CEMC Pascal, 2020

Algebra Difficulty 1.6 Junior Find the answer Canada

Ben participates in a prize draw. He receives one prize that is equally likely to be worth $5, $10 or $20. Jamie participates in a different prize draw. She receives one prize that is equally likely to be worth $30 or $40. What is the probability that the total value of their prizes is exactly $50?

$\frac{1}{6}$
$\frac{1}{3}$
$\frac{1}{2}$
$\frac{2}{5}$
$\frac{2}{3}$

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Since there are two possible prizes that Jamie can win and each is equally likely, then the probability that Jamie wins $30 is $\frac{1}{2}$ and the probability that Jamie wins $40 is $\frac{1}{2}$ .

If Jamie wins $30, then for the total value of the prizes to be $50, Ben must win $20. The probability that Ben wins $20 is $\frac{1}{3}$ , since there are three equally likely outcomes for Ben.

If Jamie wins $40, then for the total value of the prizes to be $50, Ben must win $10. The probability that Ben wins $10 is $\frac{1}{3}$ .

Since Ben’s and Jamie’s prizes come from different draws, we can assume that the results are independent, and so the probability that Jamie wins $30 and Ben wins $20 is $\frac{1}{2} \times \frac{1}{3} = \frac{1}{6}$ .

Similarly, the probability that Jamie wins $40 and Ben wins $10 is $\frac{1}{2} \times \frac{1}{3} = \frac{1}{6}$ .

Therefore, the probability that the total value of their prizes is $50 is $\frac{1}{6} + \frac{1}{6} = \frac{1}{3}$ .

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