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# Study guide help!!! Im stuck on thes 5 questions for stat class!!!:(

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1) A hat contains four balls. The balls are numbered 2, 4, 4, and 7. One ball is randomly selected and not replaced, and then a second ball is selected. The numbers on the two balls are added together.

A fair decision is to be made about which of three sizes of ice cream cone will be ordered, using the sum of the numbers on the balls. The sizes are small, medium and large.

Which description accurately explains how a fair decision can be made in this situation?
A)If the sum is 6, a small cone will be ordered.
If the sum is 8 or 11, a medium cone will be ordered.
If the sum is 9, a large cone will be ordered.

B)If the sum is 6 or 8, a small cone will be ordered.
If the sum is 9, a medium cone will be ordered.
If the sum is 11, a large cone will be ordered.

C)If the sum is 6 or 9, a small cone will be ordered.
If the sum is 8, a medium cone will be ordered.
If the sum is 11, a large cone will be ordered.

D)If the sum is 6, a small cone will be ordered.
If the sum is 8 or 9, a medium cone will be ordered.
If the sum is 11, a large cone will be ordered.

2.) Twenty eight slips of paper are numbered 1 through 28 and are distributed among a group of people. A random number generator is used to select a single number between 1 and 28, inclusively.

A fair decision is made using this process.

How many people could be in this group?
A 3

B 6

C 7

D 12

3.) In a particular game, a spinner with four equally-sized sectors labeled 1, 3, 5, and 6 is spun twice. One turn is considered 2 spins of the spinner.

If the sum of the spins is even, you move backward 6 spaces. Otherwise, you move forward 2 spaces.

What is the mathematical expectation for the number of spaces moved in one turn?

A 3 spaces forward

B 1 space backward

C 3 spaces backward

D 1 space forward

4.) 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 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 6.25 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.

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

5.)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?

\$__

Apr 12, 2019

#1
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$$1)~ \text{The possible values are }\\ 6 = (2,4)\\ 8=(4,4)\\ 9=(2,7)\\ 11=(4,7)$$

$$P[6]=\dfrac{\dbinom{1}{1}\dbinom{2}{1}}{\dbinom{4}{2}}=\dfrac 1 3\\ P[8] = \dfrac{\dbinom{2}{2}}{\dbinom{4}{2}}= \dfrac 1 6\\ P[9] = \dfrac{\dbinom{1}{1}\dbinom{1}{1}}{\dbinom{4}{2}}=\dfrac 1 6\\ P[11]=\dfrac{\dbinom{2}{1}\dbinom{1}{1}}{\dbinom{4}{2}}=\dfrac 1 3$$

$$\text{We want a distribution of }\left(\dfrac 1 3,\dfrac 1 3,\dfrac 1 3\right) \text{ on cone sizes}\\ \text{And choice D is the only one that accomplishes this}$$

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Apr 13, 2019
#2
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2) It should be obvious that the number of people in the group must be a factor of 28

3) $$E=2P[\text{odd}]-6P[\text{even}]\\ P[\text{odd}]=P[1]+P[3]+P[5]=\dfrac 3 4\\ P[\text{even}] = P[6] = \dfrac 1 4\\ E= 2\cdot \dfrac 3 4 - 6 \cdot \dfrac 1 4 = 0$$

Apr 13, 2019
edited by Rom  Apr 13, 2019
#3
+6196
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4)

$$\text{the primes that can be rolled are }2,3,5,7,11\\ P[\text{roll a prime}] = \dfrac{1+2+4+6+2}{36} = \dfrac{5}{12}\\ E_A = 15 \cdot \dfrac{5}{12} = \dfrac{25}{4}\\ \text{The multiples of 6 that can be rolled are }6,12\\ P[\text{roll a multiple of 6}]=\dfrac{5+1}{36}=\dfrac 1 6\\ E_B = 42 \cdot \dfrac 1 6 = 7\\ E_B = 7 > \dfrac{25}{4} = E_A$$

Apr 13, 2019
#4
+6196
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5) Just follow the same ideas as in 4

Apr 13, 2019