Question

Let \(C_{1}\) and \(C_{2}\) be independent events with \(P\left(C_{1}\right)=0.6\) and \(P\left(C_{2}\right)=0.3\). Compute (a) \(P\left(C_{1} \cap C_{2}\right)\), (b) \(P\left(C_{1} \cup C_{2}\right)\), and (c) \(P\left(C_{1} \cup C_{2}^{c}\right)\).

Short answer

(a) The probability of the intersection of the two events is 0.18. (b) The probability of the union of the two events is 0.72. (c) The probability of the union of \(C_{1}\) and the complement of \(C_{2}\) is 0.88.

Step-by-step solution

Computing Intersection Probability

Since \(C_{1}\) and \(C_{2}\) are independent events, the probability of their intersection is the product of their probabilities. So, \(P(C_{1} \cap C_{2}) = P(C_{1}) \times P(C_{2}) = 0.6 \times 0.3 = 0.18.

Computing Union Probability

Next, to compute the union of these two events, use the formula for the probability of union of two events, that is \(P(C_{1} \cup C_{2}) = P(C_{1}) + P(C_{2}) - P(C_{1} \cap C_{2}) = 0.6 + 0.3 - 0.18 = 0.72.

Computing Union Probability with Complement

Finally, for part (c) first compute the probability of the complement of \(C_{2}\) \(P(C_{2}^{c}) = 1 - P(C_{2}) = 1 - 0.3 = 0.7. Then, compute the union of \(C_{1}\) and \(C_{2}^{c}\) using the formula for the probability of union of two events, that is \(P(C_{1} \cup C_{2}^{c}) = P(C_{1}) + P(C_{2}^{c}) - P(C_{1} \cap C_{2}^{c}). Since \(C_{1}\) and \(C_{2}^{c}\) are also independent events (as they are based on the same original events which are independent), you can use the property of independence, the probability of their intersection is the product of their probabilities. So, \(P(C_{1} \cap C_{2}^{c}) = P(C_{1}) \times P(C_{2}^{c}) = 0.6 \times 0.7 = 0.42. Substituting all values, \(P(C_{1} \cup C_{2}^{c}) = 0.6 + 0.7 - 0.42 = 0.88.