In basketball when Monica takes her first free throw, she has a 50% chance of sc
ID: 3259262 • Letter: I
Question
In basketball when Monica takes her first free throw, she has a 50% chance of scoring. If she takes an additional throw, she has a 75% chance of scoring on this throw if she scored on the immediately previous throw , but only a 25% chance of scoring on this throw if she did not score on the immediately previous throw. Suppose she has to take a set of three free throws in a row, and she does not score on the first one, what is the probability that she scores at least once on one of the subsequent two throws?
Enter the answer as a fraction in the simplest form.
Explanation / Answer
The probability that she scores at least once on one of the subsequent two throws = 1 - probability that she does not score on both of the subsequent two throws.
Probability that she does not score on both of the subsequent two throws:
1st throw = no score
2nd throw = no score
3rd throw = no score
0.5 * 0.75 * 0.75 = 0.28125
The probability that she scores at least once on one of the subsequent two throws = 1 - 0.28125 = 0.71875 = 71875/100000 = 575/800 = 23/32
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