8 Solved Problems on Plane Trigonometry - Homework 11 | MATH 111, Assignments of Trigonometry

Material Type: Assignment; Class: Plane Trigonometry; Subject: Mathematics Main; University: University of Arizona; Term: Unknown 2000;

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Pre 2010

Uploaded on 08/26/2009

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MATH 111: HW 11 SOLUTIONS AND COMMENTS
8.1, #2. The chance is 1/2 that the tooth fairy will leave $1, and 1/2 that the tooth fairy
will leave $0.50. Therefore, the expected value of the tooth fairy’s payoff is
1
2×$1.00 + 1
2×$0.50 = $0.75 .
8.1, #3. Now the chance is 1/3 that the tooth fairy will leave $1, 1/3 that the tooth fairy
will leave $0.80, and 1/3 that the tooth fairy will leave $0.50. Therefore, the expected value
of the tooth fairy’s payoff is
1
3×$1.00 + 1
3×$0.80 + 1
3×$0.50 = 1
3×$(1.00 + 0.80 + 0.50) = $2.30
3,
which is $0.7666 . . ., or approximately 76.6 cents.
8.1, #12. The probability of rolling each number from 1 to 6 is equal to 1/6, so the expected
value of your roll is:
¡1
6×1¢+¡1
6×2¢+¡1
6×3¢+¡1
6×4¢+¡1
6×5¢+¡1
6×6¢
=1
6×(1+2+3+4+5+6)
= 21/6
= 3.5.
8.1, #13.Solution 1: The standard solution. The probability of rolling a combined
total of 2 on two dice is 1/36; the probability of rolling a combined total of 3 on two dice is
2/36; the probability of rolling a combined total of 4 on two dice is 3/36; and so forth. The
expected value of your roll of two dice is therefore:
1
36 ·2 + 2
36 ·3 + 3
36 ·4 + 4
36 ·5 + 5
36 ·6 + 6
36 ·7 + 5
36 ·8 + 4
36 ·9 + 3
36 ·10 + 2
36 ·11 + 1
36 ·12
=2
36 +6
36 +12
36 +20
36 +30
36 +42
36 +40
36 +36
36 +30
36 +22
36 +12
36
= 252/36
= 7 .
Solution 2: The short, sneaky solution. Consider your two dice separately. By 8.1
#12, the expected value of your roll on the first die is 3.5; similarly the expected value of
your roll on the second die is 3.5. Therefore, the expected value of the total on the two dice
must be 3.5+3.5 = 7.
8.1, #27. Let us separate this game into two parts. First, you pay $5 to play this
contributes a total of $5 to your expected value for the game (since you pay it no matter
what). Next, you flip three coins and you receive money: if you see no heads (TTT, a 1/8
pf3
pf4

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MATH 111: HW 11 SOLUTIONS AND COMMENTS

8.1, #2. The chance is 1/2 that the tooth fairy will leave $1, and 1/2 that the tooth fairy will leave $0.50. Therefore, the expected value of the tooth fairy’s payoff is

1 2

× $1.00 +

× $0.50 = $0. 75.

8.1, #3. Now the chance is 1/3 that the tooth fairy will leave $1, 1/3 that the tooth fairy will leave $0.80, and 1/3 that the tooth fairy will leave $0.50. Therefore, the expected value of the tooth fairy’s payoff is

1 3

× $1.00 +

× $0.80 +

× $0.50 =

× $(1.00 + 0.80 + 0.50) =

which is $0. 7666.. ., or approximately 76.6 cents.

8.1, #12. The probability of rolling each number from 1 to 6 is equal to 1/6, so the expected value of your roll is: ( 1 6 ×^1

6 ×^2

6 ×^3

6 ×^4

6 ×^5

6 ×^6

= 16 × (1 + 2 + 3 + 4 + 5 + 6)

8.1, #13. Solution 1: The standard solution. The probability of rolling a combined total of 2 on two dice is 1/36; the probability of rolling a combined total of 3 on two dice is 2 /36; the probability of rolling a combined total of 4 on two dice is 3/36; and so forth. The expected value of your roll of two dice is therefore: 1 36 ·^ 2 +^

2 36 ·^ 3 +^

3 36 ·^ 4 +^

4 36 ·^ 5 +^

5 36 ·^ 6 +^

6 36 ·^ 7 +^

5 36 ·^ 8 +^

4 36 ·^ 9 +^

3 36 ·^ 10 +^

2 36 ·^ 11 +^

1 36 ·^12 = 362 + 366 + 1236 + 2036 + 3036 + 4236 + 4036 + 3636 + 3036 + 2236 + (^1236)

= 252 / 36

= 7.

Solution 2: The short, sneaky solution. Consider your two dice separately. By 8. #12, the expected value of your roll on the first die is 3.5; similarly the expected value of your roll on the second die is 3.5. Therefore, the expected value of the total on the two dice must be 3.5 + 3.5 = 7.

8.1, #27. Let us separate this game into two parts. First, you pay $5 to play — this contributes a total of −$5 to your expected value for the game (since you pay it no matter what). Next, you flip three coins and you receive money: if you see no heads (TTT, a 1/ 8

chance) you receive $20; if you see one head (TTH, THT, HTT, a 3/8 chance) you receive $5; if you see more than one head (HHT, HTH, THH, HHH, a 4/8 chance) you receive nothing; so the amount you expect to receive is

1 8

× $20 +

× $5 +

× $0 = $35/8 = $4. 375.

Therefore, in total, the expected value for the game is

−$5.00 + $4.375 = −$0. 625

or − 62 .5 cents.

Here’s another way to do the computation. In the case where you see no heads, you gain $15 net; when you see one head, you break even; when you see two or more heads, you lose $5 net. Therefore, in total, the expected value is

1 8

× $15 +

× $0 +

× (−$5) = −$0. 625.

Notice that there is no single outcome in which you lose $0.625. The expected value is not necessarily equal to one of the possible outcomes. Rather, an expected value is an average: the average value that you expect to gain (or lose) each time you play the game, if you were to play it many times.

8.2, #3.

280 , 000 , 000 × 2% = 280, 000 , 000 × 0 .02 = 5, 600 , 000 people.

8.2, #5. A 12-year-old who dies instead of living to the age of 78 loses 66 years of life expectancy; this happens 7% of the time. Therefore, the expected loss of life expectancy is

7% × 66 = 0. 07 × 66 = 4.62 years.

8.2, #6. Using the numbers given in the problem, we can calculate the following quantities:

  • The fraction of blondes who are natural blondes, and who we believe correctly to be natural blondes: 0. 9 × 0 .8 = 0.72.
  • The fraction of blondes who are natural blondes, and who we believe incorrectly to be bleached blondes: 0. 9 × 0 .2 = 0.18.
  • The fraction of blondes who are bleached blondes, and who we believe correctly to be bleached blondes: 0. 1 × 0 .8 = 0.08.
  • The fraction of blondes who are bleached blondes, and who we believe incorrectly to be natural blondes: 0. 1 × 0 .2 = 0.02.

Therefore the total fraction of blondes who we believe to be bleached blondes (correctly or incorrectly) is 0.18 + 0.08 = 0.26.

We wish to compute the chances we are wrong in our belief that Chris is a bleached blonde. This probability is equal to:

Finally, according to the statement of the problem, the number that we want to calculate is the amount that the government is willing to spend per cell phone, which is

amount the government is willing to spend number of cell phones

c/ 8 c

or 12.5 cents.