Two independent events A and B in a sample space satisfy P(A') = \frac{2}{5} and P(A \cup … — JAMB UTME

Two independent events $A$ and $B$ in a sample space satisfy $P(A') = \frac{2}{5}$ and $P(A \cup B) = \frac{7}{10}$ , where $A'$ denotes the complement of event $A$ . What is the probability that exactly one of the two events occurs?

$\frac{11}{20}$ Answer
B
$\frac{7}{10}$
C
$\frac{17}{20}$
D
$\frac{3}{20}$

Answer

\frac{11}{20}

First, find

P(A) = 1 - \frac{2}{5} = \frac{3}{5}

. Since

A

and

B

are independent,

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

. Using the addition law

P(A \cup B) = P(A) + P(B) - P(A)P(B)

, we substitute the values to get

\frac{7}{10} = \frac{3}{5} + P(B)\left(1 - \frac{3}{5}\right)

, which gives

P(B) = \frac{1}{4}

. The probability of exactly one event occurring is

P(A)P(B') + P(A')P(B) = \frac{3}{5} \times \frac{3}{4} + \frac{2}{5} \times \frac{1}{4} = \frac{11}{20}

Step-by-Step Solution

Calculate the probability of event A

P(A) = 1 - P(A') = 1 - \frac{2}{5} = \frac{3}{5}

The sum of the probabilities of an event and its complement equals 1.

Apply the addition law for independent events to find P(B)

P(A \cup B) = P(A) + P(B) - P(A)P(B) \implies \frac{7}{10} = \frac{3}{5} + P(B)\left(1 - \frac{3}{5}\right) \implies P(B) = \frac{1}{4}

For independent events, the intersection probability is

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

Calculate the complement of event B

P(B') = 1 - P(B) = 1 - \frac{1}{4} = \frac{3}{4}

The complement of event B represents the event that B does not occur.

Compute the probability that exactly one event occurs

P(\text{exactly one}) = P(A \cap B') + P(A' \cap B) = P(A)P(B') + P(A')P(B) = \left(\frac{3}{5} \times \frac{3}{4}\right) + \left(\frac{2}{5} \times \frac{1}{4}\right) = \frac{9}{20} + \frac{2}{20} = \frac{11}{20}

The occurrence of exactly one event means either A occurs and B does not, or A does not occur and B occurs.

Key Concept

Compound Probability and Probability Laws for Independent Events

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