Two independent events A and B have probabilities P(A) = \frac{1}{4} and P(B) = \frac{2}{3… — JAMB UTME

Two independent events $A$ and $B$ have probabilities $P(A) = \frac{1}{4}$ and $P(B) = \frac{2}{3}$ . What is the probability that at least one of the events occurs, represented by $P(A \cup B)$ ?

$\frac{3}{4}$ Answer
B
$\frac{11}{12}$
C
$\frac{1}{6}$
D
$\frac{5}{6}$

Answer

The probability that at least one of the events occurs is

\frac{3}{4}

The correct answer

\frac{3}{4}

is obtained by applying the multiplication law for independent events

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

, and substituting into the general addition law

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

Step-by-Step Solution

Calculate the intersection probability

P(A \cap B)

for independent events

P(A \cap B) = P(A) \times P(B) = \frac{1}{4} \times \frac{2}{3} = \frac{2}{12} = \frac{1}{6}

For independent events, the probability of both occurring together is the product of their individual probabilities.

Apply the general addition law of probability

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

The addition law accounts for double-counting the intersection when finding the union of two events.

Substitute the values and simplify the fraction

P(A \cup B) = \frac{1}{4} + \frac{2}{3} - \frac{1}{6} = \frac{3}{12} + \frac{8}{12} - \frac{2}{12} = \frac{9}{12} = \frac{3}{4}

Finding a common denominator of 12 allows exact fractional addition and simplification.

Key Concept

Probability Laws for Independent and Compound Events

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