waecmaths question:
If a number is chosen at random from the set $\{x:4\le x\le 15\}$ , find the probability that it is a multiple of 3 or a multiple of 4
Option A:
$\tfrac{1}{12}$
Option B:
$\tfrac{5}{12}$
Option C:
$\tfrac{1}{2}$
Option D:
$\tfrac{11}{12}$
waecmaths solution:
$\begin{align} & \{x:4\le x\le 15\} \\ & x=\{4,5,6,7,8,9,10,11,12,13,14,15\} \\ & \Pr (Multiple\text{ }of\text{ }3)=\frac{4}{12} \\ & \Pr (Multiple\text{ }of\text{ 4})=\frac{3}{12} \\ & \Pr (Multiple\text{ of }3\text{ }or\text{ }4)=\frac{4}{12}+\frac{3}{12}=\frac{7}{12} \\ & \Pr (Multiple\text{ of }3\text{ and }4)=\frac{4}{12}\times \frac{3}{12}=\frac{1}{12} \\ & \Pr (multiple\text{ of 3) or }(multiple\text{ }of\text{ }4)=\frac{7}{12}-\frac{1}{12}=\frac{1}{2} \\\end{align}$
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