Given that P (A) = 0.35 and P (B) = 0.45
∵ The events A and B are mutually exclusive then P (A ⋂ B) = 0
(a) 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
(b) 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
(c) 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
(d) To find (d) P (A ⋂ B)
It is given that A and B are mutually exclusive events.
∴ P (A ⋂ B) = 0
(e) To find (e) P (A ⋂ B’)
P (A ⋂ B’) = P (A) – P (A ⋂ B)
= 0.35 – 0
= 0.35
(f) To find (f) P (A’ ⋂ B’)
P (A’ ⋂ B’) = P (A ⋃ B)’
= 1 – P (A ⋃ B)
= 1 – 0.8 [from part (c)]
= 0.2
