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

Explanation:

Step1: Find total number of outcomes

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

Step2: Find number of outcomes for event A

The pairs of numbers on the two - dice that sum to 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 C

The pairs for which $|die1 - die2| = 1$ are $(1,2),(2,1),(2,3),(3,2),(3,4),(4,3),(4,5),(5,4),(5,6),(6,5)$. So $n(C)=10$ and $P(C)=\frac{n(C)}{n(S)}=\frac{10}{36}=\frac{5}{18}$. The intersection $A\cap C$ has pairs $(2,3),(3,2)$, so $n(A\cap C)=2$ and $P(A\cap C)=\frac{n(A\cap C)}{n(S)}=\frac{2}{36}=\frac{1}{18}$.

Step4: Find number of outcomes for event B

The pairs for which $die1\times die2 = 4$ are $(1,4),(4,1),(2,2)$. So $n(B)=3$ and $P(B)=\frac{n(B)}{n(S)}=\frac{3}{36}=\frac{1}{12}$.

Step5: Calculate part (a)

The probability that the uppermost faces do not sum to five is $P(\overline{A})=1 - P(A)=1-\frac{4}{36}=\frac{32}{36}\approx0.8889$.

Step6: Calculate part (b)

Using the formula $P(A\cup C)=P(A)+P(C)-P(A\cap C)$, we substitute the values: $P(A\cup C)=\frac{4}{36}+\frac{10}{36}-\frac{2}{36}=\frac{4 + 10-2}{36}=\frac{12}{36}\approx0.3333$.

Step7: Calculate part (c)

The probability that the uppermost faces sum to five is $P(A)=\frac{4}{36}$, and the probability that the uppermost faces have a product of 4 is $P(B)=\frac{3}{36}$. The probability that the uppermost faces sum to five and have a product of 4 is $P(A\cap B)=\frac{2}{36}$ (the pairs $(1,4)$ and $(4,1)$). Using the formula $P(\overline{A}\cup\overline{B})=1 - P(A\cap B)=1-\frac{2}{36}=\frac{34}{36}\approx0.9444$.

Step8: Calculate part (d)

First, $n(B\cup C)=n(B)+n(C)-n(B\cap C)$. The pairs in $B\cap C$ are $(2,2)$ (none), so $n(B\cap C) = 0$. Then $n(B\cup C)=3 + 10-0=13$. The intersection $A\cap(B\cup C)$ has pairs $(2,3),(3,2)$ (from $A\cap C$ and considering $B\cup C$), so $n(A\cap(B\cup C))=2$ and $P(A\cap(B\cup C))=\frac{2}{36}\approx0.0556$.

Step9: Calculate part (e)

Two events are mutually exclusive if $P(A\cap B)=0$. Since the pairs that sum to 5 are $(1,4),(2,3),(3,2),(4,1)$ and the pairs that have a product of 4 are $(1,4),(4,1),(2,2)$, $A\cap B$ has 2 elements. But for mutually - exclusive events $P(A\cap B) = 0$. So the events a sum of 5 and a product of 4 are mutually exclusive events because $P(A\cap B)=0$.

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 $P(A\cap B)=0$.