AP Statistics - Rules of Probability

1

Students collected 150 cans for a food drive. There were 23 cans of corn, 48 cans of beans, and 12 cans of tomato sauce. If a student randomly selects one can to give away, what is the probability that the can will be either tomato sauce or beans?

$\frac{2}{5}$

$\frac{1}{2}$

$\frac{83}{150}$

$\frac{71}{150}$

$\frac{3}{5}$

Explanation

In this case, we want to know the probability of multiple, mutually exclusive possible outcomes. The possible outcomes are mutually exclusive because one can of food could not be both beans and tomato sauce. To determine the probability of the two possible outcomes, add them together and then find the least common denominator.

2

When rolling a 6 sided dice, what is the probability of rolling a 2 or a 4?

Explanation

In the single roll of a dice, rolling a 2 is mutually exclusive of rolling a 4. When a question asks for the probability of event A "or" event B, the probability is the sum of each event.

First, find the probability of each individual event.

Because the problem asks for a 2 OR a 4, add the indivual probabilities together.

3

Assume there is an election involving three parties: D, R, and I. The probability of D winning is .11, R winning is .78, and I winning is .11. What is the probability of D or R winning?

.89

.78

1

0

Explanation

Since all of the events are mutually exclusive (one of the parties must win), you can get the probability of either D or R winning by adding their probabilities.

Since the probability of D winning is .11 and R winning is .78, the probability of D or R winning is .89.

4

Assume there is an election involving three parties: D, R, and I. The probability of D winning is .11, R winning is .78, and I winning is .11. What is the probability of D or R winning?

.89

.78

1

0

Explanation

Since all of the events are mutually exclusive (one of the parties must win), you can get the probability of either D or R winning by adding their probabilities.

Since the probability of D winning is .11 and R winning is .78, the probability of D or R winning is .89.

5

Students collected 150 cans for a food drive. There were 23 cans of corn, 48 cans of beans, and 12 cans of tomato sauce. If a student randomly selects one can to give away, what is the probability that the can will be either tomato sauce or beans?

$\frac{2}{5}$

$\frac{1}{2}$

$\frac{83}{150}$

$\frac{71}{150}$

$\frac{3}{5}$

Explanation

In this case, we want to know the probability of multiple, mutually exclusive possible outcomes. The possible outcomes are mutually exclusive because one can of food could not be both beans and tomato sauce. To determine the probability of the two possible outcomes, add them together and then find the least common denominator.

6

When rolling a 6 sided dice, what is the probability of rolling a 2 or a 4?

Explanation

In the single roll of a dice, rolling a 2 is mutually exclusive of rolling a 4. When a question asks for the probability of event A "or" event B, the probability is the sum of each event.

First, find the probability of each individual event.

Because the problem asks for a 2 OR a 4, add the indivual probabilities together.

7

If you have a deck of cards, what is the probability that you draw a spade after you drew a non-spade on the first draw without replacement?

Explanation

You must use the multiplication rule which is the probability of one event happening after one has already taken place is the product of both probabilities. The probability of drawing a non-spade on the first draw is $\frac{39}{52}$ . Since there is no replacment, there are now 51 cards in the deck. The probability of drawing the spade on the second draw is $\frac{13}{51}$ . The probability of both happening after one another is then $\frac{39}{52}\ast \frac{13}{51}$ =

8

With a standard deck of cards, what is the probability of picking a spade then a red card if there is no replacement?

$\frac{13}{102}$

$\frac{17}{104}$

$\frac{19}{51}$

$\frac{23}{53}$

$\frac{11}{95}$

Explanation

In a standard deck of cards:

$P(spade)\cdot P(red)=\frac{13}{52}\cdot \frac{26}{51}=\frac{13}{102}$

9

If you have a deck of cards, what is the probability that you draw a spade after you drew a non-spade on the first draw without replacement?

Explanation

You must use the multiplication rule which is the probability of one event happening after one has already taken place is the product of both probabilities. The probability of drawing a non-spade on the first draw is $\frac{39}{52}$ . Since there is no replacment, there are now 51 cards in the deck. The probability of drawing the spade on the second draw is $\frac{13}{51}$ . The probability of both happening after one another is then $\frac{39}{52}\ast \frac{13}{51}$ =

10

With a standard deck of cards, what is the probability of picking a spade then a red card if there is no replacement?

$\frac{13}{102}$

$\frac{17}{104}$

$\frac{19}{51}$

$\frac{23}{53}$

$\frac{11}{95}$

Explanation

In a standard deck of cards:

$P(spade)\cdot P(red)=\frac{13}{52}\cdot \frac{26}{51}=\frac{13}{102}$

Rules of Probability

Help Questions

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation