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Last year, Mike, Ben and Rob participated in a chess tournament that

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Last year, Mike, Ben and Rob participated in a chess tournament that  [#permalink]

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19 Sep 2019, 03:06
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Last year, Mike, Ben and Rob participated in a chess tournament that crowned 1 champion. If their respective probabilities of being crowned champion were 1/4, 1/3 and 1/6, then what's the probability that either Mike or Ben won?

A. 1/6
B. 5/12
C. 7/12
D. 5/6
E. 1

If the probability that Mike can win a championship is 1/4, and that of Rob winning is 1/3 and that of Ben winning is 1/6, what is the probability that Mike or Ben will win the championship but not Ben?

a) 1/6
b) 5/12
c) 7/12
d) 5/6
e) 1

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Re: Last year, Mike, Ben and Rob participated in a chess tournament that  [#permalink]

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19 Sep 2019, 03:59
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If there can be only one champion and respective probabilities of Mike and Ben being crowned champion were 1/4 and 1/3 respectively.

probability that either Mike or Ben won= 1/3+1/4=7/12

Bunuel wrote:
Last year, Mike, Ben and Rob participated in a chess tournament that crowned 1 champion. If their respective probabilities of being crowned champion were 1/4, 1/3 and 1/6, then what's the probability that either Mike or Ben won?

A. 1/6
B. 5/12
C. 7/12
D. 5/6
E. 1

If the probability that Mike can win a championship is 1/4, and that of Rob winning is 1/3 and that of Ben winning is 1/6, what is the probability that Mike or Ben will win the championship but not Ben?

a) 1/6
b) 5/12
c) 7/12
d) 5/6
e) 1
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Joined: 08 Aug 2017
Posts: 763
Re: Last year, Mike, Ben and Rob participated in a chess tournament that  [#permalink]

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19 Sep 2019, 06:39
I did it this way.
Here three contestant M, B and R has winning crown probability 1/4 , 1/3 and 1/6 respectively.
Only one participant will win the crown.
Now we have to calculate the probability of winning the crown either by M or by B. = MB'+M'B
If M win then B will be definitely loose the crown and vice versa. In such case B' = 2/3 and M'= 3/4
So. MB'+M'B= 1/4*2/3+3/4*1/3= 5/12.
Please tell me if I am wrong? And why?
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Re: Last year, Mike, Ben and Rob participated in a chess tournament that  [#permalink]

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19 Sep 2019, 07:05
gvij2017 wrote:
I did it this way.
Here three contestant M, B and R has winning crown probability 1/4 , 1/3 and 1/6 respectively.
Only one participant will win the crown.
Now we have to calculate the probability of winning the crown either by M or by B. = MB'+M'B
If M win then B will be definitely loose the crown and vice versa. In such case B' = 2/3 and M'= 3/4
So. MB'+M'B= 1/4*2/3+3/4*1/3= 5/12.
Please tell me if I am wrong? And why?

In this method you are ignoring R and other participants in chess tournaments.

We can't ignore them since they are playing in the tournament hence probability should be calculated keeping all participants in mind.
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Re: Last year, Mike, Ben and Rob participated in a chess tournament that  [#permalink]

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19 Sep 2019, 07:16
the probability that either Mike or Ben won ; 1/4+1/3 ; 7/12
IMO C

[quote="Bunuel"]Last year, Mike, Ben and Rob participated in a chess tournament that crowned 1 champion. If their respective probabilities of being crowned champion were 1/4, 1/3 and 1/6, then what's the probability that either Mike or Ben won?

A. 1/6
B. 5/12
C. 7/12
D. 5/6
E. 1

[
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Posts: 76
Re: Last year, Mike, Ben and Rob participated in a chess tournament that  [#permalink]

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19 Sep 2019, 07:45
I logically eliminated the answer choices

Option E can not be the answer choice - According to this choice Option E can never win

Option A cab ve ruled out as well -- Since the probability of Mike winning or Ben winning should be more than the probability of the individual probability of Mike or Ben

According to me since the question statement asks the probability of Mike winning or Ben winning it should be really high -- Eliminating option B and C

But there is a 33% probability my answer is correct :P
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Re: Last year, Mike, Ben and Rob participated in a chess tournament that  [#permalink]

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24 Sep 2019, 10:27
Bunuel wrote:
Last year, Mike, Ben and Rob participated in a chess tournament that crowned 1 champion. If their respective probabilities of being crowned champion were 1/4, 1/3 and 1/6, then what's the probability that either Mike or Ben won?

A. 1/6
B. 5/12
C. 7/12
D. 5/6
E. 1

The probability that either Mike or Ben won is 1/4 + 1/3 = 3/12 + 4/12 = 7/12.

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Re: Last year, Mike, Ben and Rob participated in a chess tournament that  [#permalink]

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26 Nov 2019, 13:27
Hi.

Can anyone please explain how can we simply add the probability of mike and ben and not consider R?

Why is this approach incorrect:-

1-[Prob of Mike Win x Ben Lose x R lose + Prob of Mike Lose x Ben Win x R lose ]

Simply not considering the third person is the right approach?

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Re: Last year, Mike, Ben and Rob participated in a chess tournament that  [#permalink]

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29 Nov 2019, 06:34
@banuel sir can you offer your 2 cents. Simply adding the probability of Mike And Ben winning means only these two are playing. How can we ignore Rob ? Having raised this questions to multiple quant experts all had the opinion that we cant simply add 1/4 and 1/3 . Instead P( M win x B lose x R lose ) + P (M lose x B win x R lose ) is what they recommended. Need your help.

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Re: Last year, Mike, Ben and Rob participated in a chess tournament that  [#permalink]

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29 Nov 2019, 06:37
ScottTargetTestPrep wrote:
Bunuel wrote:
Last year, Mike, Ben and Rob participated in a chess tournament that crowned 1 champion. If their respective probabilities of being crowned champion were 1/4, 1/3 and 1/6, then what's the probability that either Mike or Ben won?

A. 1/6
B. 5/12
C. 7/12
D. 5/6
E. 1

The probability that either Mike or Ben won is 1/4 + 1/3 = 3/12 + 4/12 = 7/12.

Hi. How can you ignore that Rob is also playing and that only one can win the tournament out of Mike and Ben. If Mike wins Rob Loses and so does Ben. Why wont we consider both the cases when M win B loses R loses + M loses B wins R loses?

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Re: Last year, Mike, Ben and Rob participated in a chess tournament that  [#permalink]

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04 Dec 2019, 17:59
2
Sam10smart wrote:

Hi. How can you ignore that Rob is also playing and that only one can win the tournament out of Mike and Ben. If Mike wins Rob Loses and so does Ben. Why wont we consider both the cases when M win B loses R loses + M loses B wins R loses?

Posted from my mobile device

'
In this question, we are using the formula P(Mike or Ben) = P(Mike) + P(Ben) - P(both). Since it is stated that there can be only one champion, P(both) = 0 and thus, P(Mike or Ben) is simply the sum of the individual probabilities of success.
The reason we are ignoring Rob is that participation of Rob and his probability of success has no effect whatsoever in P(Mike or Ben). First, notice that Mike, Ben, and Rob are not the only participants in this tournament since 1/4 + 1/3 + 1/6 < 1. So not only we are ignoring Rob, we are ignoring any other participants as well. Next, the event that "Mike wins" implies that both Ben, Rob and any other participant loses. Similarly for the event "Ben wins". There are no events such as "Mike wins and Rob does not loose" or "Ben wins and Mike does not loose". That's why it is not necessary to multiply P(Mike wins) with P(Ben loses) or P(Rob loses).

In the 1-[Prob of Mike Win x Ben Lose x R lose + Prob of Mike Lose x Ben Win x R lose ] approach, you are not only assuming that Mike, Ben, and Rob are the only participants (which is not true) but also calculating as if it was possible that Mike wins and some other participant does not loose.
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Re: Last year, Mike, Ben and Rob participated in a chess tournament that  [#permalink]

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31 May 2020, 13:02
ScottTargetTestPrep wrote:
Sam10smart wrote:

Hi. How can you ignore that Rob is also playing and that only one can win the tournament out of Mike and Ben. If Mike wins Rob Loses and so does Ben. Why wont we consider both the cases when M win B loses R loses + M loses B wins R loses?

Posted from my mobile device

'
In this question, we are using the formula P(Mike or Ben) = P(Mike) + P(Ben) - P(both). Since it is stated that there can be only one champion, P(both) = 0 and thus, P(Mike or Ben) is simply the sum of the individual probabilities of success.
The reason we are ignoring Rob is that participation of Rob and his probability of success has no effect whatsoever in P(Mike or Ben). First, notice that Mike, Ben, and Rob are not the only participants in this tournament since 1/4 + 1/3 + 1/6 < 1. So not only we are ignoring Rob, we are ignoring any other participants as well. Next, the event that "Mike wins" implies that both Ben, Rob and any other participant loses. Similarly for the event "Ben wins". There are no events such as "Mike wins and Rob does not loose" or "Ben wins and Mike does not loose". That's why it is not necessary to multiply P(Mike wins) with P(Ben loses) or P(Rob loses).

In the 1-[Prob of Mike Win x Ben Lose x R lose + Prob of Mike Lose x Ben Win x R lose ] approach, you are not only assuming that Mike, Ben, and Rob are the only participants (which is not true) but also calculating as if it was possible that Mike wins and some other participant does not loose.

Hi! I dont really get the part of X does not loose... would you please elaborate? thanks
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Re: Last year, Mike, Ben and Rob participated in a chess tournament that   [#permalink] 31 May 2020, 13:02