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The probability of winning game A is X and the probability

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The probability of winning game A is X and the probability

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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

OA D

Source: Economist GMAT

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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....

Final Answer: D

GMAT assassins aren't born, they're made,
Rich

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Contact Rich at Rich.C@empowergmat.com

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BTGmoderatorDC 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

Source: Economist GMAT
$$? = P\left( {{\text{win}}\,\,{\text{just}}\,\,{\text{A}}\,\,{\text{or}}\,\,{\text{win}}\,\,{\text{just}}\,\,{\text{B}}} \right)\,\,\,\, = \,\,\,\,P\left( {{\text{win}}\,\,{\text{A}}\,{\text{or}}\,\,{\text{B}}} \right) - P\left( {{\text{win}}\,\,{\text{both}}} \right)$$
$$P\left( {{\rm{win}}\,\,{\rm{both}}} \right) = xy\,\,\,\,\,\,\,\,\left[ {\,x,y\,\,\,{\rm{constants}}\,\,\,\, \Rightarrow \,\,\,\,{\rm{independency}}\,} \right]$$
$$P\left( {{\text{win}}\,\,{\text{A}}\,{\text{or}}\,\,{\text{B}}} \right) = x + y - xy\,\,\,\,\,\left[ {\,{\text{simplifier}}\,} \right]$$
$$? = x + y - 2xy$$



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".
(Vladimir Arnold)

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Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br



Last edited by fskilnik@GMATH on Sat Oct 13, 2018 1:30 pm; edited 2 times in total

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BTGmoderatorDC 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

OA D

Source: Economist GMAT
Given:

1. The probability of winning game A is X, and
2. The probability of winning game B is Y

Thus,

The probability of NOT winning game A = 1 - X
The probability of NOT winning game B = 1 - Y

Thus, the probability of winning exactly once = the probability of winning either A or B but not both the games

= [(Probability of winning game A)*(Probability of NOT winning game B)] + [(Probability of winning game B)*(Probability of NOT winning game A)]

= X*(1 - Y) + Y*(1 - X)
= X - XY + Y - XY
= X + Y - 2XY

The correct answer: D

Hope this helps!

-Jay
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Manhattan Review GMAT Prep

Locations: New York | Bangkok | Abu Dhabi | Rome | and many more...

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BTGmoderatorDC 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

Answer: D

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