Data tables may accompany a variety of question types on the SAT, including linear equations, functions, and statistics questions that involve mean or median calculations. But the question type most likely to be accompanied by a table is probability. Remember that probability is an expression of a situation’s desired outcomes (the outcomes asked about in the question) divided by the total outcomes possible. This means that probability is expressed as a fraction or decimal, with representing probability. (If the test is making an exception and expressing probability as a percent, that change will be explicit in the language of the question.) Watch out for tables in all sorts of data-based questions, but probability in particular.
A tournament matched North American versus South American wrestlers, 45 on each squad. Each wrestler had a choice of whether to wear the dark-colored or the light-colored singlet representing his continent. The results are listed below.
Continent Singlet Color Dark Light Total North America 18 27 45 South America 16 29 45 Total 34 56 90 If a wrestler who chose a dark singlet is selected at random, what is the probability that wrestler was from South America?
A.
B.
C.
D.
The introduction laid out the definition of probability. To state it as a formula,
probability = desired outcomes /total outcomes.
One helpful strategy when assessing probability is this: start with the total outcomes–in other words, the denominator of the fraction. One benefit of this strategy is that it will assist your UnCLES-driven attention to detail about precisely what is being asked. One way to go wrong in this question is to think that all wrestlers are in view as the denominator of the fraction. But the random selection takes place from only those who wore the dark singlet. The “Total” row tells us that this number is .
What about the numerator? The “desired outcomes” in this case is those who are from South America. Remember that we are only looking in the dark singlet column, so the total is not or ; rather it is . We know that the fraction is ; we simply need to reduce it to match and answer. Dividing top and bottom by shows us the answer is .
Notice the wrong answers and how they might be mistakenly selected. If you counted North American wrestlers with the dark singlet ( of them), you might have answered ( out of ) or ( out of ). If you had the right numerator () but used as your denominator, you would have answered . Attention to detail is the watchword for questions of probability!
You have encountered the concept of frequency when learning about finding the median of a set of data. You will see it again here; if you carefully observe the Frequency column and understand that it shows how many elements of a certain kind of number are found in the set, questions about frequency should feel fairly straightforward.
A randomizer is programmed to randomly name one of the 50 U.S. States. If the states whose name begins with a “C” are California, Colorado, and Connecticut, what is the probability the randomizer will produce the name of a state starting with “C”?
A.
B.
C.
D.
The answer is . This question is about the definition of probability. Following the recommended strategy and starting with the total, we establish (for 50 states) as the denominator. The “desired outcomes” are just those states that start with “C”: of them. (Hint: if you are uncertain which number should be the numerator and which the denominator, follow this simple rule: with probability, the smaller number goes in the numerator.) The most tempting wrong answer here is probably - arrived at if you mistakenly identify the “C” states as the total outcomes and think the numerator is because only one state is selected. This is a fairly common error, so be sure you understand why it doesn’t work.
| Dog Breed | Average Weight (pounds) | |
| Female | Male | |
| Alaskan Malamute | 78 | 87 |
| Bernese Mountain Dog | 93 | 96 |
| Chinook | 73 | 73 |
A shelter has dogs for adoption, featuring six in particular. The table shows the weight of these six, three males and three females, one of each of three different large dog breeds. If a random dog is selected from among those weighing less than pounds, what is the probability that the dog is a Chinook? (Enter your answer as a fraction or decimal.) (Note: this is a free-response question.)
The answer is or . Two details make this problem less straightforward than the previous question. 1) The denominator is not all six dogs, but “those weighing less than pounds.” Five of the six dogs weigh less than , so our denominator is . The number, meanwhile, is not simply for the one dog selected, but rather must be . Why ? Notice that the question doesn’t specify whether the Chinook that is the desired outcome is male or female, so we have to consider both. Hence our answer. The decimal equivalent, , is also acceptable.
| U.S. President’s first letter of last name | Frequency |
| A | 3 |
| B | 4 |
| C | 4 |
| E | 1 |
| F | 2 |
| G | 2 |
| H | 5 |
| J | 4 |
| K | 1 |
| L | 1 |
| M | 3 |
| N | 1 |
| O | 1 |
| P | 2 |
| R | 3 |
| T | 5 |
| V | 1 |
| W | 2 |
The table above shows the frequency of each letter that occurs as the first letter of the last name of one of the 45 people who have served as president of the United States. If one letter from the list is randomly chosen, what is the probability that it is either the letter most represented or the letter (tied for) least represented?
A.
B.
C.
D.
The answer is . As is often the case, the most challenging part of this problem is identifying and distinguishing the numerator and denominator. We start with the denominator as usual: what number represents the total outcomes in this case? You might think , since the question says that people have been president. But the key phrase here is, “one letter from the list …” Not only are we only focusing on letters of the alphabet, but it’s also not all the letters of the alphabet (8 are not represented). So the total number for the denominator is .
The numerator, meanwhile, is not a simple discovery. We have to count the letters that have either the most or the least representation. The “most” is a tie between T and H, so that’s possibilities. Meanwhile, six letters (E, K, L, N, O, and V) all tie for least represented, at one president. That’s a total of more letters, for a total of . That gives us , a fraction which reduces to .
| Row # in Transition Metals | Electronegativity | ||
| 1.0-1.5 | 1.6-2.0 | 2.1-2.5 | |
| 1 | 4 | 2 | 2 |
| 2 | 6 |
5 | 4 |
| 3 | x-3 |
x |
x+1 |
The table above shows the electronegativities of the first three rows of transition metals in the periodic table. If an element is chosen at random from Row 3, the probability that its electronegativity is between 1.0 and 2.0, inclusive, is . What is the value of x? (Note: this is a free-response question.)
The answer is 3. At its most difficult, the SAT may introduce variables into a probability scenario, leading the student to create an equation in order to solve (especially when there is no multiple choice, so no possibility of working backwards from the answers). As usual, we want to start with the total outcomes to find our denominator; we are helped by the language of “chosen at random from Row 3.” So we can ignore the first two rows; the problem wants us to consider the cells containing algebraic expressions. Since we want the total of this row, we have to add the expressions: .
Meanwhile, the phrasing “the probability that …” gives us the numerator again, but in this case we are supplied with the numerator rather than asked for it. Be careful: we have to add two different cells to get a range of to ; in Row 3, those two cells are and , so the total desired outcomes is .
What now? We can’t simplify the fraction , but we don’t have to if we pay attention to all the information. We are instead told to set this fraction equal to to , because that is the probability represented by our algebraic fraction. When we set them equal and cross-multiply, we get:
| 9 or 10 | 7 or 8 | 5 or 6 | 3 or 4 | 1 or 2 | Total | |
| Part 1 | 6 | 11 | 4 | 1 | 3 | 25 |
| Part 2 | 8 | 9 | 5 | 2 | 1 | 25 |
| Part 3 | 11 | 8 | 4 | 2 | 0 | 25 |
| Total | 25 | 28 | 13 | 5 | 4 | 75 |
The table above shows the scores a teacher gives to three-part presentations given by her class of 25 students, one three-part presentation per student. If a part of a presentation is selected at random, what is the probability that either Part 2 or Part 3 received a score of 7 or 8, given that the presentation received a score of 7 or 8 on exactly one of its three parts?
A.
B.
C.
D.
The answer is . The difficulty advances here because of the language used to express the total outcomes in this case: “given that.” If you think about it, “given that” is telling you what all the possibilities are; you are called upon to assume that total number of possibilities and then use other information in the question to establish the number of desired outcomes. If we want to find a presentation that scored a or exactly once, we have to count the total number of instances when a or was given. The bottom of the “7 or 8” column tells us that the number is . Notice that only two answer choices appear possible already!
Where do we look for our desired outcomes? A dependable place is after the phrase “what is the probability of …” That sentence shows us that we’re supposed to focus only on Part 2 and Part 3 of the presentation (but remember we’re still looking only at the “7 or 8” column). That’s and , respectively, for Part 2 and Part 3, so the total desired outcomes is . The trap answer of comes from adding Parts 1 and 3 instead, while the wrong answer of would be arrived at by mistakenly using all 75 scores as the total instead of the narrower amount this question prescribes.
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