The probability that a young Canadian plays soccer in the summer is 0.57. The probability that a young Canadian plays hockey in the winter is 0.83. Assuming that these are independent events, if a young Canadian is chosen at random, what is the probability that they play hockey or soccer?

Answer 1

To find the probability that a young Canadian plays hockey or soccer, we can use the formula for the probability of the union of two independent events:

P(hockey or soccer) = P(hockey) + P(soccer) - P(hockey and soccer)

Since the events of playing hockey and playing soccer are independent, P(hockey and soccer) = P(hockey) * P(soccer)

Using the given probabilities, we can substitute them into the formula:

P(hockey or soccer) = 0.83 + 0.57 - (0.83 * 0.57)

Simplifying the expression, we get:

P(hockey or soccer) = 0.83 + 0.57 - 0.4731

P(hockey or soccer) = 0.9269

Therefore, the probability that a young Canadian plays hockey or soccer is approximately 0.927 or 92.7%.