If A and B are mutually exclusive events, P (A) = 0.35 and P (B) = 0.45, find
(a) P (A′)
(b) P (B′)
(c) P (A ∪ B)
(d) P (A ∩ B)
(e) P (A ∩ B′)
(f) P (A′∩ B′)
Given that P (A) = 0.35 and P (B) = 0.45
∵ the events A and B are mutually exclusive then P(A ⋂ B) = 0
To find: (a) P (A′)
We know that,
P (A) + P (A’) = 1
⇒ 0.35 + P(A’) = 1 [given]
⇒ P(A’) = 1 – 0.35
⇒ P(A’) = 0.65
To find: (b) P (B′)
We know that,
P (B) + P (B’) = 1
⇒ 0.45 + P(B’) = 1
⇒ P(B’) = 1 – 0.45
⇒ P(B’) = 0.55
To find: (c) P (A ⋃ B)
We know that,
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
⇒ P (A ⋃ B) = 0.35 + 0.45 – 0 [given]
⇒ P (A ⋃ B) = 0.80
To find: (d) P (A ⋂ B)
It is given that A and B are mutually exclusive events.
∴ P (A ⋂ B) = 0
To find: (e) P (A ⋂ B’)
P (A ⋂ B’) = P (A) – P (A ⋂ B)
= 0.35 – 0
= 0.35
To find: (f) P (A’ ⋂ B’)
P (A’ ⋂ B’) = P (A ⋃ B)’
= 1 – P (A ⋃ B)
= 1 – 0.8 [from part (c)]
= 0.2
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