A and B are events such that P(A) = 0.42,
P(B) = 0.48 and P(A and B) = 0.16. Determine
(i) P (not A ) (ii) P (not B) and (iii) P (A or B)

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Solution

Here P (A) = 0.42, P(B) = 0.48 and P(A $\cap$ B ) = 0.16
(i) $P (not A) = P (¯ ¯¯ ¯ A) = 1 - P (A) = 1 - 0.42 = 0.58 (ii) P (not B) = P (¯ ¯¯ ¯ B) = 1 - P (B) = 1 - 0.48 = 0.52 We know that P (A \cup B) = P (A) + P (B) - P (A \cap B) = 0.42 + 0.48 - 0.16 = 0.74.$

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Similar questions

A and B are events such that P(A) = 0.42, P(B) = 0.48 and P(A and B) = 0.16. Determine (i) P(not A), (ii) P (not B) and (iii) P(A or B).

A

and

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are events such that

P (A) = 0.42, P (B) = 0.48

and P(A and B)

= 0.16

Determine (i) P(not A), (ii) P(not B) and (iii) P(A or B)

Q. A and B are two events such that P(A) = 0.54, P(B) = 0.69 and P(A ∩ B) = 0.35. Find (i) P(A ∩ B) (ii) P(A′ ∩ B′) (iii) P(A ∩ B′) (iv) P(B ∩ A′)

A and B are two events such that P(A) = 0.54, P(B) = 0.69 and P(A \cap B) = 0.35.
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Q. Given two independent events A and B such that P (A) = 0.3, P (B) = 0.6. Find (i) P (A and B) (ii) P (A and not B) (iii) P (A or B) (iv) P (neither A nor B)

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A and B are events such that P(A) = 0.42, P(B) = 0.48 and P(A and B) = 0.16. Determine (i) P (not A ) (ii) P (not B) and (iii) P (A or B)

Here P (A) = 0.42, P(B) = 0.48 and P(A ∩ B ) = 0.16 (i) P (not A)=P(¯¯¯¯A)=1−P(A)=1−0.42=0.58(ii) P (not B) = P(¯¯¯¯B)=1−P(B)=1−0.48=0.52We know thatP(A∪B)=P(A)+P(B)−P(A∩B)=0.42+0.48−0.16=0.74.

A and B are events such that P(A) = 0.42,
P(B) = 0.48 and P(A and B) = 0.16. Determine
(i) P (not A ) (ii) P (not B) and (iii) P (A or B)

Here P (A) = 0.42, P(B) = 0.48 and P(A $\cap$ B ) = 0.16
(i) $P (not A) = P (¯ ¯¯ ¯ A) = 1 - P (A) = 1 - 0.42 = 0.58 (ii) P (not B) = P (¯ ¯¯ ¯ B) = 1 - P (B) = 1 - 0.48 = 0.52 We know that P (A \cup B) = P (A) + P (B) - P (A \cap B) = 0.42 + 0.48 - 0.16 = 0.74.$