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

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Senior Manager
Joined: 05 May 2003
Posts: 424
Location: Aus
A chess player participates in two games. He has 30 percent [#permalink]

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17 Dec 2003, 22:35
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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%

Intern
Joined: 27 Nov 2003
Posts: 33
Location: Moscow

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18 Dec 2003, 02:23
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
_________________

me

Director
Joined: 28 Oct 2003
Posts: 501
Location: 55405

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18 Dec 2003, 08:37
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
Joined: 05 May 2003
Posts: 424
Location: Aus

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18 Dec 2003, 08:59
Thank you. Where can I find more information on these kind of problems ?
18 Dec 2003, 08:59
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