Question

two fair dice are tossed and the uppermost face of each die is observed. the following events are defined from this random experiment:
a represent the event the uppermost faces sum to five
b represent the event that the product of the uppermost faces is four. for example, die1*die2 = 4
c represent the event that the absolute difference between the uppermost faces is 1. for example, |die1 - die2| = 1
part (a) find the probability that the uppermost faces do not sum to five. 0.8889 (use four decimals in your answer)
part (b) find p(a ∪ c) 0.3333 (use four decimals)
part (c) what is the probability that the uppermost faces do not sum to five or are not a product of 4? 0.9444 (use four decimals)
part (d) find p(a ∩ (b ∪ c)) 0.0556 (use four decimals)
part (e) are the events a sum of 5 and a product of 4 mutually exclusive events? select the most appropriate reason below.
a. a sum of 5 and a product of 4 are not mutually exclusive events because p(a ∩ b) = p(a)p(b).
b. a sum of 5 and a product of 4 are not mutually exclusive events because p(a ∩ b) ≠ 0.
c. a sum of 5 and a product of 4 are not mutually exclusive events because p(a ∩ b) ≠ p(a)p(b).
d. a sum of 5 and a product of 4 are mutually exclusive events because p(a ∩ b) = 0.
e. a sum of 5 and a product of 4 are mutually exclusive events because they are not independent events.

Explanation:

Step1: Calculate total number of outcomes

When two fair dice are tossed, the total number of outcomes is \(n(S)=6\times6 = 36\) since each die has 6 possible outcomes.

Step2: Find outcomes for event A

The event \(A\) (sum of uppermost - faces is 5) has the following outcomes: \((1,4)\), \((2,3)\), \((3,2)\), \((4,1)\), so \(n(A)=4\). Then \(P(A)=\frac{n(A)}{n(S)}=\frac{4}{36}=\frac{1}{9}\approx0.1111\). The probability that the uppermost faces do not sum to five is \(P(\overline{A}) = 1 - P(A)=1-\frac{1}{9}=\frac{8}{9}\approx0.8889\).

Step3: Find outcomes for event C

The event \(C\) (absolute - difference between uppermost faces is 1) has the following outcomes: \((1,2)\), \((2,1)\), \((2,3)\), \((3,2)\), \((3,4)\), \((4,3)\), \((4,5)\), \((5,4)\), \((5,6)\), \((6,5)\), so \(n(C)=10\). Then \(P(C)=\frac{n(C)}{n(S)}=\frac{10}{36}=\frac{5}{18}\). The intersection \(A\cap C=\{(2,3),(3,2)\}\), so \(n(A\cap C) = 2\) and \(P(A\cap C)=\frac{2}{36}=\frac{1}{18}\). Using the formula \(P(A\cup C)=P(A)+P(C)-P(A\cap C)\), we have \(P(A\cup C)=\frac{4}{36}+\frac{10}{36}-\frac{2}{36}=\frac{4 + 10-2}{36}=\frac{12}{36}=\frac{1}{3}\approx0.3333\).

Step4: Find outcomes for event B

The event \(B\) (product of uppermost faces is 4) has the following outcomes: \((1,4)\), \((4,1)\), \((2,2)\), so \(n(B)=3\) and \(P(B)=\frac{3}{36}=\frac{1}{12}\). The probability that the uppermost faces do not sum to five or are not a product of 4 is \(P(\overline{A}\cup\overline{B})=1 - P(A\cap B)\). Since \(A\cap B=\{(1,4),(4,1)\}\), \(n(A\cap B) = 2\) and \(P(A\cap B)=\frac{2}{36}=\frac{1}{18}\). So \(P(\overline{A}\cup\overline{B})=1-\frac{1}{18}=\frac{17}{18}\approx0.9444\).

Step5: Find \(P(A\cap(B\cup C))\)

First, find \(B\cup C\). \(B=\{(1,4),(4,1),(2,2)\}\) and \(C=\{(1,2),(2,1),(2,3),(3,2),(3,4),(4,3),(4,5),(5,4),(5,6),(6,5)\}\). Then \(B\cup C=\{(1,2),(2,1),(2,2),(2,3),(3,2),(3,4),(4,1),(4,3),(4,5),(5,4),(5,6),(6,5)\}\). \(A\cap(B\cup C)=\{(2,3),(3,2)\}\), so \(n(A\cap(B\cup C)) = 2\) and \(P(A\cap(B\cup C))=\frac{2}{36}\approx0.0556\).

Step6: Determine mutual - exclusivity

Two events \(A\) and \(B\) are mutually exclusive if \(P(A\cap B)=0\). Since the set of outcomes for \(A\) (sum is 5) and \(B\) (product is 4) have no common outcomes other than \((1,4)\) and \((4,1)\) which we have accounted for, and in general, there are no non - overlapping new outcomes, \(P(A\cap B) = 0\) means they are mutually exclusive.

Answer:

Part (a): \(0.8889\)
Part (b): \(0.3333\)
Part (c): \(0.9444\)
Part (d): \(0.0556\)
Part (e): D. A sum of 5 and a product of 4 are mutually exclusive events because \(P(A\cap B)=0\).