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A lottery game works as follows: The player draws a numbered [#permalink]
29 Aug 2009, 07:18

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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 the numbers 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?

Re: High Caliber Probability [#permalink]
29 Aug 2009, 08:32

what kind of question is this?

the player will lose in the second draw no matter what he get: 1) if he gets a even ball, he loses 2) if he gets a odd ball, (this number) + (the number he drew in the first time)= even, still loses. _________________

Re: High Caliber Probability [#permalink]
29 Aug 2009, 17:21

3

This post received KUDOS

1) I got \frac{18}{125} in the following way

Three draws of 5 balls result in 5^3=125options. Not to lose the game, a player has to draw odd numbers on the first draw and even numbers on all other draws. Therefore to lose on the third draw, a player has to draw an odd number on the first draw, an even number on the second and then an odd number on the third draw. Hence, the options to draw (odd,even,odd)=3\times 2\times 3=18 Hence the probability is \frac{18}{125}

2) I got \frac{198}{3125} in the following way.

Again, no to lose a game a play have to draw odd number at the 1st draw and then even numbers... To collect 7 points, there are 6 possible outcomes: (1,2,2,2) (1,4,2) (1,2,4) (3,2,2) (3,4) (5,2) P(To collect 7 points and lose on the next)=P(to collect 7 points on the 2st and lose on the 3rd)+P(to collect 7 points on the 3rd and lose on the 4th) +P(to collect 7 points on the 4rth and lose on the 5th) 1) To collect 7 points at the 2nd and lose at the 3rd: 2 events (3,4),(5,2) and then any of the tree odd numbers => \frac{6}{5^3} 2) To collect 7 points on the 3rd and lose on the 4th, 3 events (1,4,2),(1,2,4),(3,2,2) and then any of the tree odd => \frac{9}{5^4} 3) to collect 7 points on the 4rth and lose on the 5th, one event (1,2,2,2) and the any of the tree odd=> \frac{3}{5^5}

Re: High Caliber Probability [#permalink]
29 Aug 2009, 17:28

DenisSh wrote:

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 the numbers 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?

Re: High Caliber Probability [#permalink]
29 Aug 2009, 17:40

1

This post received KUDOS

Flyingbunny, Since these are the rules of the game, there is a certain sequence of events. Here is how I understand the rules. A player draws a ball (first ball). If the first ball is even, a player is out of the game but puts the ball back. If a player draws an odd ball, he gets his points, puts a ball back and gets another turn. And since the second draw, he has to watch out for the sum of his points not to be even, meaning he has to draw only even numbers, starting from his second draw...if at any turn (after the 1st) he draws an odd ball he will be out of the game... My solution was based on this interpretation of rules...

Re: High Caliber Probability [#permalink]
01 Mar 2011, 09:23

1

This post received KUDOS

a) to lose on the 3rd turn the sequence would be Odd-Even-Odd (OEO). the probability of getting OEO = \frac{3}{5}*\frac{2}{5}*\frac{3}{5} = \frac{18}{125}

Re: A lottery game works as follows: The player draws a numbered [#permalink]
05 Nov 2013, 06:39

FIRST QUESTION

Here's my notes from reading the problem:

Five balls numbered 1,2,3,4,5 Round 1: Player loses on even, player gets points on odd Round 2-inf: Player loses on even point total, keeps going on odd

FIND: Prob LOSE on Round 3 (Win 1, Win 2, Lose 3)

Here's my solving logic:

Round 1 Lose: 2/5 chance (two even numbers, 2 and 4) Win: 3/5 chance End of round point total: 1, 3 or 5

Round 2 Since point total is odd, if the player draws an odd they will lose, as it will make their point total even. Lose: 3/5 chance (three odd numbers, 1,3 and 5) Win: 2/5 chance End of round point total: 1, 3 or 5 plus 2 or 4

Round 3 Again, since point total must be odd, they have to draw an odd to lose. Lose: 3/5 chance Win: 2/5 chance

So the probability that the player loses on the third turn is the probability of winning the first two rounds and losing the third:

P(W1) * P(W2) * P(L3) (3/5)*(2/5)*(3/5) = 18/125

Can come back for second question.

gmatclubot

Re: A lottery game works as follows: The player draws a numbered
[#permalink]
05 Nov 2013, 06:39