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Combinatorics problem !

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Combinatorics problem !

by sreak1089 » Sat Jul 03, 2010 9:30 am
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 teh 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 toal 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 hten loses on the next turn?
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by albatross86 » Sat Jul 03, 2010 10:05 am
Cool question. Not phrased as a GMAT one obviously, but great for probability and number property concepts.

I'm not sure if my answers are correct, but I hope my approach is. Would love to see a shorter/ more elegant solution.

Two even numbers: 2 and 4, and Three odd numbers : 1,3 and 5

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(a) Loses on the 3rd turn. This means the sum of the 3 numbers drawn is even.

The sum of 3 numbers is even only when

1. They are all 3 even, i.e (EVEN + EVEN + EVEN)
2. 2 Odd and 1 even, i.e. (ODD + EVEN + ODD), (ODD + ODD + EVEN) OR (EVEN + ODD + ODD)

We only consider the second case, since in the first case the first game would have been lost.

Also, (ODD + ODD + EVEN) OR (EVEN + ODD + ODD) won't work because you would lose in the second and the first games respectively.

Thus it MUST be (ODD + EVEN + ODD)

A: Probability of the first number being odd = P(A) = 3/5
B: Probability of the second number being even = P(B) = 2/5
C: Probability of the third number being odd = P(C) = 3/5

We want A, B and C all to occur in this event.

Thus Probability that the player loses the game on the third turn is:
P(A) * P(B) * P(C)
= 3/5 * 2/5 * 3/5

= 18 / 125 .... Ans

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(b) Accumulates 7 points and then loses on the next turn.

The sequence needs to be of the following form to continue:

ODD + EVEN + EVEN + EVEN + EVEN + ....

The moment you add an odd after the first odd number, you are going to make the total even.

So first pick the odd number, and then even numbers until you receive 7 points. Finally pick an odd to lose. Here are the possibilities:

1) 1 + 2 + 4 + odd
2) 1 + 4 + 2 + odd
3) 1 + 2 + 2 + 2 + odd
4) 3 + 4 + odd
5) 3 + 2 + 2 + odd
6) 5 + 2 + odd

Take the first case and see how the probability of it happening works out:

Probability of the first number being 1 = 1/5
Probability of the second number being 2 = 1/5
Probability of the third number being 4 = 1/5
Probability of the fourth number being odd = 3/5

Thus total probability = 1/5*1/5*1/5*3/5 = 3/625

Similarly find the probability of the next 5 cases. Then add up all 6 cases since each of them is a favourable event.

Net probability = 3/625 + 3/625 + 3/3125 + 3/125 + 3/625 + 3/125

= (15 + 15 + 3 + 75 + 15 + 75) / 3125

= 198/3125 .... Ans
~Abhay

Believe those who are seeking the truth. Doubt those who find it. -- Andre Gide
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by rash.patil » Fri Jul 09, 2010 6:42 am
b) I have a doubt in the above explanation.

Till listing of various options, it is fine.
1) 1 + 2 + 4 + odd
2) 1 + 4 + 2 + odd
3) 1 + 2 + 2 + 2 + odd
4) 3 + 4 + odd
5) 3 + 2 + 2 + odd
6) 5 + 2 + odd

but how can points 1, 2, 3 and 5 be considered? Because as soon as we get an even in the second turn, the game is lost and total points is not 7.

so only 4 and 6 are to be considered.
Hence
for case 4, probability = 1/5 * 1/5 * 3/5 = 3/125
Similarly for case 6, the probabilty = 3/125.
Total probabilty of loosing after 7 points = 2 * 3/125 = 6 / 125.
correct me if i have missed out something.
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by albatross86 » Fri Jul 09, 2010 6:49 am
Because as soon as we get an even in the second turn, the game is lost and total points is not 7.
This is incorrect. The game is as follows:

1. Draw the first ball. If it is even you lose. If it is odd you get that many points and may continue.
2. Draw the second ball. Add this number to the previous number. If total = even => Lose. Total = odd => Points are added.

Remember, the first number MUST be odd, and each subsequent number must be EVEN to ensure an ODD total and thus an accumulation of points. The next ODD number will cause the TOTAL to become even and thus the game will be lost.

You are imagining that it is 3 + 4 (lose). Points = 7
Actually it is 3 + 4 + Odd (lose). Points = 7
~Abhay

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by frank1 » Sat Jul 10, 2010 9:13 pm
ok if it appears in gmat there is no way i will be cunning my head to solve it....(the reading question took almost 1 minute and understanding took around 1.5)
so
if it infact appears...
how to make an educated guess...
any one?
GMAT score is equally counted as your GPA and 78 clicks can change you life.
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