JAMB Mathematics 2013 Paper

Concept:An integer divisible by both $2$ and $3$ must be divisible by their least common multiple, which is $6$.Explanation:The integers satisfying $1 \le x \le 25$ form the sample space.There are $25$ possible integers in total, so $n(\varepsilon) = 25$.We need numbers divisible by both $2$ and $3$.A number divisible by both $2$ and $3$ is a multiple of $6$.The multiples of $6$ between $1$ and $25$ are:$$6, 12, 18, 24$$Thus, the number of favourable outcomes is $N = 4$.Probability is given by the formula:$$P = \frac{\text{favourable outcomes}}{\text{total outcomes}} = \frac{N}{n(\varepsilon)}$$Substituting the values:$$P = \frac{4}{25}$$This matches option C.Answer:$$P = \frac{4}{25}$$, so the correct option is C.