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

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

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New post 20 Mar 2018, 23:29
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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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Re: The probability that team A will not win the tournament is 80% and the  [#permalink]

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New post 20 Mar 2018, 23:29
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%


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

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New post 21 Mar 2018, 00:20
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,
Gladi
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Re: The probability that team A will not win the tournament is 80% and the  [#permalink]

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New post 22 Mar 2018, 15:00
Gladiator59 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,
Gladi



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

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New post 23 Mar 2018, 13:52
1
rocko911 wrote:
Gladiator59 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,
Gladi



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

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New post 23 Mar 2018, 14:09
1
aserghe1 wrote:
rocko911 wrote:
Gladiator59 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,
Gladi



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

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New post 26 Mar 2018, 09:42

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%.

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

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New post 26 Mar 2018, 10:09
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

Answer: D

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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New post 27 Mar 2018, 11:32
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

Answer: D
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Re: The probability that team A will not win the tournament is 80% and the &nbs [#permalink] 27 Mar 2018, 11:32
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