Since each free throw is an independent event, we can calculate the probability of John making all three free throws by multiplying the individual probabilities together.
Given that John is a 70% free throw shooter, the probability of him making one free throw is 0.7. Therefore, the probability of him missing a free throw is 0.3.
To find the probability of him making all three free throws, we multiply the probabilities together:
0.7 * 0.7 * 0.7 = 0.343
Therefore, there is a 34.3% chance that John makes all three free throws and wins the game.
John Hass to make all three free throws to win the game. Assume each free-throw is an independent event, and that the pressure of the situation has no impact on him. If John is a 70% free throw shooter, what is the percent chance that John makes all three free throws?
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