Question

The table below shows all outcomes for rolling 2 dice. The bold numbers 1 through 6 show the possible results for each die. All the other numbers show the possible sums when the 2 dice are rolled.
Part I: Circle each outcome in the table where at least 1 die is a 4 and cross out each outcome where the sum of the dice is greater than 5.
part 2 -Use the table in Part I to find each probability when 2 dice are rolled. Write all probabilities as fractions in lowest terms. Define the events A and B as:
Event A = At least 1 die is a 4.

Event B = The sum of the dice is greater than 5.
A. What is the probability that at least 1 die is a 4?
B. What is the probability that the sum of the dice is greater than 5?
C. What is the probability that at least 1 die is a 4 and that the sum of the dice is greater than 5?
D. What is the probability that at least 1 die is a 4 or that the sum of the dice is greater than 5?

Answer 1

Part I:

1. If you have to circle each outcome in the table that at least 1 die is a 4, then you have to circle row corresponding to the number 4 and column corresponding to the number 4.

2. If you have to cross out each outcome where the sum of the dice is greater than 5, then you have to cross out all not bold numbers 6, 7, 8, 9, 10, 11 and 12 in the table.

Part II:

Here you can see 36 possible outcomes.

1. The probability that at least 1 die is a 4 is:

$Pr(A)=\dfrac{11}{36}$ - (favorable outcomes are 4+1=5, 4+2=6, 4+3=7, 4+4=8, 4+5=9, 4+6=10, 1+4=5, 2+4=6, 3+4=7, 5+4=9, 6+4=10).

2. The probability that the sum of the dice is greater than 5 is:

$Pr(B)=\dfrac{26}{36}=\dfrac{13}{18}.$

3. The probability that at least 1 die is a 4 and that the sum of the dice is greater than 5 is

$Pr(A\cap B)=\dfrac{9}{36}=\dfrac{1}{4}.$

4. The probability that at least 1 die is a 4 or that the sum of the dice is greater than 5 is:

$Pr(A\cup B)=\dfrac{28}{36}=\dfrac{7}{9}.$