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Pablo plays 3 rounds of a game, in which his chances of winn

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siddhans
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Pablo plays 3 rounds of a game, in which his chances of winn [#permalink] New post   Updated on: 09 Sep 2013, 02:47
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Pablo plays 3 rounds of a game, in which his chances of winning each round are 1/3 ,1/6 , and 1/n, respectively. If n ≠ 0, what is the probability that Pablo wins the first two rounds, but loses the third?

A. 1/18n
B. (n-1)/18n
C. 1/2n
D. (n+2)/2n
E. (3n-2)/2n
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B

Originally posted by siddhans on 22 Jun 2011, 01:34.
Last edited by Bunuel on 09 Sep 2013, 02:47, edited 1 time in total.
Renamed the topic, edited the question and the OA.
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BrentGMATPrepNow
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Re: Pablo plays 3 rounds of a game, in which his chances of winn [#permalink] New post   12 Dec 2019, 08:48
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siddhans wrote:
Pablo plays 3 rounds of a game, in which his chances of winning each round are 1/3 ,1/6 , and 1/n, respectively. If n ≠ 0, what is the probability that Pablo wins the first two rounds, but loses the third?

A. 1/18n
B. (n-1)/18n
C. 1/2n
D. (n+2)/2n
E. (3n-2)/2n


GIVEN:
P(win 1st round) = 1/3
P(win 2nd round) = 1/6
P(win 3rd round) = 1/n, so P(LOSE 3rd round) = 1 - 1/n = n/n - 1/n = (n-1)/n


P(Win - Win - Lose) = P(win 1st round AND win 2nd round AND LOSE 3rd round)
= P(win 1st round) x P(win 2nd round) x P(LOSE 3rd round)
= 1/3 x 1/6 x (n-1)/n
= (n-1)/18n

Answer: B

Cheers,
Brent
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Re: PS MGMAT [#permalink] New post   22 Jun 2011, 02:39
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1/3 * 1/6 * (1-1/n)

= 1/18 * (n-1)/n

= (n-1)/18n

Answer - B
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Re: PS MGMAT [#permalink] New post   22 Jun 2011, 21:11
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p(winning) = 1/3 * 1/6 * (n-1/n)

=(n-1)/(18n)

Answer is B.

r u sure OA is C?
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Re: PS MGMAT [#permalink] New post   23 Dec 2011, 03:50
I also went to B. Why C?
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Re: PS MGMAT [#permalink] New post   27 Dec 2011, 04:57
I also got B ... why C?????? Please someone explain
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Re: PS MGMAT [#permalink] New post   28 Dec 2011, 03:13
Same here B as well, I don't understand why C
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Re: PS MGMAT [#permalink] New post   09 Jan 2012, 15:24
This is the quickest MGMAT problem I have ever solved.
1st, 2nt, no 3rd.

1/3 * 1/6 * (1 - 1/n) = (n-1)/18n
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Re: PS MGMAT [#permalink] New post   09 Jan 2012, 21:02
I also got the answer as B. Can you please check the answer
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Re: PS MGMAT [#permalink] New post   09 Jan 2012, 21:28
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B.
Please explain the solution given in MGMAT for OA=C.
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Re: Pablo plays 3 rounds of a game, in which his chances of winn [#permalink] New post   09 Sep 2013, 02:49
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siddhans wrote:
Pablo plays 3 rounds of a game, in which his chances of winning each round are 1/3 ,1/6 , and 1/n, respectively. If n ≠ 0, what is the probability that Pablo wins the first two rounds, but loses the third?

A. 1/18n
B. (n-1)/18n
C. 1/2n
D. (n+2)/2n
E. (3n-2)/2n


Edited the OA.

P = 1/3*1/6*(1-1/n) = (n-1)/18n.

Answer: B.
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Re: Pablo plays 3 rounds of a game, in which his chances of winn [#permalink] New post   01 Dec 2019, 15:35
Expert Reply
Hi All,

We're told that Pablo plays 3 rounds of a game and his chances of winning each round are 1/3, 1/6 and 1/N respectively (where n ≠ 0). We're asked for the probability that Pablo wins the first two rounds, but loses the third. This question can be solved by TESTing VALUES.

IF... N=3...then
Pablo has a 1/3 chance to win the first round
Pablo has a 1/6 chance to win the second round
Pablo has a 2/3 chance to lose the third round.

The probability of that exact event is (1/3)(1/6)(2/3) = 2/54 = 1/27. So we're looking for that answer when N=3. There's only one answer that matches...

Final Answer:
Show Spoiler
B


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Rich
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Re: Pablo plays 3 rounds of a game, in which his chances of winn [#permalink] New post   10 Dec 2019, 20:36
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siddhans wrote:
Pablo plays 3 rounds of a game, in which his chances of winning each round are 1/3 ,1/6 , and 1/n, respectively. If n ≠ 0, what is the probability that Pablo wins the first two rounds, but loses the third?

A. 1/18n
B. (n-1)/18n
C. 1/2n
D. (n+2)/2n
E. (3n-2)/2n


P(winning first two rounds but not the third) is the following:

1/3 x 1/6 x (1 - 1/n)

1/18 x (1 - 1/n)

1/18 - 1/18n

n/18n - 1/18n

(n - 1)/18n

Answer: B
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Re: Pablo plays 3 rounds of a game, in which his chances of winn [#permalink] New post   12 Feb 2021, 18:14
P(not winning in round 3) = 1 - 1/n = n - 1 / n

1/3 x 1/6 x (n-1 / n) = n-1 / 18n <---Probability of winning the first two and not winning the last

Answer is B.
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Re: Pablo plays 3 rounds of a game, in which his chances of winn [#permalink]
12 Feb 2021, 18:14
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