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by \'manpreet singh » Mon Jan 21, 2013 10:32 pm
Q.1) A lottery game works as follows: The player draws a numbered ball at random from an urn containing five balls numbered 1, 2, 3, 4, and 5. If the number on the ball is even, the player loses the game and receives no points: if the number on the ball is odd, the player receives the number of points indicated on the ball. Afterward, he or she replaces the ball in the urn and draws again. On each subsequent turn, the player loses the game if the total of all the numbers drawn becomes even, and gets another turn (after receiving the number of points indicated on the ball and then replacing the ball in the urn) each time the total remains odd.
(a) What is the probability that the player loses the game on the third turn?
(b) What is the probability that the player accumulates exactly 7 points and then loses
on the next turn?
soory don't know how question got missed!
Last edited by \'manpreet singh on Tue Jan 22, 2013 12:59 am, edited 1 time in total.
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by Brent@GMATPrepNow » Mon Jan 21, 2013 10:47 pm
'manpreet singh wrote:Q.1) A lottery game works as follows: The player draws a numbered ball at random from an urn containing five balls numbered 1, 2, 3, 4, and 5. If the number on the ball is even, the player loses the game and receives no points: if the number on the ball is odd, the player receives the number of points indicated on the ball. Afterward, he or she replaces the ball in the urn and draws again. On each subsequent turn, the player loses the game if the total of all the numbers drawn becomes even, and gets another turn (after receiving the number of points indicated on the ball and then replacing the ball in the urn) each time the total remains odd.

soory don't know how question got missed!
You've given us the set-up, but we're still missing the question.

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by \'manpreet singh » Tue Jan 22, 2013 1:00 am
I have edited the question now.
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by Brent@GMATPrepNow » Tue Jan 22, 2013 6:54 am
'manpreet singh wrote:Q.1) A lottery game works as follows: The player draws a numbered ball at random from an urn containing five balls numbered 1, 2, 3, 4, and 5. If the number on the first ball drawnis even, the player loses the game and receives no points: if the number on the ball is odd, the player receives the number of points indicated on the ball. Afterward, he or she replaces the ball in the urn and draws again. On each subsequent turn, the player loses the game if the total of all the numbers drawn becomes even, and gets another turn (after receiving the number of points indicated on the ball and then replacing the ball in the urn) each time the total remains odd.
(a) What is the probability that the player loses the game on the third turn?
(b) What is the probability that the player accumulates exactly 7 points and then loses
on the next turn?
soory don't know how question got missed!
This question is a little ambiguous as it stand. I think you mean to say, "If the very first draw is even then the player loses."
I've amended it above.

(a) What is the probability that the player loses the game on the third turn?

The only way for the player to lose on the 3rd turn is to get the following scenario:
Odd on first draw THEN even on second draw THEN odd on third draw (giving us an even sum and ending the game)

So, P(lose on 3rd turn) = P(odd on 1st draw AND even on 2nd draw AND odd on 3rd draw)
= P(odd on 1st draw) x P(even on 2nd draw) x P(odd on 3rd draw)
= 3/5 x 2/5 x 3/5
= [spoiler]18/125[/spoiler]

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by Brent@GMATPrepNow » Tue Jan 22, 2013 6:59 am
(b) What is the probability that the player accumulates exactly 7 points and then loses on the next turn?

There are too many calculations required to be a legitimate GMAT question.

In how many different ways can we accumulate 7 points and then lose on the next turn? There are 6 different cases to consider.
case a) 1-2-2-2-any odd
case b) 1-2-4-any odd
case c) 1-4-2-any odd
case d) 3-2-2-any odd
case e) 3-4-any odd
case f) 5-2-any odd

Too many calculations!
The calculations aren't difficult - just time-consuming.
If someone wants to take it from here, fill your boots. :-)

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by \'manpreet singh » Wed Jan 23, 2013 10:48 pm
Thanks for the quick response and i too felt second question was a little too time consuming to be attempted.

Br,
Manpreet.
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