<p>The integers 1 through 6 on the six faces of a cube, one on each face. If three such cubes are rolled, what is the probability that the sum of the numbers on the top faces is 17 or 18?
(A) 1/108
(B) 1/54
(C) 1/27
(D) 1/18
(E) 1/16 </p>
<p>Please explain that question.</p>
<p>Think of rolling the die in sequence.</p>
<p>What’s the probability of a 17 total? To compute: the sum of the 3 rolls is 17 if you get a 5 on the first roll and 6 on the other two, or a 5 on the second roll and 6 on the other two, or a 5 on the third roll and a 6 on the other 2. So 3 possible ways to get a 17, each with probability 1/(6<em>6</em>6) – so for a 17 the probability is 3/108</p>
<p>In how many different ways can you get a 18 total? You must get a 6 on each of the three rolls. So there is only one way, this with probability 1/(6<em>6</em>6) = 1/108.</p>
<p>Add the probabilities for a 17 and 18.</p>
<p>You have 4/(6<em>6</em>6) = 4/108 = 1/27</p>
<p>@fogcity it’s 1/54, but thank you for making it clear to me. The flaw in your approach was that for 17 there is 3 ways, (it’s actually 1 way not 3), so is for 18. Now all I have to do is calculate it like that: 1/108 + 1/108 so it yields 1/54. Thank you again!</p>
<p>Actually, the flaw is in your multiplication. 6<em>6</em>6 = 216, not 108. 4/216 = 1/54</p>
<p>there are four possible outcomes for
1-6,6,6
2-6,6,5
3-6,5,6
4-5,6,6
Now that we got what we wanted we just have to divide it to all the possible outcomes
since there are three dice all the outcomes will be 6<em>6</em>6=216
4/216=1/54
hope this helps</p>
<p>^ ^<br>
Oh, it makes sense, thank you guys!</p>