In: Statistics and Probability
1. You are going to toss one coin and roll one die. Using a tree diagram determine how many outcomes are possible and list the sample space. (6 points).
Then based on your sample space answer the following questions: (2 points each)
a. What is the probability you will toss a tail and roll a composite number?
b. What is the probability you will toss a head or roll a number greater than 3?
c. What is the probability you will toss a tail AND not roll number less than 3?
d. What is the probability you will toss a tail and roll a divisible by 3?
e. What are the odds in favor you will toss a head or roll a number less than 4?
Following is the tree diagram:
Total number of possible outcomes : 2*6 = 12
The sample space is
S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}
Since each of these 12 outcomes are equally likely so probability of each outcome is 1/12.
Composite numbers: 4, 6
Prime numbers: 1, 2, 3, 5
a)
The probability you will toss a tail and roll a composite number is
P(tail and composite number) = P(T4) + P(T6) = 1/12 + 1/12 = 2/12 = 1/6
b)
The probability you will toss a head or roll a number greater than 3 is
P(head or roll greater than 3) = P(H1) + P(H2) + P(H3) + P(H4) + P(H5) + P(H6) +P(T4) + P(T5) + P(T6) = 9 /12 = 0.75
c)
The probability you will toss a tail AND not roll number less than 3 is
P(tail and not roll number less than 3) = P(T3) + P(T4) + P(T5) + P(T6) = 4/12 = 1/3
d)
The probability you will toss a tail and roll a divisible by 3 is
P(tail and divisible by 3) = P(T3) + P(T6) = 1/12 + 1/12 = 1 /6
e)
The number of outcomes in favor of event " you will toss a head or roll a number less than 4" is 9 so
P(E) = P(head or roll a number less than 4) = P(H1) + P(H2) + P(H3) + P(H4) + P(H5) + P(H6) +P(T1) + P(T2) + P(T3) = 9/12
By the complement rule,
P(E') = 1 - P(E) = 3/12
The dds in favor you will toss a head or roll a number less than 4 is
odds in favor = P(E) : P(E') = 9/12 : 3/12 = 9 : 3 = 3 :1