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# Study guide help 2

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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?

May 2, 2019

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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. May 2, 2019 ### 2+0 Answers #1 +115 +2 Best Answer 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.

Anthrax May 2, 2019
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Nice, Anthrax.....!!!!!

May 3, 2019