Question

a triangular die is a four - sided die, each side possessing either a number 1, 2, 3, or 4. two such dice are tossed simultaneously and the bottom faces - the face that each die lands on - is observed. assume the two dice are fair. consider the following events: a the bottom - most faces sum to six b each bottom - most face shows an even number c both bottom - most faces show the same number, referred to as doubles part (a) what is the probability that the bottom - most faces do not sum to six? (use four decimals in your answer) part (b) compute $p(acup b)$ (use four decimals) part (c) compute $p(a^{c}cap c^{c})$ (use four decimals) part (d) compute $p(acap bcap c)$ (use four decimals) part (e) are the events a and c mutually exclusive events? select the most appropriate reason below. a. a and c are mutually exclusive events because $p(acap c)=0$. b. a and c are not mutually exclusive events because $p(acap c)=p(a)p(c)$. c. a and c are not mutually exclusive events because $p(acap c)
eq0$. d. a and c are mutually exclusive events because they are not independent events. e. a and c are not mutually exclusive events because $p(acap c)
eq p(a)p(c)$.

Explanation:

Step1: Find total number of outcomes

When two four - sided dice are tossed, the total number of outcomes is $n(S)=4\times4 = 16$ since each die has 4 possible outcomes.

Step2: Define event A

Event A: The bottom - most faces sum to six. The pairs that sum to six are $(2,4),(3,3),(4,2)$. So $n(A)=3$. Then $P(A)=\frac{n(A)}{n(S)}=\frac{3}{16} = 0.1875$.

Step3: Answer part (a)

The probability that the bottom - most faces do not sum to six is $P(A^C)=1 - P(A)=1 - 0.1875=0.8125$.

Step4: Define event B

Event B: Each bottom - most face shows an even number. The pairs are $(2,2),(2,4),(4,2),(4,4)$. So $n(B) = 4$ and $P(B)=\frac{n(B)}{n(S)}=\frac{4}{16}=0.25$. And $A\cap B=\{(2,4),(4,2)\}$, so $n(A\cap B)=2$ and $P(A\cap B)=\frac{n(A\cap B)}{n(S)}=\frac{2}{16}=0.125$.

Step5: Answer part (b)

Using the formula $P(A\cup B)=P(A)+P(B)-P(A\cap B)$, we substitute the values: $P(A\cup B)=0.1875 + 0.25-0.125=0.3125$.

Step6: Define event C

Event C: Both bottom - most faces show the same number. The pairs are $(1,1),(2,2),(3,3),(4,4)$. So $n(C)=4$ and $P(C)=\frac{n(C)}{n(S)}=\frac{4}{16}=0.25$. Also, $A^C$ is the event that the sum is not six and $C^C$ is the event that the numbers are not doubles.
The number of elements in $A^C\cap C^C$: First, find the number of non - six sum and non - double pairs. The total number of pairs is 16. The number of six - sum pairs is 3 and the number of double pairs is 4, and the pair $(3,3)$ is counted in both. So $n(A^C\cap C^C)=16-(3 + 4-1)=10$. Then $P(A^C\cap C^C)=\frac{n(A^C\cap C^C)}{n(S)}=\frac{10}{16}=0.6250$.

Step7: Define $A\cap B\cap C$

$A\cap B\cap C$ is the intersection of all three events. Since there is no pair that is a double, sums to six and has both numbers even simultaneously, $n(A\cap B\cap C)=0$ and $P(A\cap B\cap C)=\frac{n(A\cap B\cap C)}{n(S)}=0$.

Step8: Answer part (e)

Two events A and C are mutually exclusive if $P(A\cap C) = 0$. The pair $(3,3)$ is in both A and C, so $P(A\cap C)
eq0$. So A and C are not mutually exclusive events because $P(A\cap C)
eq0$.

Answer:

Part (a): $0.8125$
Part (b): $0.3125$
Part (c): $0.6250$
Part (d): $0$
Part (e): C. A and C are not mutually exclusive events because $P(A\cap C)
eq0$.