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JAMB Mathematics 2024 Paper

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Question : 38

Total: 50

A bag contains 7 red and 4 black identical balls. Two balls were picked at random from the bag and replaced each time. Find the probability the two balls were of same colour.

$\frac{54}{121}$
$\frac{65}{121}$
$\frac{29}{55}$
$\frac{27}{55}$

Solution:

Concept:

Probability of same colour means picking red twice or black twice, with replacement.

Explanation:

Total balls

= 7 + 4 = 11

.

Since each ball is replaced, the probability remains the same for both picks.

P(\text{red}) = \frac{7}{11}

, so

P(\text{red, red}) = \frac{7}{11} \times \frac{7}{11} = \frac{49}{121}

.

P(\text{black}) = \frac{4}{11}

, so

P(\text{black, black}) = \frac{4}{11} \times \frac{4}{11} = \frac{16}{121}

.

Required probability

= \frac{49}{121} + \frac{16}{121} = \frac{65}{121}

.

Answer:

\frac{65}{121}

(Option B).

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