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Bunuel
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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Gladiator59
Bunuel
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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Bunuel
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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aserghe1
rocko911
Gladiator59


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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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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Bunuel
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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Probability of A winning the tournament = 1 - 0.8 = 0.2
Similarly, Probability of B winning the tournament = 0.4

Thus,probability that either team A or team B wins the tournament = 0.2 + 0.4 = 0.6
Or, the probability that either team A or team B wins the tournament = 60%

Thus, the correct option is D.
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