two fair dice are tossed, and the uppermost face of each die is observed. the following events are defined…

two fair dice are tossed, and the uppermost face of each die is observed. the following events are defined from the random experiment:\na represent the event the uppermost faces sum to five\nb represent the event that the product of the uppermost faces is four. for example, die1*die2 = 4\nc represent the event that the absolute difference between the uppermost faces is 1. for example, |die1 - die2| = 1\npart (a) find the probability that the uppermost faces do not sum to five. 0.8889 (use four decimals in your answer)\npart (b) find p(a ∪ c) 0.3333 (use four decimals)\npart (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)\npart (d) find p(a ∩ (b ∪ c)) 0.1111 (use four decimals)\npart (e) are the events a sum of 5 and a product of 4 mutually exclusive events? select the most appropriate reason below\na. a sum of 5 and a product of 4 are not mutually exclusive events because p(a ∩ b) = p(a)p(b).\nb. a sum of 5 and a product of 4 are not mutually exclusive events because p(a ∩ b) ≠ 0.\nc. a sum of 5 and a product of 4 are not mutually exclusive events because p(a ∩ b) ≠ p(a)p(b).\nd. a sum of 5 and a product of 4 are mutually exclusive events because p(a ∩ b) = 0.\ne. a sum of 5 and a product of 4 are mutually exclusive events because they are not independent events.

two fair dice are tossed, and the uppermost face of each die is observed. the following events are defined from the random experiment:\na represent the event the uppermost faces sum to five\nb represent the event that the product of the uppermost faces is four. for example, die1*die2 = 4\nc represent the event that the absolute difference between the uppermost faces is 1. for example, |die1 - die2| = 1\npart (a) find the probability that the uppermost faces do not sum to five. 0.8889 (use four decimals in your answer)\npart (b) find p(a ∪ c) 0.3333 (use four decimals)\npart (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)\npart (d) find p(a ∩ (b ∪ c)) 0.1111 (use four decimals)\npart (e) are the events a sum of 5 and a product of 4 mutually exclusive events? select the most appropriate reason below\na. a sum of 5 and a product of 4 are not mutually exclusive events because p(a ∩ b) = p(a)p(b).\nb. a sum of 5 and a product of 4 are not mutually exclusive events because p(a ∩ b) ≠ 0.\nc. a sum of 5 and a product of 4 are not mutually exclusive events because p(a ∩ b) ≠ p(a)p(b).\nd. a sum of 5 and a product of 4 are mutually exclusive events because p(a ∩ b) = 0.\ne. a sum of 5 and a product of 4 are mutually exclusive events because they are not independent events.

Answer

Explanation:

Step1: Find total number of outcomes

When two dice are tossed, the total number of outcomes is $n(S)=6\times6 = 36$.

Step2: Find number of outcomes for event A

The pairs for event A (sum is 5) are $(1,4),(2,3),(3,2),(4,1)$, so $n(A)=4$. Then $P(A)=\frac{n(A)}{n(S)}=\frac{4}{36}=\frac{1}{9}$.

Step3: Find number of outcomes for event B

The pairs for event B (product is 4) are $(1,4),(2,2),(4,1)$, so $n(B)=3$. Then $P(B)=\frac{n(B)}{n(S)}=\frac{3}{36}=\frac{1}{12}$.

Step4: Find number of outcomes for event C

The pairs for event C (absolute - difference is 1) are $(1,2),(2,1),(2,3),(3,2),(3,4),(4,3),(4,5),(5,4),(5,6),(6,5)$, so $n(C)=10$. Then $P(C)=\frac{n(C)}{n(S)}=\frac{10}{36}=\frac{5}{18}$.

Step5: Find $P(A\cap C)$

The common pairs of A and C are $(2,3),(3,2)$, so $n(A\cap C)=2$. Then $P(A\cap C)=\frac{n(A\cap C)}{n(S)}=\frac{2}{36}=\frac{1}{18}$.

Step6: Find $P(A\cup C)$

Using the formula $P(A\cup C)=P(A)+P(C)-P(A\cap C)$, we substitute the values: $P(A\cup C)=\frac{1}{9}+\frac{5}{18}-\frac{1}{18}=\frac{2 + 5-1}{18}=\frac{6}{18}=\frac{1}{3}\approx0.3333$.

Step7: Find $P(A\cap B)$

The common pairs of A and B are $(1,4),(4,1)$, so $n(A\cap B)=2$. Then $P(A\cap B)=\frac{n(A\cap B)}{n(S)}=\frac{2}{36}=\frac{1}{18}\neq0$.

Step8: Find $P(A\cap(B\cup C))$

First, find $n(B\cup C)=n(B)+n(C)-n(B\cap C)$. The common pairs of B and C are $(1,4),(4,1)$, so $n(B\cap C)=2$. Then $n(B\cup C)=3 + 10-2=11$. The common pairs of A and $(B\cup C)$ are $(1,4),(4,1),(2,3),(3,2)$, so $n(A\cap(B\cup C)) = 4$. Then $P(A\cap(B\cup C))=\frac{n(A\cap(B\cup C))}{n(S)}=\frac{4}{36}\approx0.1111$.

Step9: Answer part (e)

Two events are mutually - exclusive if $P(A\cap B)=0$. Since $P(A\cap B)=\frac{2}{36}\neq0$, a sum of 5 and a product of 4 are not mutually exclusive events because $P(A\cap B)\neq0$.

Answer:

Part (a): $0.8889$ Part (b): $0.3333$ Part (c): $0.9444$ Part (d): $0.1111$ Part (e): B. A sum of 5 and a product of 4 are not mutually exclusive events because $P(A\cap B)\neq0$.