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The probability of Team X beating Team Y
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11 Oct 2021, 05:21
11 Oct 2021, 05:21
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Question Stats:
96% (00:51) correct
3% (00:35) wrong based on 32 sessions
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The probability of Team X beating Team Y is 2/3. The probability of Team X beating Team Z is 1/4. Team X plays one game with Team Y and one game with Team Z.
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
The probability that Team X wins both games |
The probability that Team X loses both games |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Re: The probability of Team X beating Team Y
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12 Oct 2021, 20:22
12 Oct 2021, 20:22
1
\(X\) beating \(Y = \frac{2}{3}\)
\(X\) beating \(Z = \frac{1}{4}\)
Qt A
\(X\) winning both: \(\frac{2}{3} * \frac{1}{4} = \frac{1}{6}\)
Qt B
\(X\) losing both: \(\frac{1}{3} * \frac{3}{4} = \frac{1}{4}\)
Qt B > Qt A
Answer B
\(X\) beating \(Z = \frac{1}{4}\)
Qt A
\(X\) winning both: \(\frac{2}{3} * \frac{1}{4} = \frac{1}{6}\)
Qt B
\(X\) losing both: \(\frac{1}{3} * \frac{3}{4} = \frac{1}{4}\)
Qt B > Qt A
Answer B
gmatclubot
Re: The probability of Team X beating Team Y [#permalink]
12 Oct 2021, 20:22
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