91Ó°ÊÓ

Since the sum of all probabilities in a distribution table must be equal to 1, we can find the missing probability for outcome 'e' using the following equation: \(1 - (P(a) + P(b) + P(c) + P(d)) = P(e)\) Given values: \(P(a) = 0.1, P(b) = 0.05, P(c) = 0.6, P(d) = 0.05\) Now, we can find the missing probability: \(P(e) = 1 - (0.1 + 0.05 + 0.6 + 0.05)= 0.2\) Now, our completed probability distribution table looks like this: $$ \begin{array}{|r|c|c|c|c|c|} \hline \text { Outcome } & \mathrm{a} & \mathrm{b} & \mathrm{c} & \mathrm{d} & \mathrm{e} \\\ \hline \text { Probability } & .1 & .05 & .6 & .05 & .2 \\\ \hline \end{array} $$

Now, let's calculate the required probabilities: a. \(P(\{a, c, e\})\) To find this probability, we will sum the individual probabilities for outcomes 'a','c' and 'e': \(P(\{a, c, e\}) = P(a) + P(c) + P(e) = 0.1 + 0.6 + 0.2 = 0.9\) b. \(P(E \cup F)\), where \(E=\{a, c, e\}\) and \(F=\{b, c, e\}\) The union of events E and F means the probability of at least one of them happening. We will sum the individual probabilities for outcomes 'a', 'b', 'c', and 'e' as they are present in both events: \(P(E \cup F) = P(a) + P(b) + P(c) + P(e) = 0.1 + 0.05 + 0.6 + 0.2 = 0.95\) c. \(P\left(E^{\prime}\right)\), where E is as in part (b) The complement of event E, denoted by \(E'\), means the probability of everything that is not in E. In our table, outcomes 'b' and 'd' are not in E. So, we will sum the individual probabilities for outcomes 'b' and 'd': \(P(E') = P(b) + P(d) = 0.05 + 0.05 = 0.1\) d. \(P(E \cap F)\), where E and F are as in part (b) The intersection of events E and F means the probability of both events happening simultaneously. In our table, outcomes 'c' and 'e' are present in both events. So, we will sum the individual probabilities for outcomes 'c' and 'e': \(P(E \cap F) = P(c) + P(e) = 0.6 + 0.2 = 0.8\)