smodak wrote:

Johnny the gambler tosses 6 plain dice. In order to win the jackpot, he needs a 5 or 6 on exactly three of the dice. What are Johnny's chances to win?

A. 20×(2^2/3^6)

B. 10×(2^3/3^6)

C. 20×(2^3/3^5)

D. 20×(2^3/3^6)

E. 20×(2^2/3^5)

A nice question. I think it is worth of discussing again. Here I have the official answers:

This looks like a tough nugget, but let's break it down. First of all, look at one possible arrangement (h = 5 or 6, l = 1,2,3 or 4):

h,h,h,l,l,l (h=high, l=low)

The probability of getting 5 or 6 in a die toss is 1/6+1/6 = 1/3.

The probability of getting 1,2,3 or 4 in a die toss is 2/3.

So the odds of getting that particular arrangement are:

1/3*1/3*1/3*2/3*2/3*2/3 = 2^3/3^6

This is the chance for any given arrangement of three h and three l. However, this is only one possible scenario of having high results on three rolls. How many such scenarios exist?

What you need now is the number of possible arrangements: the number of ways of picking 3 high results out of 6 tosses. Order of choice doesn't matter, because you only care about which rolls are chosen - not the order you chose them in. For example, if you chose tosses 1,2,3 to yield high results, it does not matter in what order you chose them. Now that we have reduced the problem to a simple case of choosing k=3 out of n=6, order doesn't matter, use the Combinations formula:

6!/3!*3! = 20

So the probability that you're looking for is the number of arrangements times the probability of each arrangement.

Answer D

I am happy to help.

_________________

Saving was yesterday, heat up the gmatclub.forum's sentiment by spending KUDOS!

PS Please send me PM if I do not respond to your question within 24 hours.