# SAT Counting and Probability

To successfully solve problems about counting and probability on the SAT, you need to know:

- the rule of sum, when counting
- how to count integers in a range
- the rule of product
- how to find the probability of equally likely outcomes
- how to find 1-dimensional and 2-dimensional geometric probabilities

## Examples on Counting

\(X\) = number of integers between 45 and 75, 45 and 75 not included

\(Y\) = number of integers between 45 and 75, 45 included

\(Z\) = number of integers between 45 and 75, 45 and 75 includedWhich of the following is the value of \(X + Y + Z?\)

(A) \(\ \ 60\)

(B) \(\ \ 87\)

(C) \(\ \ 90\)

(D) \(\ \ 93\)

(E) \(\ \ 100\)

Correct Answer: C

Solution 1:

Tip: If \(a<b\) are two integers, the number of integers between \(a\) and \(b\) when one endpoint is included is \(b-a.\)

From the givens, we know that \(X\) includes one of the endpoints and \(Z\) excludes one of the endpoints. So, \(X\) counts one less integers than \(Y,\) whereas \(Z\) counts one more integers than \(Y.\) Since That is, \(X = Y - 1\) and \(Z = Y + 1.\) It follows that \(X + Z = 2Y.\) Since \(Y = 75-45 = 30,\) we get \(X + Y + Z = 30 + 2\cdot 30 = 90.\)

Solution 2:

Tip: If \(a<b\) are two integers, the number of integers between \(a\) and \(b,\) endpoints NOT included is \(b-a-1.\)

Tip: If \(a<b\) are two integers, the number of integers between \(a\) and \(b\) when one endpoint is included is \(b-a.\)

Tip: If \(a<b\) are two integers, the number of integers between \(a\) and \(b\) when both endpoints are included is \(b-a+1.\)\(X = 75-45-1 = 29\)

\(Y = 75-45 = 30\)

\(Z = 75-45+1 = 31\)So, \(X + Y + Z = 29 + 30 + 31 = 90.\)

Incorrect Choices:

(A)

This answer is just offered to confuse you.

(B)

If you think that \(X, Y,\) and \(Z\) are equal to \(75-45-1=29,\) you will get this wrong answer.

(D)

If you think that \(X, Y\) and \(Z\) are equal to \(75-45+1=31,\) you will get this wrong answer.

(E)

This answer is just offered to confuse you.

## Examples on Probability

Two fair dice are rolled. What is the probability that the sum of the numbers on top of the two dice is 8?

(A) \(\ \ \frac{1}{12}\)

(B) \(\ \ \frac{1}{9}\)

(C) \(\ \ \frac{1}{8}\)

(D) \(\ \ \frac{5}{36}\)

(E) \(\ \ \frac{31}{36}\)

Correct Answer: D

Solution 1:

Tip: Assuming that all the possible outcomes of an event \(A\) are equally likely, the probability that \(A\) will occur is \(P(A) = \frac{\text{ # of favorable outcomes}}{\text{total # of outcomes}}.\)

When a die is rolled, the number on the top face can be an integer from 1 to 6. Let's think of the outcome of rolling both dice as an ordered pair \((x,y),\) where \(x\) is the number displayed on the first die, and \(y\) is the number displayed on the second die.Because there are two dice, there are \(6\cdot 6 = 36\) possible outcomes when rolling both of them, and therefore 36 possible sums.

There are 5 ways in which we could get the sum 8, represented by the ordered pairs \((2, 6), (3, 5), (4,4), (6, 2), (5, 3).\)

So, \(P(\text{sum of two top numbers is 8}) = \frac{5}{36}.\)

Solution 2:

Tip: Assuming that all the possible outcomes of an event \(A\) are equally likely, the probability that \(A\) will occur is \(P(A) = \frac{\text{ # of favorable outcomes}}{\text{total # of outcomes}}.\)

If we think of the possible outcomes from rolling both dice as pairs of numbers, then all the possibilities are:\[\begin{array}{c c c c c c c} (1, 1) & (1, 2) & (1, 3) & (1, 4) & (1, 5) & (1, 6)\\ (2, 1) & (2, 2) & (2, 3) & (2, 4) & (2, 5) & \boxed{(2, 6)}\\ (3, 1) & (3, 2) & (3, 3) & (3, 4) & \boxed{(3, 5)} & (3, 6)\\ (4, 1) & (4, 2) & (4, 3) & \boxed{(4, 4)} & (4, 5) & (4, 6)\\ (5, 1) & (5, 2) & \boxed{(5, 3)} & (5, 4) & (5, 5) & (5, 6)\\ (6, 1) & \boxed{(6, 2)} & (6, 3) & (6, 4) & (6, 5) & (6, 6)\\ \end{array}\]

We have boxed up the outcomes that yield a sum of 8. Out of 36 possibilities, there are 5 such outcomes.

So, \(P(\text{sum of two top numbers is 8}) = \frac{5}{36}.\)

Incorrect Choices:

(A)

If you consider only the pairs \((2, 6), (3, 5), (4, 4)\) or \((6, 2), (5, 3), (4, 4)\), instead of \((2, 6), (3, 5), (4,4), (6, 2), (5, 3),\) you will get this wrong answer.

(B)

If you forget to account for the event (4,4), you will get this wrong answer.

(C)

This wrong choice is offered to confuse you. It uses the number in the prompt.

(E)

This is the probability that the sum is NOT 8.

A hat contains tickets marked 1, 2, ..., n. A ticket is drawn from the hat, and then, without the first ticket being replaced, a second ticket is drawn from the hat. What is the probability that the first ticket drawn is the number 1 and the second ticket drawn is the number 3?

(A) \(\ \ \frac{1}{n}\)

(B) \(\ \ \frac{1}{n-1}\)

(C) \(\ \ \frac{1}{n^2 -n}\)

(D) \(\ \ \frac{2n-1}{n^2-n}\)

(E) \(\ \ \frac{1}{n^2}\)

## Review

If you thought these examples difficult and you need to review the material, these links will help:

## SAT Tips for Counting and Probability

- If \(a<b\) are two integers, the number of integers between \(a\) and \(b\) when one endpoint is included is \(b-a.\)
- If \(a<b\) are two integers, the number of integers between \(a\) and \(b\) when both endpoints are included is \(b-a+1.\)
- If \(a<b\) are two integers, the number of integers between \(a\) and \(b,\) endpoints NOT included is \(b-a-1.\)
- If there are \(n\) ways for an event to happen and \(m\) ways for another event to happen, then the number of ways for both events to happen is \(m\cdot n.\)
- If \(P(A)\) is the probability that event \(A\) will occur, then \(0 \leq P(A) \leq 1.\)
- If \(P(B)\) is the probability that event \(A\) does NOT occur, then \(P(A) = 1- P(B).\)
- Assuming that all the possible outcomes of an event \(A\) are equally likely, the probability that \(A\) will occur is \(P(A) = \frac{\text{ # of favorable outcomes}}{\text{total # of outcomes}}.\)
- Two events are independent if the outcome of one does not affect the outcome of the other.
- If events \(A\) and \(B\) are independent, then \(P(A\ \text{and}\ B) = P(A) \cdot P(B).\)
- If events \(A\) and \(B\) are mutually exclusive, then \(P(A\ \text{or}\ B) = P(A) + P(B).\)
- If a point is chosen at random in a geometric figure, the probability that the point lies in a particular region is: \(\frac{\text{area of region}}{\text{area of whole figure}}.\)
- SAT General Tips

**Cite as:**SAT Counting and Probability.

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