Question

Given two independent events A and B such that P(A)=0.3, P(B)=0.6, Find .$P\left(AandB\right)$
$P\left(AandnotB\right)$
$P\left(AorB\right)$
$P\left(neitherAnorB\right)$

Answer 1

Given P(A) = 0.3, P (B) = 0.6.

(i): Finding P (A and B):

If A and B are independent events, P(A∩B)=P(A)P(B)P(A∩B)=P(A)P(B)

⇒⇒ P(A∩B)=P(A)P(B)P(A∩B)=P(A)P(B) = 0.3 ×× 0.6 = 0.18

(ii): Finding P (A and not B):

If A and B are independent events, A¯A¯ and B¯B¯ are also independent.

⇒P(A∩B¯)=P(A)×P(B¯)=P(A)×(1−P(B))=0.3×(1−0.6)=0.3×0.4=0.12⇒P(A∩B¯)=P(A)×P(B¯)=P(A)×(1−P(B))=0.3×(1−0.6)=0.3×0.4=0.12

(iii): Finding P (A or B):

P (A ∪∪ B) = P(A) + P(B) - P(A ∩∩ B)

⇒P(A∪B)=0.3+0.6−0.18=0.72.⇒P(A∪B)=0.3+0.6−0.18=0.72.

(iv): Finding P (neither A nor B):

P (neither A nor B) = P (A¯∩B¯A¯∩B¯) = 1 - P (A ∪∪ B) = 1 - 0.72 = 0.28.