<p>A coin is tossed three times. Given that at least one head appears, what is the probability that exactly two heads will appear?
Barron's solves this by using a sample space. Is there any way I can solve it without having to list the possibilities? Imagine the numbers were bigger...</p>
<p>What is the probability of getting at least three heads when flipping four coins?</p>
<p>How do you solve these?
Answers, btw, are: 3/7, 5/16 (respectively).</p>
<p>A sample works but you can reason the following way:</p>
<p>For the first one, there are 8 total possibilities because 2^3 = 8
Given that at least one head appears means you have to subtract the case in which all tails appear, so since there is one case of all tails, there are 7 possibilities to consider. Now within these 7 possibilities, given that two heads appear means there is exactly one tail appearing, and there are 3 possibilities for the 1 tail, either the first flip, second flip, or third flip. So the answer is 3 out of 7. </p>
<p>Similar reason for the second one. Four coins means 16 possibilities since 2^4=16. Getting at least 3 heads means it can be all four heads (1 possibility) or the cases for exactly three heads. Exactly three heads means exactly one tails, and there are 4 possibilities for the one tail, either the 1st flip, 2nd flip, 3rd flip, and 4th flip. So There are (4+1) / 16 or 5/16 probability total.</p>
<p>If the question involved larger numbers, they would have to choose numbers that lend themselves to the same type of reasoning or else it would be a huge sample space or messy combinatorics. A question like what is the probability of getting at least 7 heads flipping 8 coins? Similar procedure.</p>