You roll a fair die three times. What is the probability of each of the following? a) You roll all 4's. b) You roll all even numbers. c) None of your rolls gets a number divisible by 2. d) You roll at least one 2. e) The numbers you roll are not all 2's.
Accepted Solution
A:
Answer:(a) [tex]\frac{1}{216}[/tex](b) [tex]\frac{1}{8}[/tex](c) [tex]\frac{1}{8}[/tex](d) [tex]\frac{191}{216}[/tex](e) [tex]\frac{215}{216}[/tex]Step-by-step explanation:It is given that we roll a fair die three times. We need to find the following probability.Total possible value = 1,2,3,4,5,6(a) You roll all 4's.Probability of getting a 4 = [tex]\frac{1}{6}[/tex]Probability of getting 4 on all rolls = [tex]\frac{1}{6}\times \frac{1}{6}\times \frac{1}{6}=\frac{1}{216}[/tex]Therefore, the probability of getting 4 on all rolls is [tex]\frac{1}{216}[/tex].(b) You roll all even numbers.Even number = 2,4,6Probability of getting an even number [tex]=\frac{3}{6}=\frac{1}{2}[/tex]Probability of getting even number on all rolls = [tex]\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}=\frac{1}{8}[/tex]Therefore, the probability of getting even number on all rolls is [tex]\frac{1}{8}[/tex].(c) None of your rolls gets a number divisible by 2. odd number = 1,3,5Probability of getting an odd number [tex]=\frac{3}{6}=\frac{1}{2}[/tex]Probability of getting odd number on all rolls = [tex]\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}=\frac{1}{8}[/tex]Therefore, the probability of getting odd number on all rolls is [tex]\frac{1}{8}[/tex].(d) You roll at least one 2. Probability of getting a 2 = [tex]\frac{1}{6}[/tex]Probability of getting any number except 2 = [tex]\frac{5}{6}[/tex]Probability of getting any number except 2 in all rolls = [tex]\frac{5}{6}\times \frac{5}{6}\times \frac{5}{6}=\frac{125}{216}[/tex]Probability of getting at least one 2 = 1 - P(getting no 2)Probability of getting at least one 2 = [tex]1-\frac{125}{216}=\frac{191}{216}[/tex]Therefore, the probability of getting at least one 2 is [tex]\frac{191}{216}[/tex].(e) The numbers you roll are not all 2's.Probability of getting a 2 = [tex]\frac{1}{6}[/tex]Probability of getting all 2 = [tex]\frac{1}{6}\times \frac{1}{6}\times \frac{1}{6}=\frac{1}{216}[/tex]Probability of getting all not all 2's = [tex]1-\frac{1}{216}=\frac{215}{216}[/tex]Therefore, the probability of getting not all 2's is [tex]\frac{215}{216}[/tex].