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# High Caliber Probability

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High Caliber Probability [#permalink]  29 Aug 2009, 08:18
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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?

OA
[Reveal] Spoiler:
\frac{18}{125}

If you can solve that question, then try even harder one:

(b) What is the probability that the player accumulates exactly 7 points and then loses on the next turn?

OA
[Reveal] Spoiler:
\frac{198}{3,125}
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Re: High Caliber Probability [#permalink]  29 Aug 2009, 09:32
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.
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Re: High Caliber Probability [#permalink]  29 Aug 2009, 13:01
flyingbunny wrote:
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

Odd + Even = Odd. The player is in the game.
Odd + Odd = Even. The player loses.

The problem is still solvable.
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Re: High Caliber Probability [#permalink]  29 Aug 2009, 18:21
3
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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}

Probability= \frac{6}{5^3} + \frac{9}{5^4} + \frac{3}{5^5}= \frac{198}{3125}

Thanks for the question. The second part was very interesting.
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Re: High Caliber Probability [#permalink]  29 Aug 2009, 18:28
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?

OA
[Reveal] Spoiler:
\frac{18}{125}

If you can solve that question, then try even harder one:

(b) What is the probability that the player accumulates exactly 7 points and then loses on the next turn?

OA
[Reveal] Spoiler:
\frac{198}{3,125}

The player can't draw any even numbers at any time? or this rule only applies to the first draw?
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Re: High Caliber Probability [#permalink]  29 Aug 2009, 18:40
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...
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Re: High Caliber Probability [#permalink]  30 Aug 2009, 02:13
LenaA wrote:
1) I got ...

Yes, you are right. Perhaps the most challenging thing is to realize how to interpret rules and convert them into Odd & Even sequences.
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Re: High Caliber Probability [#permalink]  30 Aug 2009, 07:53
the question is so twisted
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Re: High Caliber Probability [#permalink]  06 Dec 2009, 14:41
half the GMAT is paying attention to the wording. i think it's worded clearly. 1st part is very easy. 2nd . . . not so much.
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Re: High Caliber Probability [#permalink]  01 Mar 2011, 10:23
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}

b) the player will lose after getting a 2nd Odd.

O---E---E---E
1---2---2---2---Odd = \frac{1}{5^4}*\frac{3}{5} = \frac{3}{5^5}

1---2---4-------Odd = \frac{1}{5^3}*\frac{3}{5} = \frac{3}{5^4} = \frac{3*5}{5^5}

1---4---2-------Odd = \frac{1}{5^3}*\frac{3}{5} = \frac{3}{5^4} = \frac{3*5}{5^5}

3---2---2-------Odd = \frac{1}{5^3}*\frac{3}{5} = \frac{3}{5^4} = \frac{3*5}{5^5}

3---4-----------Odd = \frac{1}{5^2}*\frac{3}{5} = \frac{3}{5^3} = \frac{3*5^2}{5^5}

5---2-----------Odd = \frac{1}{5^2}*\frac{3}{5} = \frac{3}{5^3} = \frac{3*5^2}{5^5}

total probability = \frac{3}{5^5} + 3*\frac{3*5}{5^5} + 2*\frac{3*5^2}{5^5} = \frac{198}{3125}

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Re: High Caliber Probability [#permalink]  31 Dec 2011, 02:25
I normally hate probability question, but this one was really easy:

3/5 * 2/5 * 2/5 = 18/125
Re: High Caliber Probability   [#permalink] 31 Dec 2011, 02:25
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# High Caliber Probability

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