# Can someone please check this for me?

• Feb 9th 2010, 04:56 PM
bm203
Can someone please check this for me?
I am sure I know the answer to this question but it is very important that it is correct. Can someone please double check for me to make sure there are no mistakes?

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QUESTION: If three (fair) dice are rolled what is the probability of rolling at most two matching numbers? what is the probability of rolling at least two matching numbers?

ANSWER: total possible outcomes of rolling 3 dice = 6*6*6= 216

two matching pair can be achieved by rolling
[x, x, 1]
[x, x, 2]
[x, x, 3]
[x, x, 4]
[x, x, 5]
[x, x, 6]
where x = any number from 1-6

Notice that at x=1 the first set is [1,1,1] which is more than two matching numbers. likewise at x=2 the second set would be [2, 2, 2] and so on for each number, x. therefore there are 5 possibilities in this case for each number in the set.

six numbers each with five possible outcomes = 30 possible outcomes in this set

Also two matching numbers could be achieved by

[x, 1, x]
[x, 2, x]
[x, 3, x]
[x, 4, x]
[x, 5, x]
[x, 6, x]

or

[1, x, x]
[2, x, x]
[3, x, x]
[4, x, x]
[5, x, x]
[6, x, x]

again each number, x has 5 possible outcomes giving us an additional 30 outcomes for each of these two sets.
30*3 = 90
divide by total possible outcomes
90/216 = .4167

AT LEAST two matching numbers includes the possiblity of rolling triples which can occur in six ways [1, 1, 1], [2, 2, 2], [3, 3, 3] and so on
for a total probability of 6/216

adding this to the previous answer we get 6/216 + 90/216 = 96/216 = .4444
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again it is extremely important that these answers and the explanations are 100% correct. if anyone could double check the solution for me it would be greatly appreciated.

Thank you
• Feb 9th 2010, 06:25 PM
Soroban
Hello, bm203!

Let me show you a way to avoid making long lists . . .

Quote:

If three (fair) dice are rolled what is the probability of rolling:

(a) at most two matching numbers?

(b) at least two matching numbers?

There are: . $6^3 = 216$ possible outcomes.

(a) At most two matching numbers. .This means 0 matching numbers,
. . .1 matching number (whatever that means?!), or 2 matching numbers.

We do not want 3 matching numbers.
. . There are: $6$ ways to have 3 matching numbers.

Hence, there are: . $216 - 6 \,=\,210$ ways to have at most 2 matching numbers.

Therefore: . $P(\text{at most 2 matches}) \:=\:\frac{210}{216} \:=\:\frac{35}{36}$

(b) At least 2 matching numbers.
. . .This means: 2 matches or 3 matches.

For 2 matches, there are 6 choices of for the value of the Pair.
We want: $\{X,\:X,\:\text{other}\}$ in some order.
. . $P(\text{2 matches}) \:=\:6\cdot {3\choose1}\left(\frac{1}{6}\right)^2\left(\frac{5 }{6}\right) \;=\;\frac{90}{216}$

For 3 matches: there are 6 choices for the value of the Triple.
. . $P(\text{3 matches}) \;=\;6\cdot\left(\frac{1}{6}\right)^6 \;=\;\frac{6}{216}$

Therefore: . $P(\text{at least 2 matches}) \;=\;\frac{90}{216} + \frac{6}{216} \;=\;\frac{96}{216} \;=\;\frac{4}{9}$

• Feb 9th 2010, 06:30 PM
bm203
thank you. i did not mean to frame the question is such a way that would imply zero matches should be counted among the solutions. i will rephrase the first part of the question to say "at least two but not three matching numbers" or perhaps i'll simply make it a three part question.

thanks again.