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?
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%.
To find the probability that a young Canadian plays hockey or soccer, we can add the probabilities of each event:
P(Hockey or Soccer) = P(Hockey) + P(Soccer)
Given that the probability of playing soccer in the summer (P(Soccer)) is 0.57 and the probability of playing hockey in the winter (P(Hockey)) is 0.83, we have:
P(Hockey or Soccer) = 0.83 + 0.57
P(Hockey or Soccer) = 1.40
However, probabilities cannot be greater than 1. Therefore, the maximum probability of playing hockey or soccer is 1.
Therefore, the probability that a young Canadian plays hockey or soccer is 1.