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Data Session 8: Solutions
 
Session 8 Part A Part B Part C Part D Homework
 
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A B C D 
Homework

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Solutions for Session 8, Part B

See solutions for Problems: B1 | B2 | B3 | B4 | B5 | B6| B7| B8| B9


Problem B1

Here is the probability table:

Value

Frequency

Probability

1

1

1/5

2

1

1/5

3

1

1/5

4

1

1/5

5

1

1/5

Note that each of the five regions is equally likely to appear.

<< back to Problem B1


 

Problem B2

The probability does not change; it is still one-half, or 50%. The same would be true regardless of the outcomes of the previous three tosses.

<< back to Problem B2


 

Problem B3

a. 

The possible outcomes are HH, HT, TH, and TT. There are four possible outcomes, and each has a one-fourth (25%) probability of occurring.

b. 

The possible outcomes are HHH, HTH, THH, TTH, HHT, HTT, THT, and TTT. There are eight possible outcomes, and each has a one-eighth probability of occurring.

<< back to Problem B3


 

Problem B4

a. Here are the tables of all 36 possible outcomes and their sums:

Red Die

Blue Die

+

1

2

3

4

5

6

1

1 + 1

1 + 2

1 + 3

1 + 4

1 + 5

1 + 6

2

2 + 1

2 + 2

2 + 3

2 + 4

2 + 5

2 + 6

3

3 + 1

3 + 2

3 + 3

3 + 4

3 + 5

3 + 6

4

4 + 1

4 + 2

4 + 3

4 + 4

4 + 5

4 + 6

5

5 + 1

5 + 2

5 + 3

5 + 4

5 + 5

5 + 6

6

6 + 1

6 + 2

6 + 3

6 + 4

6 + 5

6 + 6

Red Die

Blue Die

+

1

2

3

4

5

6

1

2

3

4

5

6

7

2

3

4

5

6

7

8

3

4

5

6

7

8

9

4

5

6

7

8

9

10

5

6

7

8

9

10

11

6

7

8

9

10

11

12

b. The table will have 36 entries, the same number as we found in the answer to part (a). Note that the way the table is constructed, with six rows and six columns, guarantees that there will be 6 x 6 = 36 entries.

<< back to Problem B4


 

Problem B5

Here is the table of sums. Sums where Player A wins are highlighted in purple, and sums where Player B wins are highlighted in green:

Red Die

Blue Die

+

1

2

3

4

5

6

1

2

3

4

5

6

7

2

3

4

5

6

7

8

3

4

5

6

7

8

9

4

5

6

7

8

9

10

5

6

7

8

9

10

11

6

7

8

9

10

11

12

a. 

There are 12 outcomes (highlighted in purple) that produce sums of 2, 3, 4, 10, 11, or 12. These are equally likely outcomes because the dice are fair, so the probability that Player A wins is 12/36, or 33%.

b. 

There are 24 outcomes (highlighted in green) that produce sums of 5, 6, 7, 8, or 9. These are equally likely outcomes because the dice are fair, so the probability that Player B wins is 24/36, or 67%.

c. 

Player B will win this game two-thirds of the time.

<< back to Problem B5


 

Problem B6

One potential change is to change the sums that each players wins with. Here's one possible solution:

 

Player A wins when the sum is 2, 3, 4, 7, 10, 11, or 12.

 

Player B wins when the sum is 5, 6, 8, or 9.

It may seem surprising that this is a fair game, but with this change each player will win one-half (18/36) of the time.

<< back to Problem B6


 

Problem B7

Here is the completed probability table:

Sum

Frequency

Probability

2

1

1/36

3

2

2/36

4

3

3/36

5

4

4/36

6

5

5/36

7

6

6/36

8

5

5/36

9

4

4/36

10

3

3/36

11

2

2/36

12

1

1/36

<< back to Problem B7


 

Problem B8

This can be found by adding the probabilities of the sums that Player A will win with:

1/36 + 2/36 + 3/36 + 3/36 + 2/36 + 1/36 = 12/36

Player A wins with probability 12/36, or one-third of the time.

<< back to Problem B8


 

Problem B9

Since one of the two players has to win, the sum of both probabilities -- that of Player A winning and that of Player B winning -- is 1. So a faster way to find the probability that Player B wins is to subtract the probability that Player A wins (12/36) from 1:

1 - (12/36) = (36/36) - (12/36) = 24/36

Player B wins with probability 24/36, or two-thirds of the time.

<< back to Problem B9


 

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