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# The probability of winning game A is X and the probability of winning

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The probability of winning game A is X and the probability of winning  [#permalink]

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29 Sep 2016, 04:38
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The probability of winning game A is X and the probability of winning game B is Y. When playing a single round of each of the two games, what is the probability of winning exactly once?

A. X-YX
B. XY
C. X+Y
D. X+Y-2XY
E. X^2Y^2-XY^2-X^2Y

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Re: The probability of winning game A is X and the probability of winning  [#permalink]

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29 Sep 2016, 07:21
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Probability of winning exactly once --> A wins and B loses or A loses and B wins
= X(1 - Y) + (1 - X)Y
= X - XY + Y - XY
= X + Y - 2XY

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Re: The probability of winning game A is X and the probability of winning  [#permalink]

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29 Sep 2016, 12:54
I picked numbers

X=1/3
Y=1/4

Probability of winning A and losing B (1/3)*(3/4)=1/4
Probability of losing A and winning B (2/3)*(1/4)=1/6

Because it is either option A OR option B we add the two probabilities together

1/4+1/6=5/12. If you plug the numbers back into the answer choices you will get D.
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Re: The probability of winning game A is X and the probability of winning  [#permalink]

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27 Feb 2017, 04:24
P(game A win) = X, P(game A lose) = 1-X
P(game B win) = Y, P(game B lose) = 1-Y
P(exactly one game win = P(game A win) P(game B lose) + P(game B win) P(game A lose) = X(1-Y) + Y(1-X) = X +Y -2XY. Option D
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Re: The probability of winning game A is X and the probability of winning  [#permalink]

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13 Mar 2017, 21:16
Bunuel wrote:
The probability of winning game A is X and the probability of winning game B is Y. When playing a single round of each of the two games, what is the probability of winning exactly once?

A. X-YX
B. XY
C. X+Y
D. X+Y-2XY
E. X^2Y^2-XY^2-X^2Y

Hi,

I used the following approach :
P(Winning A) = X
P(Winning B) = Y
p(Winning Both) = X.Y

Total = X+Y-XY

Kindly suggest where I'm going wrong.
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Re: The probability of winning game A is X and the probability of winning  [#permalink]

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24 Jun 2018, 13:33
ByjusGMATapp wrote:
P(game A win) = X, P(game A lose) = 1-X
P(game B win) = Y, P(game B lose) = 1-Y
P(exactly one game win = P(game A win) P(game B lose) + P(game B win) P(game A lose) = X(1-Y) + Y(1-X) = X +Y -2XY. Option D

Confused why is this not X+y following gmatclub_dot_com/forum/the-probability-that-team-a-will-not-win-the-tournament-is-80-and-the-261771.html
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The probability of winning game A is X and the probability of winning  [#permalink]

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24 Jun 2018, 13:55
Sguha0305 wrote:
ByjusGMATapp wrote:
P(game A win) = X, P(game A lose) = 1-X
P(game B win) = Y, P(game B lose) = 1-Y
P(exactly one game win = P(game A win) P(game B lose) + P(game B win) P(game A lose) = X(1-Y) + Y(1-X) = X +Y -2XY. Option D

Confused why is this not X+y following gmatclub_dot_com/forum/the-probability-that-team-a-will-not-win-the-tournament-is-80-and-the-261771.html

See in this question, two games are seperate. Each game have a seperate win and loss.
But in that link question its a tournament, there will be only a single win and multiple failures i.e. If team A is winning that automatically means that B is loosing

whereas in this question the win of team A doesnt effect team B. If team A wins, team B can also win.

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Re: The probability of winning game A is X and the probability of winning  [#permalink]

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12 Oct 2018, 07:59
Bunuel wrote:
The probability of winning game A is X and the probability of winning game B is Y. When playing a single round of each of the two games, what is the probability of winning exactly once?

A. X-YX
B. XY
C. X+Y
D. X+Y-2XY
E. X^2Y^2-XY^2-X^2Y

P(winning exactly once) = 1 - P(winning zero times) - P(winning both times)

P(wWinning exactly once) = 1 - (1 - X)(1 - Y) - XY = 1 - (1 - X - Y + XY) - XY = X + Y - 2XY

Alternate Solution:

There are two ways of winning exactly once, either (Win A, Lose B) OR (Lose A, Win B). The probability of winning A is X, so the probability of losing A is (1 - X). Similarly, the probability of winning B is Y, and so the probability of losing B is (1 - Y).

P(Win A, Lose B) = X*(1 - Y) = X - XY

P(Lose A, Win B) = (1 - X)*Y = Y - XY

The probability that either of these events happens is the sum of their individual probabilities, so P(Winning exactly once) = X - XY + Y - XY = X + Y - 2XY

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The probability of winning game A is X and the probability of winning  [#permalink]

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12 Oct 2018, 13:31
Bunuel wrote:
The probability of winning game A is X and the probability of winning game B is Y. When playing a single round of each of the two games, what is the probability of winning exactly once?

A. X-YX
B. XY
C. X+Y
D. X+Y-2XY
E. X^2Y^2-XY^2-X^2Y

$$? = P\left( {{\text{win}}\,\,{\text{just}}\,\,{\text{A}}\,\,{\text{or}}\,\,{\text{win}}\,\,{\text{just}}\,\,{\text{B}}} \right)\,\,\,\, = \,\,\,\,P\left( {{\text{win}}\,\,{\text{A}}\,{\text{or}}\,\,{\text{B}}} \right) - P\left( {{\text{win}}\,\,{\text{both}}} \right)$$

$$P\left( {{\rm{win}}\,\,{\rm{both}}} \right) = xy$$

$$\left[ {\,x,y\,\,{\rm{constants}}\,\,\,\, \Rightarrow \,\,\,\,{\rm{win}}\,A\,\,{\rm{and}}\,\,{\rm{win}}\,B\,\,{\rm{are}}\,\,{\rm{independent}}\,\,{\rm{events}}\,} \right]$$

$$P\left( {{\text{win}}\,\,{\text{A}}\,{\text{or}}\,\,{\text{B}}} \right) = x + y - xy\,\,\,\,\,\left[ {\,{\text{simplifier}}\,} \right]$$

$$? = x + y - 2xy$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

"Kolmogorov - Poincaré - Gauss - Euler - Newton, are only five lives separating us from the source of our science."
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Re: The probability of winning game A is X and the probability of winning  [#permalink]

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14 Oct 2018, 13:20
Hi All,

We're told that the probability of winning Game A is X and the probability of winning Game B is Y. We're asked when playing a SINGLE round of EACH of the two games, what is the probability of winning EXACTLY ONCE. This question can be solved in a couple of different ways, including by TESTing VALUES.

IF... X = .4 AND Y = .5
then there is a 40% chance to win Game A and a 60% chance to lose Game A
and a 50% chance to win Game B and a 50% chance to lose Game B.

The probability of winning A and losing B = (.4)(.5) = .2
The probability of winning B and losing A = (.5)(.6) = .3
Total probability of winning just one Game = .2 + .3 = .5

Thus, we're looking for an answer that equals .5 when X = .4 and Y = .5 There's only one answer that matches....

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Re: The probability of winning game A is X and the probability of winning   [#permalink] 14 Oct 2018, 13:20
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