∪ B) = 1 – 0⋅1 = 0⋅9
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
⇒ 0.9 = 0⋅3 + 0⋅6 – P(A ∩ B)
⇒ P(A ∩ B) = 0
⇒ A and B are mutually exclusive.
Maths Statistics Watch
- Thread Starter
- 10-01-2015 15:22
- 11-01-2015 10:48
For two events to be mutually exclusive, they can't both happen. So the probability of both occurring has to be zero, and this is what you are trying to prove, that P(A ∩ B) = 0.
To do this, you can use the addition rule: P(A ∪ B) = P(A) + P(B) – P(A ∩ B).
So to show P(A ∩ B) = 0, you need the values of P(A), P(B), and P(A ∪ B). To work out P(A ∪ B), it would be best to draw a Venn diagram of A and B intersecting. Then shade P(A′∩ B′), and it should be fairly obvious to you why P(A′∩ B′) = 1 – P(A ∪ B)