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Probability Theory Only....No Combinatorics Please

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by knight247 » Thu Oct 06, 2011 10:54 pm
4 dices are thrown at the same time. What is the probability of getting ONLY 2 dice showing the same face?

Total number of possible outcomes is 6^4= 1296

We could have Same Same Different1 Different2 which can be arranged in 4!/2!=12 Ways

For the first dice---> 6 Outcomes
For the second dice--> 1 Outcome
For the third dice---> 5 Outcomes
For the fourth dice--> 4 Outcomes

Desired outcomes=6*1*5*4*12= 1440

The desired outcomes are greater than that the possible outcomes. Why is the extra *2 coming? What am I doing wrong? No combinatorics please, as I'm able to solve this using combinatorics. Thanks.
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by shankar.ashwin » Thu Oct 06, 2011 11:07 pm
Hi, I think you have considered order twice here.
The 6*1*5*4 part is correct and already includes ordering.

And from here any 2 could repeat, so we have 4C2, which is 6.

So 6*6*1*5*4
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by GmatMathPro » Fri Oct 07, 2011 9:18 am
Agreed. It's the same as the mistake you were making on the MATHEMATICS question.

6*1*5*4=120 will count cases like

2265 AND 2256. But when you multiply by 12 to get the different arrangements, you're essentially saying, "2265 can be arranged 12 different ways, and 2256 can be arranged 12 different ways" But these are identical groups. We don't want to count them twice. This will happen with every unique sequence, hence your answer is two times too big.

Another way: represent the out come with 4 blanks: _ _ _ _ . We have to choose two slots to put the matching dice in. 4 choices for the first, 3 choices for the second: 4*3=12. Divide by 2 to eliminate duplicates: 12/2=6.

We have the following possibilities with S representing the ones that are the same:

S S _ _
S _ S _
S _ _ S
_ S S _
_ S _ S
_ _ S S

6 choices for the format. 6 choices for the value of S. 5 choices for the first blank. 4 choices for the second blank: 6*6*5*4=720
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