Let S = {S1, S2, S3, S4, S5} be a sample space with events A = {S1, S2, S4, S5}, B = {S2, S3}, and D = {S1, S4, S5}. Use the partial probability distribution of outcomes to compute... Let S = {S1, S2, S3, S4, S5} be a sample space with events A = {S1, S2, S4, S5}, B = {S2, S3}, and D = {S1, S4, S5}. Use the partial probability distribution of outcomes to compute each of the probabilities: (a) P(S2) = ? (b) P(Ac) = ? (c) P(Ac ∪ D) = ? (d) P(A ∩ B ∩ D) = ? Read more

Steps to Solve

1. Find $P(S_2)$

From the table, the probability of $S_2$ is $\frac{1}{10}$. Convert this to a fraction with a denominator of 100. $$\frac{1}{10} = \frac{1 \times 10}{10 \times 10} = \frac{10}{100}$$

2. Find $P(A^c)$

$A^c$ is the complement of $A$, which means it includes all outcomes in $S$ that are not in $A$. Since $A = {S_1, S_2, S_4, S_5}$, then $A^c = {S_3}$. So, we need to find $P(S_3)$. From the table, $P(S_3) = \frac{3}{20}$. Convert this to a fraction with a denominator of 100. $$\frac{3}{20} = \frac{3 \times 5}{20 \times 5} = \frac{15}{100}$$

3. Find $P(A^c \cup D)$

$A^c = {S_3}$ and $D = {S_1, S_4, S_5}$. The union of these two sets $A^c \cup D = {S_1, S_3, S_4, S_5}$. So, $P(A^c \cup D) = P(S_1) + P(S_3) + P(S_4) + P(S_5)$. $P(S_1) = \frac{31}{100}$, $P(S_3) = \frac{3}{20} = \frac{15}{100}$, $P(S_4) = \frac{7}{50} = \frac{14}{100}$, and $P(S_5) = \frac{7}{50} = \frac{14}{100}$. Therefore, $P(A^c \cup D) = \frac{31}{100} + \frac{15}{100} + \frac{14}{100} + \frac{14}{100} = \frac{31+15+14+14}{100} = \frac{74}{100}$.

4. Find $P(A \cap B \cap D)$

$A = {S_1, S_2, S_4, S_5}$, $B = {S_2, S_3}$, and $D = {S_1, S_4, S_5}$. $A \cap B \cap D$ means the intersection of all three sets, i.e., the elements that are present in all three sets. The only element in all three sets is $S_2$ is not in $D$, $S_3$ is not in $A$ or $D$, $S_1$ is not in $B$, $S_4$ is not in $B$, and $S_5$ is not in $B$. Therefore, $A \cap B \cap D = \emptyset$, which means $P(A \cap B \cap D) = 0$.