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dghosh2602
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Guide 4 - Probability - Chapter 12 - Page 191

by dghosh2602 Sun May 23, 2010 11:40 pm

A gambler rolls three fair 6 sided dice. What is the probability that 2 of the dice show the same number but the third shows a different number?

I don't see the flaw in my reasoning:

Lets say the gambler played the dice the first time: 1/6 is the probability for any number to come up, then for the second to match there is 1/6 again and the third not to match, its 5/6.

So the probability for the first chain of events becomes 1/6 * 1/6 * 5/6.

Now to figure out how many such chains exist:
Either the 1st and 3rd are same or 1st and the 2nd are the same or 2nd and the 3rd are the same, which gives us 3 possibilities.
This is also confirmed by the distinguishability logic: So if we are looking for all possibilities of arranging the alphabets: S, S and D => its 3! / 2! = 3.

So the total probability becomes: 1/6 * 1/6 * 5/6 * 3. But the answer is 15/36.

I don't understand the flaw in this reasoning. Please help! Thanks...
jthomasma
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Re: Guide 4 - Probability - Chapter 12 - Page 191

by jthomasma Mon May 24, 2010 6:40 am

I thought of this problem this way.

We roll the first two dice and end up with 1,1
The third die ends up anything other than 1 [2:6]
5 possible outcomes
1,1,2
1,1,3
1,1,4
1,1,5
1,1,6

You have 5 outcomes for each duplicate digit start:

2,2, x where x not = to 2 [5 solutions]

So we have 30 outcomes. With matching 1st and 2nd die.

Then as you mentioned order is not important so we have 2 additional sets of 30 outcomes

So we have 90 outcomes that satisfy x,x,y where y not = x

Total outcomes 6 x 6 x 6

90:216

reduces to 15:36

As far as your reasoning I think you start off with the wrong foot

The first die really doesn't matter, it is the second die that must match the first. Using your method you should start

6/6 x 1/6 x 5/6

In words it does not matter if I get a 1 or a 6 on the first die, it is the second die that must match what I receive on the first die.
tim
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Re: Guide 4 - Probability - Chapter 12 - Page 191

by tim Tue Jun 01, 2010 2:09 pm

Perfect explanation! Yes, the original poster neglected to use the fact that the first die roll can be anything, and then the constraints are applied only to the second and third dice..
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saurabhbanerjeeiimk
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Re: Guide 4 - Probability - Chapter 12 - Page 191

by saurabhbanerjeeiimk Mon Dec 26, 2011 9:58 am

Q1.

While calculating the probability, why are we EXCLUDING the first dice in each of the three possibilities?

Q2 Why are we not considering the first dice in the final answer?

Q3. I went through the alternate explanation on the coloured dice, which shows the calculation as 1x 1/6x 5/6 for Dice1, Dice2 and Dice 3 respectively. Are we designating "1" to "Dice 1" because the outcome of "Dice 1" could be ANY number (6/6=1), and hence the probability of "Dice 2" and "Dice 3" is based on "Dice 1" probability. And if that is the case, is that why "1" has not been mentioned in the original explanation, because 1X either of the three possibilities = either of the possibilities.

I.e. Possibility 1= 1X1/6X5/6= 5/36
Possibility 2= 1X5/6X1/6= 5/36
Possibility 3= 1X5/6X1/6 = 5/36
Which is 3x5/36 = 15/36

Thanks
tim
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Re: Guide 4 - Probability - Chapter 12 - Page 191

by tim Sat Dec 31, 2011 1:50 pm

All three of your questions appear to be the same question. The reason we exclude the first die is that anything works. In other words, the first die is acceptable with 100% probability. If you don’t feel comfortable ignoring the first die entirely, you can factor it into your calculations by multiplying the 100% by what you get for the other dice. So in other words the discussion you bring up in question 3 is perfectly valid..
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