Question

a pair of fair dice is tossed. define the events a and b as follows. complete parts a through d below. a {a 7 is rolled} (the sum of the numbers of dots on the upper - faces of the two dice is equal to 7.) b {at least one of the two dice is showing a 4} identify the sample points in the event a^c. choose the correct answer below. a. a^c ={(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)} b. a^c ={(3,4),(4,3)} c. a^c ={(1,1),(1,2),(1,3),(1,4),(1,5),(2,1),(2,2),(2,3),(2,4),(2,6),(3,1),(3,2),(3,3),(3,5),(3,6),(4,1),(4,2),(4,4),(4,5),(4,6),(5,1),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3),(6,4),(6,5),(6,6)} d. a^c ={(1,4),(1,6),(2,4),(2,5),(3,4),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,2),(5,4),(6,1),(6,4)} b. find p(a), p(b), p(a ∩ b), p(a ∪ b), and p(a^c) by summing the probabilities of the appropriate sample points. since the probability of each sample point in a is 1/6 and there are 6 sample point(s) in a, p(a)= 1/6. since the probability of each sample point in b is 11/36 and there is/are 11 sample point(s) in b, p(b)= 11/36. since the probability of each sample point in a ∩ b is 1/36 and there is/are sample point(s) in a ∩ b, p(a ∩ b)=

Explanation:

Step1: Recall sample - space size

When two dice are rolled, the sample - space $n(S)=6\times6 = 36$.

Step2: Calculate $P(A)$

Event $A$: at least one of the two dice shows a 4. The number of elements in $A$ can be found by considering the cases: $(1,4),(2,4),(3,4),(4,4),(5,4),(6,4),(4,1),(4,2),(4,3),(4,5),(4,6)$. So $n(A)=11$, and $P(A)=\frac{n(A)}{n(S)}=\frac{11}{36}$.

Step3: Calculate $P(B)$

Event $B$: the sum of the numbers on the upper - faces of the two dice is equal to 7. The elements of $B$ are $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$. So $n(B)=6$, and $P(B)=\frac{n(B)}{n(S)}=\frac{6}{36}=\frac{1}{6}$.

Step4: Calculate $P(A\cap B)$

The elements in $A\cap B$ are the elements that are in both $A$ and $B$. The elements of $A\cap B$ are $(3,4),(4,3)$. So $n(A\cap B)=2$, and $P(A\cap B)=\frac{n(A\cap B)}{n(S)}=\frac{2}{36}=\frac{1}{18}$.

Step5: Calculate $P(A\cup B)$

Use the formula $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. Substitute $P(A)=\frac{11}{36}$, $P(B)=\frac{1}{6}=\frac{6}{36}$, and $P(A\cap B)=\frac{1}{18}=\frac{2}{36}$ into the formula. Then $P(A\cup B)=\frac{11 + 6-2}{36}=\frac{15}{36}=\frac{5}{12}$.

Step6: Calculate $P(A^{C})$

Use the formula $P(A^{C})=1 - P(A)$. Since $P(A)=\frac{11}{36}$, then $P(A^{C})=1-\frac{11}{36}=\frac{36 - 11}{36}=\frac{25}{36}$.

Answer:

$P(A)=\frac{11}{36}$, $P(B)=\frac{1}{6}$, $P(A\cap B)=\frac{1}{18}$, $P(A\cup B)=\frac{5}{12}$, $P(A^{C})=\frac{25}{36}$