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On every game, there are only two cases: the player wins or loses

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On every game, there are only two cases: the player wins or loses  [#permalink]

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New post 11 Dec 2016, 05:50
2
1
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

43% (02:50) correct 57% (02:30) wrong based on 84 sessions

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On every game, there are only two cases: the player wins or loses. The probability that Michael wins on each game is 0.4. How many games does Michael have to play at least so that the probability that Michael wins on at least one game is greater than 0.95?

A. 4
B. 5
C. 6
D. 7
E. 8

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Re: On every game, there are only two cases: the player wins or loses  [#permalink]

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New post 11 Dec 2016, 09:22
2
Hi

P(winning at least one game)>0,95 is the same as P(loosing all)<0,05

\(0,6^n < 0,05\)

\(n=5 ----> 0,0777...\)

\(n=6 ----> 0,04665...\)

Hence answer is C.
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Re: On every game, there are only two cases: the player wins or loses  [#permalink]

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New post 11 Dec 2016, 23:07
Another way to calculate:

\(0.6^n < 0.05\)

\(ln (6^n) < ln(0.05)\)

\(n*ln(0.6) < ln(0.05)\)

\(n*(-0.5108) < -2.9957\)

\(n > 5.864..\)

Min integer value of \(n\) is \(6\).

Although it's hard to be done without calculator :(
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Re: On every game, there are only two cases: the player wins or loses  [#permalink]

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New post 13 Dec 2016, 13:27
Probability of wining AT LEAST one game is 1 - probability of wining no games:

1 - P(0) = P(>=1)

If we want the probability of wining at least 1 games to be at least 95% (>=0.95); then probability of wining no games has to be less than or equal to 5% (<=.05).

Probability of losing a game [P(0)] is (6/10)^ n, where n is the number of games played.

We start with option c to know if we should go up or down if c is false.

(6/10)^6 = 45,792/1000000= 0.045792 chance that you lose 6 games in a row

The probability of winning at least 1 game is 1 - P(0)
1 - 0.045792 > .95

So the answer is C
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Re: On every game, there are only two cases: the player wins or loses  [#permalink]

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New post 18 Aug 2017, 07:16
A similar tough and tricky question is here

https://gmatclub.com/forum/tough-and-tr ... 85179.html
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Re: On every game, there are only two cases: the player wins or loses  [#permalink]

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New post 10 Mar 2019, 09:59
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: On every game, there are only two cases: the player wins or loses   [#permalink] 10 Mar 2019, 09:59
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