Jambmaths question:
$\begin{align} & \text{What is the probability that an integer }x\text{ (1}\le \text{x}\le \text{25) chosen at random is divisible by both 2 and 3} \\ & \text{(A) }\tfrac{3}{4}\text{ (B) }\tfrac{1}{25}\text{ (C) }\tfrac{1}{5}\text{ (D) }\tfrac{4}{25} \\\end{align}$
Jamb Maths Solution:
$\begin{align} & \text{The total possible outcomes is 25} \\ & \text{Numbers divisible by both 2 and 3 are 6, 12, 18, 24} \\ & \text{Therefore number of expected outcomes is 4} \\ & \text{Probability }=\frac{\text{Number of expected outcomes}}{\text{Number of total possible outcomes}} \\ & \text{Probability of number both divisible by 2 and 3 is} \\ & =\frac{4}{25}=\frac{4}{25} \\\end{align}$
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