GMAT Problem Dissection: Probability Part 2
Alright, you’ve been studying the GMAT basics. You understand what is really being tested along with the problem types, methods, and strategies. You’ve even memorized all the basic formulas and taken a dive into some of the more challenging content areas. So what’s next? That’s simple, practice…over and over and over again. If you do this, you are likely to start hitting tough probability questions at some point. Along with combinations and permutations, this is a content area that the GMAT can use to up the degree of difficulty quite a bit. So let’s take a look at a tough probability question and break it down:
A fair coin is tossed five times. What is the probability that it lands heads up at least twice?
The key to solving this is the phrase ‘at least twice.’ This means that out of our 5 tosses, two, three, four and five heads are all desired outcomes (heads). The question to ask here is: “is it easier to come up with the likelihood of getting two, three, four, or five heads or the likelihood of getting zero or one heads. Here it’s important to remember that the sum of all possible outcomes must be 1. Thus, we can find the probability that something does NOT happen and subtract that from 1, in order to find the probability it DOES happen. In this case, if we flip zero heads or one head, we will NOT have at least two heads. Finding these two probabilities, adding them together (remember, that ‘or’ becomes addition in probability) and subtracting from 1 will be the fastest way to get to the answer.
Probability means desired outcomes over possible outcomes. Thus, we will need to find each to solve this problem. Let’s look at the denominator (possible outcomes) first. Each flip has two possible outcomes: heads and tails. Just like in a GMAT permutations question when we are trying to determine the total number of codes possible or 4-digit numbers, we would multiply these individual probabilities together in order to find total possible outcomes. Therefore, there are 2x2x2x2x2 = 2^5 = 32 total possibilities.
Next we look at our numerator (desired outcomes). The first outcome we need to consider is zero heads. Only one way exists for this to happen: all of the flips come up as tails. Thus, we have ONE desired outcome in which zero heads appear. The second outcome we are looking for is exactly one head. Five outcomes would provide one head, as the head could be first, with all other flips coming up tails, the head could be second, with all other flips coming up tails, and so on.
Therefore, in total we have six outcomes that do not give us at least two heads.
When we put these together we have a 6/32, or 3/16, probability of not flipping at least two heads. Since we found the probability of what we do not want to happen, we still need to subtract our result from 1 to find the probability it does happen. The math for this is as follows:
1 – 3/16 = 16/16 – 3/16 = 13/16
13/16 is answer choice (D) and is the correct answer.