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WAEC Further Mathematics 2023 Paper

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Question : 33

Total: 40

If

X

Y

are two independent events such that

P (X) = \;\frac{1}{8}

P (X \cup Y) = \;\frac{5}{8}

P (Y)

.

$\;\frac{1}{6}$
$\;\frac{4}{7}$
$\;\frac{4}{21}$
$\;\frac{3}{7}$

Solution:

Concept:

For independent events,

P(X \cap Y) = P(X) \cdot P(Y)

.

So the union formula becomes

P(X \cup Y) = P(X) + P(Y) - P(X)P(Y)

.

Explanation:

Given

P(X) = \frac{1}{8}

and

P(X \cup Y) = \frac{5}{8}

.

Substitute these values into the union formula:

\frac{5}{8} = \frac{1}{8} + P(Y) - \frac{1}{8}P(Y)

.

Group the terms containing

P(Y)

:

\frac{5}{8} = \frac{1}{8} + P(Y)\left(1 - \frac{1}{8}\right)

.

So

\frac{5}{8} - \frac{1}{8} = P(Y)\left(\frac{7}{8}\right)

.

This gives

\frac{4}{8} = \frac{7}{8}P(Y)

.

Divide both sides by

\frac{7}{8}

:

P(Y) = \frac{4}{8} \times \frac{8}{7} = \frac{4}{7}

.

Answer:

P(Y) = \frac{4}{7}

, which is option B.

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