# Statistics with R

## Student Resources

# Chapter 4: Introduction to Probability

1. How many unique 7-card hands can be drawn from a deck of 104 cards?

- 18,210,410,721
- 21,243,342,120 X
- 13,378,456,038
- 22,108,543,112

**Solution:**

> choose(104, 7)

[1] 21243342120

2. The state of California has roughly 15,000,000 registered motor vehicles. Suppose that when designing new license plates, the California Department of Motor Vehicles is considered using 7 characters to be filled with 4 numbers and 3 letters. Assuming the Department has decided not to use either the letters “i” or “o” or the numbers “1” or “0” (since these can be confused with one another), how many unique license plates are possible?

- 509,607,939
- 169,869,312
- 56,623,104 X
- 18,874,368

**Solution:**

> 8 * 8 * 8 * 8 * 24 * 24 * 24

[1] 56623104

or

> (8) ^ 4 * (24) ^ 3

[1] 56623104

3. An urn has 25 marbles that are numbered 1 through 25. How many unique samples of 12 can be drawn from 25? Ignore the order of selection.

- 2,042,975
- 4,457,400
- 3,268,760
- 5,200,300 X

**Solution:**

> choose(25, 12)

[1] 5200300

4. An urn has 25 marbles that are numbered 1 through 25. Assuming the order of selection matters, how many unique samples of 12 can be drawn from 25?

- 2,490,952,020,480,000 X
- 2,740,047,222,528,000
- 2,241,856,818,432,000
- 1,992,761,616,384,000

**Solution:**

> factorial(25) / factorial(25 - 12)

[1] 2490952020480000

or

> prod(25 : 14)

[1] 2490952020480000

5. An experiment consists of a sequence of 3 steps: on the first step, a single card is selected from a deck of 52; on the second step, a coin is tossed; and on the third step, a die is cast. How many experimental outcomes are possible for the overall experiment?

- 624 X
- 588
- 644
- 528

**Solution:**

> 52 * 2 * 6

[1] 624

6. Four friends meet for dinner during a recent holiday trip to France. When the dessert menu is presented, they note that there are 7 choices possible. Assuming each person wants to have dessert, how many possible orderings are there?

- 2,268
- 2,980
- 2,660
- 2,401 X

**Solution:**

> 7 * 7 * 7 * 7

[1] 2401

7. Over the course of a recent semester, data were collected on the region-of-origin for the 40 graduate students enrolled in a statistics class: 15 are from China, 4 are from Europe, 10 are from India, 3 are from Latin America, and 8 are from the US. What is the probability that a randomly selected student is from either China or India? That is, what is p(China È India) = p(China) + p(India) – p(China Ç India)?

- 0.575
- 0.625 X
- 0.675
- 0.525

**Solution:**

> 15 / 40 + 10 / 40 - 0

[1] 0.625

8. The following data from a sample of 1,051 families show the record of university attendance by fathers and their oldest sons: in 231 families, both father and son attended university; in 326, neither father nor son attended; in 126 families, the father attended while the son did not; and in 368 families, the son attended but the father did not. What is the probability that a son attended university given that his father attended university? That is, what is p(son_yes|father_yes)?

- 0.4697
- 0.3529
- 0.5303
- 0.6471 X

**Solution:**

(231 / 1051) / (357 / 1051)

[1] 0.6470588

9. The following data from a sample of 1,051 families show the record of university attendance by fathers and their oldest sons: in 231 families, both father and son attended university; in 326, neither father nor son attended; in 126 families, the father attended while the son did not; and in 368 families, the son attended but the father did not. What is the probability that a son attended university given that his father did not attended university? That is, what is p(son_yes|father_no)?

- 0.4697
- 0.3529
- 0.5303 X
- 0.6471

**Solution:**

(368 / 1051) / (694 / 1051)

[1] 0.5302594

10. Events E_{1} and E_{2} have probabilities p(E_{1})=0.75 and p(E_{2})=0.25; for another event, E_{3}, p(E_{3}|E_{1})=0.30 and p(E_{3}|E_{2})=0.50. What are p(E_{1}|E_{3}) and p(E_{2}|E_{3})?

- p(E
_{1}|E_{3})=0.7000 and p(E_{2}|E_{3})=0.3000 - p(E
_{1}|E_{3})=0.6429 and p(E_{2}|E_{3})=0.3571 X - p(E
_{1}|E_{3})=0.7500 and p(E_{2}|E_{3})=0.2500 - p(E
_{1}|E_{3})=0.3571 and p(E_{2}|E_{3})=0.6429

**Solution:**

> ((0.75) * (0.30)) / ((0.75) * (0.30) + (0.25) * (0.50))

[1] 0.6428571

and

> ((0.25) * (0.50)) / ((0.75) * (0.30) + (0.25) * (0.50))

[1] 0.3571429