Two events A and B are defined in a sample space such that P(A) = \frac{2}{3}, P(B) = \fra… — JAMB UTME

Two events $A$ and $B$ are defined in a sample space such that $P(A) = \frac{2}{3}$ , $P(B) = \frac{1}{4}$ , and $P(A \cup B) = \frac{3}{4}$ . If $B'$ denotes the complement of event $B$ , what is the value of $P(A \cap B')$ ?

$\frac{1}{2}$ Answer
B
$\frac{1}{6}$
C
$\frac{2}{3}$
D
$\frac{1}{4}$

Answer

The probability

P(A \cap B')

\frac{1}{2}

Using the addition law of probability,

P(A \cap B) = P(A) + P(B) - P(A \cup B) = \frac{2}{3} + \frac{1}{4} - \frac{3}{4} = \frac{1}{6}

. Since

P(A) = P(A \cap B) + P(A \cap B')

, we find

P(A \cap B') = \frac{2}{3} - \frac{1}{6} = \frac{1}{2}

. Alternatively, since

P(A \cap B) = \frac{1}{6} = P(A)P(B)

, events

A

and

B

are independent, so

P(A \cap B') = P(A)P(B') = \frac{2}{3} \times \left(1 - \frac{1}{4}\right) = \frac{2}{3} \times \frac{3}{4} = \frac{1}{2}

Step-by-Step Solution

Apply the addition law of probability to calculate

P(A \cap B)

P(A \cap B) = P(A) + P(B) - P(A \cup B) = \frac{2}{3} + \frac{1}{4} - \frac{3}{4} = \frac{11}{12} - \frac{9}{12} = \frac{2}{12} = \frac{1}{6}

The general addition law relates the probabilities of the union and intersection of two compound events.

Calculate the probability that event

A

occurs while event

B

does not occur,

P(A \cap B')

P(A \cap B') = P(A) - P(A \cap B) = \frac{2}{3} - \frac{1}{6} = \frac{4}{6} - \frac{1}{6} = \frac{3}{6} = \frac{1}{2}

Event

A

can be decomposed into two mutually exclusive parts:

A \cap B

and

A \cap B'

Key Concept

Addition law of probability and complement of compound events

Estimated Time:2m 0s

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