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AP Statistics : Statistical Patterns and Random Phenomena

Study concepts, example questions & explanations for AP Statistics

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Example Questions

Example Question #1 : Statistical Patterns And Random Phenomena

The probability that it will rain today is 0.35. What is the probability that it will not rain?

Possible Answers:

0.35

0.5

0.65

Correct answer:

0.65

Explanation:

The answer is 0.65 because Pr(~Rain) is the complement of Pr(Rain) and both events are mutually exclusive.

When two events are mutually exclusive, $Pr(A\ and\ B)=Pr(A)+Pr(B)$ . Since probabilities must sum up to 1, this implies that Pr(B)=1-Pr(A) .

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Example Question #2 : Statistical Patterns And Random Phenomena

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?

Possible Answers:

.78

.89

Correct answer:

.89

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.

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Example Question #62 : Ap Statistics

If 1 card is chosen at random from a deck of cards, what is the probability that it will be a heart or a king?

Possible Answers:

17/52

29/52

13/52

26/52

4/13

Correct answer:

4/13

Explanation:

In a deck of cards, there are 52 total cards, 13 hearts, 4 kings, and 1 king that is a heart.

So, 13/52 + 4/52 - 1/52 = 16/52 = 4/13.

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Example Question #61 : Ap Statistics

Using a standard deck of cards, what is the probability of choosing a single face card?

Possible Answers:

$\frac{19}{26}$

$\frac{7}{13}$

$\frac{9}{13}$

$\frac{7}{26}$

$\frac{3}{13}$

Correct answer:

$\frac{3}{13}$

Explanation:

There are cards in a standard deck: cards in suits. There are face cards (King, Queen, Jack) in each suit, so there are total face cards.

Thus the probability of choosing a single face card is $\frac{12}{52}$ or $\frac{3}{13}$ .

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Example Question #62 : Ap Statistics

A student randomly selected a highlighter from her desk. There were five highlighters on the desk, each of a different color--blue, green, yellow, red, and orange. What is the probability that the student selected either the red or the yellow highlighter?

Possible Answers:

4/5

2/5

3/5

1/5

Correct answer:

2/5

Explanation:

In this case, we want to know the probability of multiple, mutually exclusive possible outcomes. To determine the probability of the two possible outcomes, simply add them together. This is called the addition rule.

1/5+1/5=2/5

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Example Question #64 : Ap Statistics

What is the probability of obtaining at most heads when tossing a fair coin times?

Possible Answers:

$p = \frac{56}{1024}$

$p = \frac{17}{20}$

$p = \frac{45}{1024}$

$p = \frac{3}{20}$

$p = \frac{1013}{1024}$

Correct answer:

$p = \frac{56}{1024}$

Explanation:

First find the number of ways to get 0, 1, or 2 heads.

Remember, $\binom{n}{k}=\frac{n!}{k!(n-k)!}$ .Also, $\binom{n}{0}=1$ and $\binom{n}{1}=n$ .

heads: $\binom{10}{0} = 1$

head: $\binom{10}{1} = 10$

heads: $\binom{10}{2} = 45$

$\textup{number of possible outcomes of 10 tosses}= 2^{10} = 1024$

$p\left ( x\right )=\frac{\textup{number\ of\ desired\ outcomes}}{\textup{number\ of\ all\ possible\ outcomes}}$

$p(\textup{0 heads in 10 coin tosses}) = \frac{1}{1024}$

$p(\textup{1 head in 10 coin tosses}) = \frac{10}{1024}$

$p(\textup{2 heads in 10 coin tosses}) = \frac{45}{1024}$

Now combine these to find the probability of seeing at most 2 heads in 10 coin tosses:

$p(\textup{at most 2 heads in 10 coin tosses})$

$=p(\textup{0 heads})+p(\textup{1 head})+p(\textup{2 heads})$

$=\frac{1}{1024} +\frac{10}{1024} +\frac{45}{1024}$

$=\frac{56}{1024}$

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Example Question #3 : Statistical Patterns And Random Phenomena

150 students are athletes at the school. 65 play baseball, 15 play basketball, and 10 play both basketball and baseball.

What is the probability that a randomly selected athlete will play either baseball or basketball or both sports?

Possible Answers:

8/15

80/150

3/5

7/15

65/150

Correct answer:

7/15

Explanation:

We want to know the probability of multiple possible outcomes that are not mutually exclusive. To do this, we use the addition rule with one step that we would not use if the possible outcomes were mutually exclusive. Add the probabilities of each possible outcome, subtract from that sum the number counted twice, then reduce the answer to the least common denominator.

65/150 + 15/150 - 10/150= 70/150 = 7/15

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Example Question #2 : Probability

At a business conference, one participant was selected to give a presentation. There were total participants. had business degrees and did not. were entrepreneurs and were not. of those with business degrees were entrepreneurs. What is the probability that the person selected will have a business degree or be an entrepreneur?

Possible Answers:

2/15

11/15

13/15

2/3

1/3

Correct answer:

2/3

Explanation:

Add the probabilities of each possible outcome, subtract from that sum the number counted twice, then reduce the answer to the least common denominator. In our case, we are told that 2 of those who have a business degree are also entrepreneurs therefore we need to subtract to from the total to get the correct probability.

7/15+5/15=12/15

12/15-2/15=10/15

10/15=2/3

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Example Question #1 : Probability

In a standard deck of playing cards, what is the probability that the first card drawn is a face card (a king, queen or jack) or a diamond?

Possible Answers:

$\frac{25}{52}$

$\frac{3}{52}$

$\frac{3}{13}$

$\frac{11}{26}$

$\frac{1}{4}$

Correct answer:

$\frac{11}{26}$

Explanation:

Bear in mind how the addition rule applies here. This will be a lengthy application, but applied properly, it will see you through this problem. It helps to think of this as four events that are not mutually exclusive - drawing a queen, drawing a jack, drawing a king or drawing a diamond. We then realize that we have to avoid counting the king, queen and jack of diamonds twice.

$P(face \bigcup diamond)$

$P(king \bigcup queen \bigcup jack \bigcup diamond)$

$P(king) + P(queen) + P(jack) + P(diamond) - P(diamond \bigcap face)$

$\frac{1}{13} + \frac{1}{13} + \frac{1}{13} + \frac{1}{4} - \frac{3}{52}$

$\frac{4}{52} + \frac{4}{52} + \frac{4}{52} + \frac{13}{52} - \frac{3}{52} = \frac{22}{52} = \frac{11}{26}$

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Example Question #2 : Statistical Patterns And Random Phenomena

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?

Possible Answers:

$\frac{2}{5}$

$\frac{83}{150}$

$\frac{3}{5}$

$\frac{1}{2}$

$\frac{71}{150}$

Correct answer:

$\frac{2}{5}$

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.

12/150 + 48/150 = 60/150 = 2/5

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AP Statistics : Statistical Patterns and Random Phenomena

Study concepts, example questions & explanations for AP Statistics

All AP Statistics Resources

Example Questions

Example Question #1 : Statistical Patterns And Random Phenomena

Example Question #2 : Statistical Patterns And Random Phenomena

Example Question #62 : Ap Statistics

Example Question #61 : Ap Statistics

Example Question #62 : Ap Statistics

Example Question #64 : Ap Statistics

Example Question #3 : Statistical Patterns And Random Phenomena

Example Question #2 : Probability

Example Question #1 : Probability

Example Question #2 : Statistical Patterns And Random Phenomena

All AP Statistics Resources

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