i think 1 is B and 2 is $4.05

1) Kyle 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, Kyle receives 15 points.

OPTION B: If the sum of the rolls is a multiple of 6, Kyle receives 42 points.

Which statement best explains the option he should choose?

Kyle should choose Option A. The mathematical expectation of this option is 6.25 and is the greater mathematical expectation of the two options.

Kyle should choose Option B. The mathematical expectation of this option is 6.25 and is the greater mathematical expectation of the two options.

Kyle should choose Option B. The mathematical expectation of this option is 7 and is the greater mathematical expectation of the two options.

Kyle should choose Option A. The mathematical expectation of this option is 7 and is the greater mathematical expectation of the two options.

Anya 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, 2, 5, and 5. If the sum of the two spins is greater than 6, Anya is awarded $8. Otherwise, she must pay $2.

OPTION B: Flip a coin three times. If heads appears 3 times, Anya is awarded $50. Otherwise, she must pay $1.

Anya chooses the option with the greater mathematical expectation.

How much more money can Anya expect to make by choosing this option over the other option?

jjennylove May 1, 2019

#1**+1 **

\(\text{The distribution of a sum of 2 6 sided dice is}\\ \left( \begin{array}{cc} 2 & \frac{1}{36} \\ 3 & \frac{1}{18} \\ 4 & \frac{1}{12} \\ 5 & \frac{1}{9} \\ 6 & \frac{5}{36} \\ 7 & \frac{1}{6} \\ 8 & \frac{5}{36} \\ 9 & \frac{1}{9} \\ 10 & \frac{1}{12} \\ 11 & \frac{1}{18} \\ 12 & \frac{1}{36} \\ \end{array} \right)\)

\(\text{Option A:}\\ P[\text{roll a prime}]=P[2,3,5,7,11] = \dfrac{1}{36}+\dfrac{1}{18}+\dfrac{1}{9}+\dfrac{1}{6}+\dfrac{1}{18}=\dfrac{5}{12}\\ E[A] = 15 \cdot \dfrac{5}{12} = \dfrac{75}{12}=\dfrac{25}{4}=6.25\\ \text{Option B:}\\ P[6,12] = \dfrac{5}{36}+\dfrac{1}{36}= \dfrac{1}{6}\\ E[B] = \dfrac 1 6 \cdot 42 = 7\)

(C) is the correct answer

Rom May 1, 2019

#3**0 **

oh okay now I see where I made my mistakes Thank you for your explaination I will take notes on this. Could yu help me with the second question?

jjennylove
May 1, 2019

#2**+1 **

\(\text{The distribution of the sum of spins is }\\ \left( \begin{array}{cc} 2 & \frac{1}{16} \\ 3 & \frac{1}{8} \\ 4 & \frac{1}{16} \\ 6 & \frac{1}{4} \\ 7 & \frac{1}{4} \\ 10 & \frac{1}{4} \\ \end{array} \right)\)

\(P[sum>6]=P[7,10]=\dfrac 1 4 + \dfrac 1 4 = \dfrac 1 2\\ E[A] = 8 \cdot \dfrac 1 2 - 2 \cdot \dfrac 1 2 = 3\)

\(P[\text{3 heads in 3 flips}] = \left(\dfrac 1 2\right)^3 = \dfrac 1 8\\ E[B] = 50 \cdot \dfrac 1 8 - 1 \cdot \dfrac 7 8 = \dfrac{43}{8} = 5\dfrac 3 8\\ 5\dfrac 3 8 - 3 = 2 \dfrac 3 8\)

.Rom May 1, 2019

#7**0 **

CPhill got this answer. I am confused which is correct.

Option A.....possible outcomes

(1, 1) (2, 2)

(1, 2) ( 2,5)

(2, 1) (5,2)

(1, 5) (5, 5)

(5, 1)

Expected vlalue = (3/9)(8) - (6/9)(2) = 8/3 - 4/3 ≈ $1.33

Option B

Expected value (50/((1/8) - (1)(7/8) ≈ $5.38

Ptiob B is better by $ [ 5.38 - 1.33] = $4.05

jjennylove
May 2, 2019