<p>I throw 8 dice up in the air. What is the probability 4 are even and 4 are ones?</p>
<p>My answer: </p>
<p>(1/2)^4 * (1/6)^4 * (8 choose 4)</p>
<p>Yes, that’s correct. By (8 choose 4) you mean in how many distinct ways can you select 4 objects out of 8 – i.e. 70.</p>
<p>Frankly I doubt that a question this hard would appear on a standard test like the SAT.</p>
<p>Perhaps a simpler version – such as involving two dice … probability that one would land even, and the other 1. Here either the first is even+second is 1, or the first is 1+second even. So you need to multiple the probability [(1/2] * [(1/6)] by 2. The reason for the 2 factor is that it is the number of distinct ways the event can occur.</p>