+642 1) Frank wants to earn as many points as possible in one turn in a game. Two number cubes whose sides are numbered 1 through 6 are rolled. He is given two options for the manner in which points are awarded in the turn.
OPTION A: If the sum of the rolls is a prime number, Frank receives 12 points.
OPTION B: If the sum of the rolls is a multiple of 4, Frank receives 24 points.
Which statement best explains the option he should choose?
A) Frank should choose Option A. The mathematical expectation of this option is 6 and is the greater mathematical expectation of the two options.
B) Frank should choose Option B. The mathematical expectation of this option is 5 and is the greater mathematical expectation of the two options.
C) Frank should choose Option B. The mathematical expectation of this option is 6 and is the greater mathematical expectation of the two options.
D) Frank should choose Option A. The mathematical expectation of this option is 5 and is the greater mathematical expectation of the two options.
2) Danielle may choose one of two options for the method in which she may be awarded a money prize.
OPTION A: Spin a spinner twice. The spinner is divided into four equally-sized sectors numbered 1, 4, 4, and 5. If the sum of the two spins is greater than 6, Danielle is awarded $8. Otherwise, she must pay $2.
OPTION B: Flip a coin three times. If heads appears once, Danielle is awarded $6. Otherwise, she must pay $1.
Danielle chooses the option with the greater mathematical expectation.
How much more money can Danielle expect to make by choosing this option over the other option?
+150 1) Considering Option A: prime numbers up to 12 are 2, 3, 5, 7, and 11. The probability of obtaining these is \(\frac{15}{36}\) (sum probabilities: P(2) + P(3) + P(5) + P(7) + P(11) = \(\frac{1}{36} + \frac{2}{36} + \frac{4}{36} + \frac{6}{36} + \frac{2}{36}\)). Multiplying the probability of obtaining a prime number by the given point value of 12 gives us 5 as the expected value of Option A.
Considering Option B: multiples of 4 up to 12 are 4, 8, and 12. The probability of obtaining these is \(\frac{9}{36}\) (sum probabilities: P(4) + P(8) + P(12) = \(\frac{3}{36} + \frac{5}{36} +\frac{1}{36}\)). Multiplying the probability of obtaining a multiple of 4 by the given point value of 24 gives us 6 as the expected value of Option B.
Therefore, the answer is C with Option B and the expectation being 6.
Here's a good table for two-dice sum probabilities: http://lh4.ggpht.com/-RSwGhQhT9JI/UblGzz7HNvI/AAAAAAAABRA/8mTpypdmFZA/TwoDiceTable9.jpg?imgmax=800
2) Considering Option A: Sum chart:
| 1 | 4 | 4 | 5 | |
|---|---|---|---|---|
| 1 | 2 | 5 | 5 | 6 |
| 4 | 5 | 8 | 8 | 9 |
| 4 | 5 | 8 | 8 | 9 |
| 5 | 6 | 9 | 9 | 10 |
So, \(P(2) = \frac{1}{16}, P(5) = \frac{4}{16}, P(6) = \frac{2}{16}, P(8) = \frac{4}{16}, P(9) = \frac{4}{16}, P(10) = \frac{1}{16}\). Then, \(P(sum > 6) = \frac{4}{16} + \frac{4}{16} + \frac{1}{16} = \frac{9}{16}\) and \(P(sum \leq 6) = 1 - \frac{9}{16} = \frac{7}{16}\). Multiplying the winning probability of \(\frac{9}{16}\) by $8, we get an expected value of $4.50. Multiplying the losing probability of \(\frac{7}{16}\) by $2, we get an expected value of $0.875. Subtracting these two expected values, our overall expectation is $3.625.
Considering Option B: the probability of flipping a coin three times and heads appearing exactly once is \(\frac{3}{8}\). The probability of losing is then \(1 - \frac{3}{8} = \frac{5}{8}\). Multiplying the probability of winning \(\frac{3}{8}\) by $6 gives us an expected value of $2.25. Multplying the probability of losing \(\frac{5}{8}\) by $1 gives us an expected value of $0.625. Subtracting these two expected values, our overall expectation is $1.625.
Good display of flipping a coin three times results: http://web.mnstate.edu/peil/MDEV102/U3/S25/Cartesian3.PNG
Therefore, Danielle would choose Option A. Danielle would expect to make $2 more choosing Option A over Option B.
+150 1) Considering Option A: prime numbers up to 12 are 2, 3, 5, 7, and 11. The probability of obtaining these is \(\frac{15}{36}\) (sum probabilities: P(2) + P(3) + P(5) + P(7) + P(11) = \(\frac{1}{36} + \frac{2}{36} + \frac{4}{36} + \frac{6}{36} + \frac{2}{36}\)). Multiplying the probability of obtaining a prime number by the given point value of 12 gives us 5 as the expected value of Option A.
Considering Option B: multiples of 4 up to 12 are 4, 8, and 12. The probability of obtaining these is \(\frac{9}{36}\) (sum probabilities: P(4) + P(8) + P(12) = \(\frac{3}{36} + \frac{5}{36} +\frac{1}{36}\)). Multiplying the probability of obtaining a multiple of 4 by the given point value of 24 gives us 6 as the expected value of Option B.
Therefore, the answer is C with Option B and the expectation being 6.
Here's a good table for two-dice sum probabilities: http://lh4.ggpht.com/-RSwGhQhT9JI/UblGzz7HNvI/AAAAAAAABRA/8mTpypdmFZA/TwoDiceTable9.jpg?imgmax=800
2) Considering Option A: Sum chart:
| 1 | 4 | 4 | 5 | |
|---|---|---|---|---|
| 1 | 2 | 5 | 5 | 6 |
| 4 | 5 | 8 | 8 | 9 |
| 4 | 5 | 8 | 8 | 9 |
| 5 | 6 | 9 | 9 | 10 |
So, \(P(2) = \frac{1}{16}, P(5) = \frac{4}{16}, P(6) = \frac{2}{16}, P(8) = \frac{4}{16}, P(9) = \frac{4}{16}, P(10) = \frac{1}{16}\). Then, \(P(sum > 6) = \frac{4}{16} + \frac{4}{16} + \frac{1}{16} = \frac{9}{16}\) and \(P(sum \leq 6) = 1 - \frac{9}{16} = \frac{7}{16}\). Multiplying the winning probability of \(\frac{9}{16}\) by $8, we get an expected value of $4.50. Multiplying the losing probability of \(\frac{7}{16}\) by $2, we get an expected value of $0.875. Subtracting these two expected values, our overall expectation is $3.625.
Considering Option B: the probability of flipping a coin three times and heads appearing exactly once is \(\frac{3}{8}\). The probability of losing is then \(1 - \frac{3}{8} = \frac{5}{8}\). Multiplying the probability of winning \(\frac{3}{8}\) by $6 gives us an expected value of $2.25. Multplying the probability of losing \(\frac{5}{8}\) by $1 gives us an expected value of $0.625. Subtracting these two expected values, our overall expectation is $1.625.
Good display of flipping a coin three times results: http://web.mnstate.edu/peil/MDEV102/U3/S25/Cartesian3.PNG
Therefore, Danielle would choose Option A. Danielle would expect to make $2 more choosing Option A over Option B.
