Hi VJesus12,

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: B

GMAT assassins aren't born, they're made,
Rich

VJesus12 wrote:Pablo plays \(3\) rounds of a game, in which his chances of winning each round are \(\dfrac13, \dfrac16\), and \(\dfrac1n\), respectively. If \(n ≠ 0\), what is the probability that Pablo wins the first two rounds, but loses the third?

A. \(\dfrac1{18n}\)
B. \(\dfrac{n-1}{18n}\)
C. \(\dfrac1{2n}\)
D. \(\dfrac{n+2}{2n}\)
E. \(\dfrac{3n-2}{2n}\)

[spoiler]OA=B[/spoiler]

Source: Manhattan GMAT

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

VJesus12 wrote:Pablo plays \(3\) rounds of a game, in which his chances of winning each round are \(\dfrac13, \dfrac16\), and \(\dfrac1n\), respectively. If \(n ≠ 0\), what is the probability that Pablo wins the first two rounds, but loses the third?

A. \(\dfrac1{18n}\)
B. \(\dfrac{n-1}{18n}\)
C. \(\dfrac1{2n}\)
D. \(\dfrac{n+2}{2n}\)
E. \(\dfrac{3n-2}{2n}\)

[spoiler]OA=B[/spoiler]

Source: Manhattan GMAT

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