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SAT Math Skill Review: Probability & Statistics

The SAT math section will most certainly invoke your ability to calculate the likelihood (probability) of an event occurring as well as test your ability to work with basic statistical measures (mean, median, and mode). Here, we’ll review a few of the  concepts that you will need to be successful on these questions.

Probability

Probability can be any number from 0 to 1, inclusive. If an event is not possible (meaning it can never occur), the probability is 0. If an event is certain (meaning it will always occur), the probability is 1. Thus, you will never have a probability that is less than 0 or greater than 1.

The probability formula can be used to calculate the probability of an event occurring:

The probability is simply the ratio of the number of successful events divided by the total number of possible events. For example, in a deck of 52 playing cards, the probability of pulling one of the sixteen face cards is 16/52 or 4/13. In contrast, the probability of drawing one of the four jacks is 4/52 or 1/13.

Dependent and Independent Events

Calculating the probability of a single event is typically pretty simple. However, you may be asked to calculate the probability that two (or possibly more) events will both occur. In order to answer such a question, you will need to understand the difference between dependent and independent events.

Events are considered dependent if the outcome of one event affects the probability of the other event. Events are considered independent if the outcome of one event does not affect the outcome of the other event. Let’s take a look at two examples:

Example 1: Jack flips a quarter three times. What is the probability that the coin lands heads up each time?
Example 2: There are 3 white marbles and 4 black marbles in a bag. Jill takes two marbles out of the bag. What’s the probability that both marbles are white?

In the first example, the events are considered independent. The probability of a coin landing heads up is 1/2 each time Jack flips the coin. To find the probability of Jack getting a heads three times in a row would simply be: (1/2) ∙ (1/2) ∙ (1/2), or 1/8.

In the second example, the first event affects the second event. This is called conditional probability. The probability of Jill selecting a white marble the first time is 3/7 as there are 3 white marbles and 7 total marbles in the bag. When it comes time to choose the second marble, however, there are only 2 white marbles and 6 total marbles left in the bag. So the likelihood of selecting a white marble drops to 2/6 or 1/3. To find the probability of both being white, you multiply them together: (3/7) ∙ (1/3) = (1/7). Note that you assume the first event is a success in this example. This can be written as the following expression:

In the above expression, P(A) is simply the probability that event A will occur. P(B/A) is the probability that event B will occur given that event A has occurred and P(C/A,B) is the probability that event C will occur given that events A and B have occurred. You can add on as many events as you want or need.

Statistics

Problems involving statistics are relatively rare on the exam, but there is usually at least one per test. It is highly unlikely that the question(s) will cover topics beyond the three most basic statistical measures: average (arithmetic mean), median and mode.

Arithmetic Mean (Average)

The mean of a set of n numbers is just the sum of the numbers divided by n (the number of items in the set.). For example, to find the average of 15, 21 and 39, you would simply add the three together and divide by 3. You may be given an average and asked to find a missing number in a set. As long as you know the number of items in the set, this isn’t a problem. Just solve for the missing information using the following formula:

Median

The median of a set of n numbers is just the middle number in the set. If there is an even number of items in the set, then the median is the average of those two numbers. For example, let’s take a look at two sets of numbers:

Set 1: {2,3,7,8,12}
Set 2: {-1,-1/2,0,0,15,21,22,200}

In Set 1, the median is easy to find. The middle number here is 7. Remember, to find the median, the numbers in a set must be in order from smallest to largest.

In Set 2, you are given an even number of items. Thus, to find the median, you must find the average of the two middle numbers: 0 and 15. Thus, the median would be 7.5.

Mode

The mode of a set of n numbers is just the item in the set that appears most frequently. If we use the same examples from our discussion of medians, above, we say that Set 1 has no mode, meaning no item in Set 1 appears more frequently than any other. In Set 2, the mode would be 0 since it appears twice, while every other item appears once.

Examples

1. In a class of 16 girls and 13 boys, the boys averaged 75 on the last exam and the girls averaged 83. What was the average (arithmetic mean) for the class as a whole, rounded to the nearest tenth?

(A) 72.0
(B) 75.0
(C) 79.0
(D) 79.4
(E) 83.2

2. If a pair of six-sided dice, with sides numbered 1 through 6, is tossed, what is the probability that the sum of the two numbers is 4?

(A) 1/12
(B) 1/8
(C) 1/6
(D) 1/2
(E) 1

1. The correct answer is D. Before we even get to doing the math here, let’s take a look at our answer choices. Right away, you should be able to eliminate three of the five choices. The correct answer must be between 75 (the boys’ average) and 83 (the girls’ average). Choice A can’t possibly be right since it’s lower than the boys’ average. Choice E is higher than the girls’ average, so that’s out for the same reason. Choice B is the boys’ average, so that can’t be correct. Thus, even if you can’t figure out the exact math here, or you are running out of time, you could get it down to two possible choices. If deciding between C and D, choice D is the better choice since it’s closer to the girl’s average and there are more girls than boys in the class. Thus, the class average will be weighted in favor of the girls’ average. Indeed choice C is simply a straight average of the boys’ average score and the girls’ average score—a distracter choice if there ever was one! To solve the problem, we need to find a weighted average. In this case, there are 13 boys and 16 girls, so we know there are 29 total students in the class. We don’t need to know each student’s individual score, just the total of all the scores for boys and girls. We find this by multiplying the # of boys times the boys’ average score and multiplying the # of girls times the girls’ average score and then, adding them together. Divide this sum by the # of total students for the class average.

2. The correct answer is A. Right away you can eliminate choice E. A probability of 1 means that the event is guaranteed to occur. Simple logic tells us this can’t be true. You could roll the dice and get a pair of ones or a pair of fives or many other combinations that won’t add up to 4. You are still left with four possible answer choices, but four is better than five. Indeed, if you were stuck, you could still probably eliminate one more choice (choice D) as too high, and guess from the remaining three possibilities. To solve the problem however, you need to first figure out how many possible combinations there are of rolls of the dice. This will be the denominator of your fraction. Since there are 6 possible rolls of each dice, the total possible outcomes is simply 6 times 6, or 36. Next, we need to figure out what outcomes give us the desired result (a sum of four):

The table above shows us all possible pairs of rolls that will give us a sum of four. There are 3 of them, so that is our numerator. Reduce this fraction and we get our answer, choice A.

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