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# The probability that team A will not win the tournament is 80% and the

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Math Expert
Joined: 02 Sep 2009
Posts: 61302
The probability that team A will not win the tournament is 80% and the  [#permalink]

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20 Mar 2018, 22:29
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15% (low)

Question Stats:

74% (01:07) correct 26% (01:18) wrong based on 142 sessions

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The probability that team A will not win the tournament is 80% and the probability that team B will not win the tournament is 60%. If there is only one tournament winner, what is the probability that either team A or team B wins the tournament?

(A) 20%
(B) 40%
(C) 48%
(D) 60%
(E) 80%

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Posts: 61302
Re: The probability that team A will not win the tournament is 80% and the  [#permalink]

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20 Mar 2018, 22:29
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Status: It always seems impossible until it's done.
Joined: 16 Sep 2016
Posts: 727
GMAT 1: 740 Q50 V40
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Re: The probability that team A will not win the tournament is 80% and the  [#permalink]

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20 Mar 2018, 23:20
Bunuel wrote:
The probability that team A will not win the tournament is 80% and the probability that team B will not win the tournament is 60%. If there is only one tournament winner, what is the probability that either team A or team B wins the tournament?

(A) 20%
(B) 40%
(C) 48%
(D) 60%
(E) 80%

IMO D.

Let P(A) be probability that Team A wins the tournament.
P(A) = 1 - 0.8 = 0.2
P(B) = 1 - 0.6 = 0.4

P(A or B) = P(A) + P(B) .... ( as only one of them can win the tournament which means there is no overlap in the two events)
P(A or B) = 0.2 + 0.4
= 0.6 or 60%

Hence D.

Best,
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Joined: 11 Feb 2017
Posts: 179
Re: The probability that team A will not win the tournament is 80% and the  [#permalink]

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22 Mar 2018, 14:00
Bunuel wrote:
The probability that team A will not win the tournament is 80% and the probability that team B will not win the tournament is 60%. If there is only one tournament winner, what is the probability that either team A or team B wins the tournament?

(A) 20%
(B) 40%
(C) 48%
(D) 60%
(E) 80%

IMO D.

Let P(A) be probability that Team A wins the tournament.
P(A) = 1 - 0.8 = 0.2
P(B) = 1 - 0.6 = 0.4

P(A or B) = P(A) + P(B) .... ( as only one of them can win the tournament which means there is no overlap in the two events)
P(A or B) = 0.2 + 0.4
= 0.6 or 60%

Hence D.

Best,

How is it different from my way?

Prob. of A win and B not win = 20/100 * 60 /100

Prob. of B win and A not win = 40/100 * 80 /100

Any one can win so we need (+)

20/100 * 60 /100 + 40/100 * 80 /100

The result is not 60%

why?
Manager
Joined: 30 Mar 2017
Posts: 113
GMAT 1: 200 Q1 V1
Re: The probability that team A will not win the tournament is 80% and the  [#permalink]

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23 Mar 2018, 12:52
1
rocko911 wrote:
Bunuel wrote:
The probability that team A will not win the tournament is 80% and the probability that team B will not win the tournament is 60%. If there is only one tournament winner, what is the probability that either team A or team B wins the tournament?

(A) 20%
(B) 40%
(C) 48%
(D) 60%
(E) 80%

IMO D.

Let P(A) be probability that Team A wins the tournament.
P(A) = 1 - 0.8 = 0.2
P(B) = 1 - 0.6 = 0.4

P(A or B) = P(A) + P(B) .... ( as only one of them can win the tournament which means there is no overlap in the two events)
P(A or B) = 0.2 + 0.4
= 0.6 or 60%

Hence D.

Best,

How is it different from my way?

Prob. of A win and B not win = 20/100 * 60 /100

Prob. of B win and A not win = 40/100 * 80 /100

Any one can win so we need (+)

20/100 * 60 /100 + 40/100 * 80 /100

The result is not 60%

why?

I dont understand the statement "Prob. of A win and B not win" because if A wins, then by default B does not win since there can only be 1 winner. Team A winning implies that every other team does not win, so (Prob. of A win and B not win) = (Prob. of A win) = 20%.
Senior PS Moderator
Status: It always seems impossible until it's done.
Joined: 16 Sep 2016
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GMAT 1: 740 Q50 V40
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Re: The probability that team A will not win the tournament is 80% and the  [#permalink]

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23 Mar 2018, 13:09
1
aserghe1 wrote:
rocko911 wrote:

IMO D.

Let P(A) be probability that Team A wins the tournament.
P(A) = 1 - 0.8 = 0.2
P(B) = 1 - 0.6 = 0.4

P(A or B) = P(A) + P(B) .... ( as only one of them can win the tournament which means there is no overlap in the two events)
P(A or B) = 0.2 + 0.4
= 0.6 or 60%

Hence D.

Best,

How is it different from my way?

Prob. of A win and B not win = 20/100 * 60 /100

Prob. of B win and A not win = 40/100 * 80 /100

Any one can win so we need (+)

20/100 * 60 /100 + 40/100 * 80 /100

The result is not 60%

why?

I dont understand the statement "Prob. of A win and B not win" because if A wins, then by default B does not win since there can only be 1 winner. Team A winning implies that every other team does not win, so (Prob. of A win and B not win) = (Prob. of A win) = 20%.

Hi rocko911,

precisely.. this is why the final answer is incorrect as rightly pointed out by aserghe1.

When we are applying the probability of A winning.. B losing is implicit - as these are mutually exclusive events. ( or in other words .. there can only be one winner)

Does that make sense?

Best,
_________________
Regards,

“Do. Or do not. There is no try.” - Yoda (The Empire Strikes Back)
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3239
Re: The probability that team A will not win the tournament is 80% and the  [#permalink]

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26 Mar 2018, 08:42

Solution:

Given:

• P (Team A not winning the tournament) = 80% = 80/100

• P (Team B not winning the tournament) = 60% = 60/100

• The tournament has only 1 winner.

Working out:

We need to find that either team A or team B wins the tournament.

Per our conceptual knowledge, we know the following:

• P (Non-event) = 1 – P(Event)

P (Team A winning the tournament) = 1 – 80/100 = 20/100

P (Team B winning the tournament) = 1- 60/100 = 40/100

Since it is already stated in the question stem that the tournament has only a single winner, the above two events must be independent.

Hence, P (Either Team A wins or Team B wins) = 20 /100 + 40/100 = 60/100, or, 60%.

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Posts: 4331
Re: The probability that team A will not win the tournament is 80% and the  [#permalink]

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26 Mar 2018, 09:09
Top Contributor
Bunuel wrote:
The probability that team A will not win the tournament is 80% and the probability that team B will not win the tournament is 60%. If there is only one tournament winner, what is the probability that either team A or team B wins the tournament?

(A) 20%
(B) 40%
(C) 48%
(D) 60%
(E) 80%

The probability that team A will NOT win the tournament is 80%
In other words, the probability that team A will NOT win the tournament is 0.8
So P(team A WILL win the tournament) = 1 - 0.8 = 0.2

The probability that team B will NOT win the tournament is 60%
So P(team B WILL win the tournament) = 1 - 0.6 = 0.4

If there is only one tournament winner, what is the probability that either team A or team B wins the tournament?
This is an OR probability. So, we'll apply the OR probability formula:

P(event X occurs OR event Y occurs ) = P(X occurs ) + P(Y occurs ) - P(X and Y BOTH occur)

So, we get: P(A or B wins tournament) = P(A wins tournament) + P(B wins tournament) - P(A AND B both win tournament)
= 0.2 + 0.4 - 0
= 0.6

ASIDE: P(A AND B both win tournament) = 0, because we are told that "there is only one tournament winner." So, both teams cannot win.

Cheers,
Brent
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Re: The probability that team A will not win the tournament is 80% and the  [#permalink]

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27 Mar 2018, 10:32
Bunuel wrote:
The probability that team A will not win the tournament is 80% and the probability that team B will not win the tournament is 60%. If there is only one tournament winner, what is the probability that either team A or team B wins the tournament?

(A) 20%
(B) 40%
(C) 48%
(D) 60%
(E) 80%

Recall that P(A or B) = P(A) + P(B) - P(A and B). The probability that team A wins is 1 - 0.8 = 0.2 and the probability that team B wins is 1 - 0.6 = 0.4. Since they can’t both be the winner, P(A and B) = 0. Thus,

P(A or B) = 0.2 + 0.4 - 0 = 0.6

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Re: The probability that team A will not win the tournament is 80% and the   [#permalink] 27 Mar 2018, 10:32
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