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Data Session 8: Solutions
 
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Solutions for Session 8, Part C

See solutions for Problems: C1 | C2 | C3 | C4 | C5 | C6 | C7 | C8 | C9 | C10 | C11


Problem C1

Two branches of the tree end with one head out of two tosses (HT and TH), and only one branch ends with zero heads (TT). Therefore, it is more likely to get one head than no heads.

<< back to Problem C1

 

Problem C2

Here is the tree diagram for three tosses of a fair coin:

<< back to Problem C2

 

Problem C3

Here is the completed probability table:

Number of Heads

Frequency

Probability

0

1

1/8

1

3

3/8

2

3

3/8

3

1

1/8

<< back to Problem C3

 

Problem C4

Here is the completed probability table:

Number of Heads

Frequency

Probability

0

1

1/16

1

4

4/16

2

6

6/16

3

4

4/16

4

1

1/16

<< back to Problem C4

 

Problem C5

No, it would not be feasible to plot the necessary branches. There would be a total of 210 = 1,024 branches to this tree diagram, and it would be far too cumbersome to count all the outcomes.

<< back to Problem C5

 

Problem C6

The fifth row is 1, 5, 10, 10, 5, 1, and the sixth row is 1, 6, 15, 20, 15, 6, 1, as shown below:

<< back to Problem C6

 

Problem C7

Here are the completed probability tables:

Five Tosses

Number of Heads

Frequency

Probability

0

1

1/32

1

5

5/32

2

10

10/32

3

10

10/32

4

5

5/32

5

1

1/32

Six Tosses

Number of Heads

Frequency

Probability

0

1

1/64

1

6

6/64

2

15

15/64

3

20

20/64

4

15

15/64

5

6

6/64

6

1

1/64

<< back to Problem C7

 

Problem C8

If you say that P = the number of possible outcomes, and n is the number of tosses, then P = 2n.

<< back to Problem C8

 

Problem C9

The seventh row is 1, 7, 21, 35, 35, 21, 7, 1. The eighth row is 1, 8, 28, 56, 70, 56, 28, 8, 1. The ninth row is 1, 9, 36, 84, 126, 126, 84, 36, 9, 1. The 10th row is 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1. The frequency of five heads in 10 coin tosses is the sixth number in this row, which is 252 (note that it is the center number in the row). Since there are 210 = 1,024 possible outcomes in this row, the probability of getting five heads out of 10 tosses is 252/1,024, or about 24.6%.

<< back to Problem C9

 

Problem C10

a. 

The most probable score is two correct. It has a probability of 6/16.

b. 

The least probable scores are zero correct and four correct. Each has a probability of 1/16.

c. 

The probability of getting at least two answers correct is 6/16 + 4/16 + 1/16 = 11/16.

d. 

The probability of getting at least three answers correct is 4/16 + 1/16 = 5/16.

<< back to Problem C10

 

Problem C11

The simplest way to approach this problem is to find the probability of getting less than two correct, then subtracting this from one. The probability of getting less than two correct is 1/1,024 + 10/1,024 = 11/1024, so the alternate probability is

1 - 11/1,024 = (1,024/1,024) - 11/1,024 = 1,013/1,024, or approximately 98.9%.

<< back to Problem C11

 

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