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A chess player participates in two games. He has 30 percent

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Senior Manager
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A chess player participates in two games. He has 30 percent [#permalink] New post 17 Dec 2003, 21:35
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A
B
C
D
E

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(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 2 sessions
A chess player participates in two games. He has 30 percent chance of winning the first game and 20 percent chance of winning the second game. what is the probability that he wins exactly one ?

(A)50%
(B)38%
(C)28%
(D)14%
(E)16%

Please explain.
Intern
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 [#permalink] New post 18 Dec 2003, 01:23
It is B:

1) Probability of winning exactly one game is:

sum of two probabilities:
(winning the first and not winning the second) + (NOT winning the first and winning The second) = 0.3 (1-0.2) + (1-0.3) 0.2 = 0.38 or 38%


2) In order to check, we can solve the other way:

100%- (probability of winning both games + probability of not winning both games) = 1- ( 0.3*0.2+ (1-0.3)(1-0.2)) = 0.38
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 [#permalink] New post 18 Dec 2003, 07:37
I used the same method as anvar did in his check example:

Odds of going 1W-1L are 1-(two wins)-(two losses)

1-(.2*.3)-(.8*.7)
Senior Manager
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 [#permalink] New post 18 Dec 2003, 07:59
Thank you. Where can I find more information on these kind of problems ?
  [#permalink] 18 Dec 2003, 07:59
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A chess player participates in two games. He has 30 percent

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