Question

A game involves rolling a fair six-sided die. If the number obtained on the die is a multiple of three, the player wins an amount equal to the number on the die times $20. If the number is not a multiple of three, the player gets nothing.

How will you model the simulation for the roll of a die?
Opt
Use the numbers 1–20 to represent the numbers rolled when a player wins.
Use the numbers 1–6 to represent the unfavorable outcomes.
Use the numbers 1–3 to represent all the outcomes.
Use the numbers 1–10 to represent the unfavorable outcomes.
The problem cannot be solved using a simulation.

Answer 1

will medal please help!

Answer 2

Extra 1.1 Let A be an event that happens 40% of the time. Let B be an event that happens 75% of the
time. Answer the following 4 questions. What is the smallest probability the intersection
of A and B can have? What is the largest probability the intersection of A and B can
have? What is the smallest probability the union of A and B can have? What is the largest
probability the union of A and B can have?
.15, .4, .75, and 1 respectively.
Extra 1.2 Suppose we are rolling 2 independent, fair 10-sided die. Let A be the event that the sum of
the rolls is a prime number. Let B be the event that the sum of the rolls is odd. Let C be
the event that the sum of the rolls is even. Find the following sets: BC, A ∩ B, A ∪ C, A
∩ C, (A ∩ C)C, A ∩ (B ∩ C)C, (A ∩ (B ∩ C))C?
B
C = C,
A ∩ B = A \ {2},
A ∪ C = Ω \ {9, 15},
A ∩ C = {2},
(A ∩ C)C =Ω \ {2},
A ∩ (B ∩ C)C = A,
and (A ∩ (B ∩ C))C = Ω.
Extra 1.3 Suppose a lottery has balls numbered 1-20. 4 balls are picked at random and without
replacement. Let A be the event that all 4 balls are even. Let B be the event that all 4
balls are less than 10. Let C be the event that all 4 balls are primes. (Allow 1 to be a
prime.) Find P(A), P(B), and P(C). (Hint an extended general multiplication rule could be
helpful.)
P(all 4 even) = (10)4
(20)4
= .0433.
P(all less than 10) = (9)4
(20)4
= .0260.
What numbers are prime? 1, 2, 3, 5, 7, 11, 13, 17, and 19. That is 9 numbers total. P(all
prime) = (9)4
(20)4
=.0260.
Extra 1.4 A school has 100 students. The school offers only 3 language classes, namely Spanish, Italian
and Russian. 50 students do not take a language. The Spanish, Italian, and Russian classes
have 28, 26, and 16 students respectively. However, 12 students take both Spanish and
Italian, 4 students take both Spanish and Russian, and 6 students take both Italian and
Russian.
• How many students take all 3 language classes? 2
• What is the probability a randomly chosen student takes exactly 1 language class? .14
+ .1 + .08 = .32.
• You randomly draw 2 students. What is the probability that they are taking at least
1 language class between them? 1 - P(neither in a language) = 1 - .5∗
49
99 = .7525.


this may help

Answer 3

@Mateaus