GMAT Probability 101, Part 5
January 24, 2011
In our GMAT Probability 101—Part 3 article, we were in the middle of discussing how to find the probability of getting at least two heads when flipping five fair, two-sided coins. In our Part 4 article, we went over combinations and permutations, which are essential to answering questions such as the one above. Specifically, we will need to use the combinations formula, n!/(k!(n-k)!)
As we went over last time, to answer this problem most efficiently we will find the probability of NOT getting at least two heads, and subtract this total from one. This means the desired outcomes are zero heads or one head.
The only way we can get zero heads is to get all tails, so there is one possible outcome for zero heads.
In order to determine the number of ways we can get exactly one head, we must use the combinations formula, using the number of flips (5) as our n value and the number of heads we want (1) as our k value. We do this because we want to select one flip to be heads. When we plug these numbers into the combinations formula we end up with a result of five. Listed out, these outcomes are HTTTT, THTTT, TTHTT, TTTHT and TTTTH (where T is Tails and H is Heads). The five outcomes that give us one head plus the one outcome that gives us zero heads make six total desired outcomes. But remember, these are the outcomes we do NOT want.
In our coin toss scenario, the total possible outcomes (denominator for our overall probability formula) can be calculated by multiplying the possible outcomes for each individual flip together. Since we have five flips and each flip has two possible outcomes, we have 2x2x2x2x2=32 possible outcomes. If six of these are outcomes we do not want, then 26 are desired. This makes the probability of getting at least two heads 26/32, which can be simplified to 13/16.
If you follow the rules in this article, along with those in the previous articles in this series, you should have the foundation you will need to succeed on GMAT probability problems.