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The probability that A wins a game is p and B wins the game is q then  [#permalink]

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7 00:00

Difficulty:   55% (hard)

Question Stats: 39% (01:04) correct 61% (01:03) wrong based on 74 sessions

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The probability that A wins a game is p and B wins the game is q then what is the probability that only one of them wins the game?
A) p+q-pq
B) 1-p-q
C) p+q-2pq
D) p+q
E) pq

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The probability that A wins a game is p and B wins the game is q then  [#permalink]

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GMATinsight wrote:
The probability that A wins a game is p and B wins the game is q then what is the probability that only one of them wins the game?
A) p+q-pq
B) 1-p-q
C) p+q-2pq
D) p+q
E) pq

P of A win = p and not winning = 1-p
P of B win= q and not winning = 1-q
so one of them winning
p*(1-q)+q*(1-p)
=> p-pq+q-pq
=> p+q-2pq
IMO C

[url=https://gmatclub.com:443/forum/memberlist.php?mode=viewprofile&un=GMATinsight]GMATinsight sir the given OA options seems to be incorrect

Originally posted by Archit3110 on 19 Mar 2020, 04:29.
Last edited by Archit3110 on 21 Mar 2020, 23:00, edited 2 times in total.
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Re: The probability that A wins a game is p and B wins the game is q then  [#permalink]

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question wants us to use probability & sets to answer

I don't understand the questions specially when its not given if Probability of A & Probability of B are independent of each other or not.

Can someone clarify this?
Target Test Prep Representative V
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Re: The probability that A wins a game is p and B wins the game is q then  [#permalink]

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GMATinsight wrote:
The probability that A wins a game is p and B wins the game is q then what is the probability that only one of them wins the game?
A) p+q-pq
B) 1-p-q
C) p+q-2pq
D) p+q
E) pq

Solution:

We can use the formula:

P(only one wins the game) = P(only A wins the game) + P(only B wins the game)

P(only one wins the game) = [P(A wins the game) - P(both win the game)] + [P(B wins the game) - P(both win the game)]

P(only one wins the game) = [p - P(both win the game)] + [q - P(both win the game)]

Assuming each of them wins the game independently from the other, then P(both win the game) = pq. Therefore, we have:

P(only one wins the game) = [p - pq] + [q - pq]

P(only one wins the game) = p + q - 2pq

Alternate Solution:

The probability that A wins the game is p, so the probability that A loses the game is (1- p). The probability that B wins the game is q, so the probability that B loses the game is (1 - q).

We have 4 possible outcomes: (A_B), (A_notB), (notA_B), and (notA_notB). The event “only one wins the game” occurs with the two middle outcomes.

P(A_notB) = P(A) x P(notB) = p x (1 - q)

P(notA_B) = P(notA) x P(B) = (1 - p) x q

Because either event satisfies our requirement, we add the two probabilities:

p x (1 - q) + q x (1 - p) = p - pq + q - pq = p + q - 2pq

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Re: The probability that A wins a game is p and B wins the game is q then  [#permalink]

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Posted from my mobile device Re: The probability that A wins a game is p and B wins the game is q then   [#permalink] 09 Apr 2020, 15:34

# The probability that A wins a game is p and B wins the game is q then  