Suppose we roll three fair six-sided dice. What is the probability that two of them show the same value,?

Question

but the third does not.

The answer should be 25/216.
Please show your work. Thanks.

Answer 1

Say the value we are comparing is a 6.

The chances of a six on each dice are independent of each other.

so you have P(6 AND 6 AND not 6) = P(6)*P(6)*P(not 6)

P(6) is 1/6
P(not 6) = 5/6

P(6 AND 6 AND not 6) = (1/6)*(1/6)*(5/6) = 5/216.

If it doesn't have to be the third dice thrown is not the same number then it can be the first or the second so it's 3 times more likely to happen i.e. 15/216

Answer 2

In order for the greatest number showing to be a 5, you need to throw between 1 and 5 on each of the three dice, out of 6 possible outcomes for each. So the probability is: (5/6)^3 = 5^3 / 6^3 = 125 / 216 Answer: 125/216, which is about 0.5787

Answer 3

Probability of all three dice showing the same value = 6/6 * 1/6 * 1/6 = 1/36

Probability of all three dice showing different values = 6/6 * 5/6 * 4/6 = 20/36

Probability that two dice will show the same value and the third will be different = (1 - 1/36 - 20/36) = 15/36 = 5/12

I got 5/12 and not 25/216.

Answer 4

3 dice by 6 sides 6 X 6 X 6 = 216 Number of possibilities. Requirement for to two show same number, and third not. 6 (First die number doesn't matter so 6 possibilities) X 1 (second die must have same as first - only one possibility) X 5 (Third die must not have have same as first, so 5 possibilities). 6 X 1 X 5 = 30. So answer is 30 to 216 not 25/216.
At least that's how I calculate it. Hope I explained my thinking well.

Answer 5

Hmm i got the answer of 15/216.

Suppose the common number is 1.
The probability of getting 1 on the first dice is 1/6.
The probability of getting another 1 on the second dice is 1/6.
Probability of not getting 1 on the third dice is 5/6. (get either 2,3,4,5 or 6)
Since they are independent of each other, we multiply the probabilities. 1/6*1/6*5/6=5/216
Because we have 3 groups of objects (3 dice) and 2 identical objects (both dice showing the same number), 5/216 *3!/2!=15/216

hope this helps.

Answer 6

the total number of possibilities is 6 ^3 = 216

Lets name 3 events:
A: all 3 dice are the same
B: all 3 dice are different
C: 2 dice are the same 1 is different

P(C) = 1-P(A U B)

A and B are mutually exclusive, clearly therefore: P(A U B) = P(A) + P(B)
Note: the intersetction of A and B is 0 because you can t have 3 the same and 3 different at the same time ;)

P(A) = 6/216 = 1/36
P(B) = (6*5*4)/216 = 20/36

Thus: P(C)=15/36

Answer 7

Actulally all the answers you got are wrong.
the correct answer is 15/36
explanation: We could get 2 ones and something else. This event has probability
3(1/6)(1/6)(5/6) of happenning,(multiply by 3 since any of the 2 dice could have shown a 1)
The event exactly 2 dice show the same number is 2ones and something eslse or 2 twos and something else.... or 2 sixes and something else. Ie. multiply the aboe answeer by 6, giving us 15/36
statistics teacher

Answer 8

It should be-:

1/6 x 5/6 x 5/6 = 25 / 216

i.e. 1/6 chance of rolling any number, each next dice has a 5/6 chance of not matching that particular 1st number.

Hope this helps.

Answer 9

I disagree with your answer and most of the other answers (one person has it right but they didn't reduce their fraction)...

6*C(3,2)*C(5,1) / 6^3 = 5/12

there are 6 different numbers, there are 3 different ways the 2 (C(3,2)) can arise in outcome and there are 5 ways the 3rd value can arise.

Answer 10

http://en.wikipedia.org/wiki/Probability