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In basketball, when Monica takes her first free throw, she h

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huda
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In basketball, when Monica takes her first free throw, she h [#permalink] New post   09 Nov 2019, 03:47
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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 didn't score on the immediately previous throw. Suppose she has to take a set of three free throws in a row, and she doesn't score on the first one. What is the probability that she scores at least once on one of the subsequent two throws?

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\(\frac{7}{16}\)
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DMilwaukee
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Re: In basketball, when Monica takes her first free throw, she h [#permalink] New post   16 Jan 2020, 05:40
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Probability of making at-least one is (1-(probability of making none))

probability that 1st and 2nd shot is missed = 3/4 * 3/4 = 9/16

1-9/16 = 7/16
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Nishan
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Re: In basketball, when Monica takes her first free throw, she h [#permalink] New post   27 Nov 2019, 08:03
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The question is poorly worded.
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Re: In basketball, when Monica takes her first free throw, she h [#permalink] New post   27 Nov 2019, 08:22
Nishan wrote:
The question is poorly worded.


Better solve it rather than judging the magoosh Very Hard Type questions.
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Re: In basketball, when Monica takes her first free throw, she h [#permalink] New post   17 Apr 2020, 07:02
DMilwaukee wrote:
Probability of making at-least one is (1-(probability of making none))

probability that 1st and 2nd shot is missed = 3/4 * 3/4 = 9/16

1-9/16 = 7/16


What about the no. of shoots she gets ? the statement says that she misses the first shot and their is a probability of 50% on that. And their are only 3 shots she gets, out of which one is a fail so there has to be one success from the next two ?
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Re: In basketball, when Monica takes her first free throw, she h [#permalink] New post   17 Apr 2020, 08:46
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Prc1995 wrote:
DMilwaukee wrote:
Probability of making at-least one is (1-(probability of making none))

probability that 1st and 2nd shot is missed = 3/4 * 3/4 = 9/16

1-9/16 = 7/16


What about the no. of shoots she gets ? the statement says that she misses the first shot and their is a probability of 50% on that. And their are only 3 shots she gets, out of which one is a fail so there has to be one success from the next two ?

let " not scored" be designated as NS
let " scored" be designated as S
we have 3 throws,
Possible outcomes are as follows
1.NS ,S, S = ( 1/2 , 1/4 ,3/4)
2.NS ,NS, S = ( 1/2 ,3/4 ,1/4)
3.NS , S , NS = (1/2 ,1/4 ,1/4)
however, we are only interested in the 2 subsequent throws
therefore,probability of the two subsequent throws = 1/4 *3/4 +3/4 *1/4 +1/4 *1/4 = 7/16
hence, ans= 7/16
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Re: In basketball, when Monica takes her first free throw, she h [#permalink] New post   17 Apr 2020, 21:31
Thank you so much...

Posted from my mobile device Image
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atharva1996
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Re: In basketball, when Monica takes her first free throw, she h [#permalink] New post   17 Apr 2020, 21:48
dare90 wrote:
Prc1995 wrote:
DMilwaukee wrote:
Probability of making at-least one is (1-(probability of making none))

probability that 1st and 2nd shot is missed = 3/4 * 3/4 = 9/16

1-9/16 = 7/16


What about the no. of shoots she gets ? the statement says that she misses the first shot and their is a probability of 50% on that. And their are only 3 shots she gets, out of which one is a fail so there has to be one success from the next two ?

let " not scored" be designated as NS
let " scored" be designated as S
we have 3 throws,
Possible outcomes are as follows
1.NS ,S, S = ( 1/2 , 1/4 ,3/4)
2.NS ,NS, S = ( 1/2 ,3/4 ,1/4)
3.NS , S , NS = (1/2 ,1/4 ,1/4)
however, we are only interested in the 2 subsequent throws
therefore,probability of the two subsequent throws = 1/4 *3/4 +3/4 *1/4 +1/4 *1/4 = 7/16
hence, ans= 7/16


Why is this combination not possible? 4. (1/2,3/4,3/4)
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Re: In basketball, when Monica takes her first free throw, she h [#permalink] New post   18 Apr 2020, 01:08
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Your combination (1/2 , 3/4 ,3/4)= NS ,NS ,NS
However, we were told to find the probability that she scores at least once in the 2 subsequent throws. Hence that combination is not possible.
The combination is only useful if you apply this formular stated below
Probability that she scored at least once in the 2 subsequent throws = 1- (probability of scoring none)
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Re: In basketball, when Monica takes her first free throw, she h [#permalink] New post   04 Nov 2021, 10:34
(and she doesn't score on the first one)
I wander ho did you get 7/16 it should be only 2/16
First shout lose so it is :0.5
second one it will be :0.25 wither it succeed or not
third one if the second one succeed will be 0.75
third one if the second one failed is 0.25
therefore we will have only two scenarios not three
1 :0.5*.25*.75
2:0.5*.25*.25
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Re: In basketball, when Monica takes her first free throw, she h [#permalink] New post   04 Nov 2021, 10:36
dare90 wrote:

2.NS ,NS, S = ( 1/2 ,3/4 ,1/4)


and she doesn't score on the first one
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Re: In basketball, when Monica takes her first free throw, she h [#permalink] New post   12 Nov 2022, 12:37
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