Land Your Score: Probability Problems Primer
On the GMAT, probability problems appear more frequently as high-difficulty questions than in low- or even medium-difficulty questions. Therefore, it should be fairly low on your priority list of content areas to brush up on. However, if you are scoring (or hoping to score) in or above the mid-600s, you should spend a little time becoming reacquainted with P.
GMAT probability problems
Probability is a stated as a percent less than 100 or a fraction less than 1; it is found by dividing the number of desired outcomes by the number of possible outcomes. So if you are tossing a coin, there are two possible outcomes. If you want heads, there is only one way to get heads (the coin lands heads up). So the probability is 1 (desired) over 2 (possible), which is 1/2, or 50%.
When multiple events occur, such as multiple coin tosses, each event adds to the total possible outcomes. For the coin example, each toss has 2 possible outcomes. So the denominator for one toss is 2, for two tosses is 4 (2 x 2), for three tosses is 8 (2 x 2 x 2), etc. To quickly calculate total possible outcomes, raise the number possible for one event to the power of the number of events. For example, the total possible outcomes for 4 coin tosses is 24= 16 possible outcomes.
If you want to know the probability of two things happening, you multiply the probabilities of the two events. So the probability of a coin landing heads up two times is 1/2 times 1/2, which is 1/4. Note that the probability of two outcomes both occurring is less than the probability of either outcome occurring alone. That’s one way to remember that you multiply to find the probability of multiple items happening; every time you multiply a fraction by a fraction, the value decreases.
Probabilities are always 100% or less (1.0 or less, if using decimals). If the probability of x occurring is 70%, the probability of x not occurring is 100% minus 70%, or 30%. Sometimes on the GMAT, calculating the probability of a desired outcome is complicated. These probability problems are usually solved much more quickly by calculating the likelihood of NOT getting the desired outcome, then subtracting that probability from 1. Here’s an example:
A fair coin is tossed 4 times. What is the probability of getting heads at least twice?
Begin approaching this probability problem by calculating the denominator, the total possible outcomes. As noted above, each toss of the coin yields 2 possible outcomes, so 2 x 2 x 2 x 2 = 16 total possible outcomes.
The four outcomes could be any combination of H (heads) and T (tails): HTHT, THTH, HHTT, TTHH, HHTH, TTHT, etc. Counting all possible ways to get two or more heads (the number of desired outcomes) would take too long on Test Day, and inexperienced test-takers will waste time doing that. Kaplan-trained test-takers, however, will recognize that the number of UNDESIRED outcomes is much easier to compute: Either no heads at all (TTTT), or heads only once (HTTT, THTT, TTHT, TTTH). That means there are 5 undesired outcomes and a probability of 5/16 for not getting the desired results.
To calculate the probability of heads at least twice, subtract the probability of NOT getting heads at least twice (5/16) from 100% (16/16). The probability of getting heads at least twice is 11/16.
You’ll be pretty well set for Test Day if you remember what probability is (desired/total), how to calculate multiple probabilities (multiply them), and the shortcut for solving difficult probability problems (subtract undesired from 1).
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