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On the Line (Posted on 2003-08-15) |
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A basketball player who shoots 80% from the free-throw line goes to the charity stripe with just 1.7 seconds remaining in a basketball game.
He has two shots, and his team is trailing by two points.
What is the probability that he will make only one of the two, and why?
(Assume that he is trying, of course, for his team to win the game.)
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Submitted by DJ
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Rating: 4.0000 (12 votes)
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Solution:
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The probability that he will make only one shot is slightly more than 4/25.
The probability of making only one shot is the sum of the probability that he makes the first shot added to the probability of making the second shot.
There is a 4/5 (80%) chance that he makes the first shot to tie the game. Then, there is a 1/5 chance that he misses the second shot, for a joint probability of 4/25.
If, however, he misses the first shot, he will have one free throw with his team down by two. Even if he makes it, they trail by one, the other team gets the ball, and there are only 1.7 seconds left in the game, hardly enough time to force a turnover and score.
In this scenario, the player would intentionally miss the shot, so that one of his teammates could possibly get the rebound and put the ball back in, for two points to tie the game, not enough for a turnaround.
Therefore, there is only the slightest chance that he misses the first shot and makes the second (since he could accidentally put the ball in while trying to give a teammate the rebound).
Therefore, the overall probability of making only one shot is 4/25, plus the small chance that he accidentally makes the second shot, for a probability of a little more than 4/25 (16%). |
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