When students, even those who consider themselves strong in math, get to the final two problems of the SAT, many begin to sweat like they are about to embark on some epic journey from which they may never return. The hard probability problem makes students very uncomfortable, but in reality most harder math problems simply require one or two more steps than less difficult problems. Probability questions are actually some of the simplest to solve.
The “Hard” Probability Question
All probability questions are the same. The definition of probability is the number of desired outcomes divided by the total number of outcomes (sometimes this is multiplied by 100 to come up with a percent probability). Every problem requires finding the total number of possibilities, the desired possibilities, or both.
Here is an example:
“Four paintings, A, B, C, and D are being hung in four adjacent display cases. What is the probability that paintings A and D will be either first or last?”
At first glance this may seem intimidating, but it is just like any other probability question. The first step is to figure out the total number of possible outcomes. This is solved by thinking about this like a counting problem. If there are four “slots” that these painting could occupy let’s imagine that each slot has a number of possible paintings that could occupy it.
For the first slot, the number of possible paintings that could occupy it is four since there are four paintings. For slot two, the number of possible paintings is three because one painting is already in slot one. This may seem difficult to understand at first, but remember that these are the number of possibilities to choose from in each slot, so there will be one less painting to choose from in slot two because one possibility is gone. For slot three, it is one fewer painting still, and there will be just one painting left by the time the fourth is selected. In order to find the total number of possible outcomes, the possibilities in each slot must simply be multiplied together: 4 x 3 x 2 x 1 = 24. This process is the same for any problem where possibilities are calculated based non repeating possibilities on discreet “slots”.
Half the problem is done (total number of possibilities); now, all that is left is to find the number of desired possibilities. For this part of the problem we can calculate desired possibilities in a very similar way as we did above. The only tricky part is this calculation will require two steps.
Imagine painting A is selected to go in slot one (one possibility for the desired outcomes). If A is in slot one, then in order for A and D to be either first or last, D will have to go in the last slot. This leaves just 2 paintings to choose from in slot two and only one in slot three. The number of possibilities when A is 1st then is 1 x 2 x 1 x 1 = 2. The other possibility is for D to be first which leads to similar constraints. Painting A must be last, two possibilities are left for slot two, and just one is left for slot three. The total number of possibilities when D is first is also 1 x 2 x 1 x 1 = 2. The total desired outcome is all the possibilities when A is first (2) added to all the possibilities when D is first (also 2), which leaves a total desired outcomes of four.
With the total desired outcomes and the total possible outcomes found, the final step is to create a fraction with desired outcomes on the top and total outcomes on the bottom or 4/24 which reduces to 1/6.
Hard probability problems and counting problems, like most hard problems on the SAT, are not really “hard”. The main thing to keep in mind is the technique that is discussed above for calculating possibilities in different contexts. If students can master this technique and remember the definition of probability (desired outcomes over total outcomes) the hard probability problem becomes a piece of cake. Happy Studying!
David Greenslade is a Veritas Prep SAT instructor based in New York. His passion for education began while tutoring students in underrepresented areas during his time at the University of North Carolina. After receiving a degree in Biology, he studied language in China and then moved to New York where he teaches SAT prep and participates in improv comedy. Read more of his articles here, including How I Scored in the 99th Percentile and How to Effectively Study for the SAT.