Question

Let the three mutually independent events \(C_{1}, C_{2}\), and \(C_{3}\) be such that \(P\left(C_{1}\right)=P\left(C_{2}\right)=P\left(C_{3}\right)=\frac{1}{4} .\) Find \(P\left[\left(C_{1}^{c} \cap C_{2}^{c}\right) \cup C_{3}\right]\)

Short answer

The probability of the event \( [(C_1^c \cap C_2^c) \cup C_3] \) is 0.8125.

Step-by-step solution

Determining the Value of Complements

To solve this problem, recall that the probability of an event and its complement must add up to 1, so the complement of \(C_1\) and \(C_2\) would be \(P(C_1^c)=1-P(C_1)=1-0.25=0.75\) and \(P(C_2^c)=1-P(C_2)=1-0.25=0.75\). At this point, \(P(C_1^c) = P(C_2^c) = 0.75\), and \(P(C_3) = 0.25\).

Using Independent Events Property

Since the three events are mutually independent, we find the intersection of the two complements by multiplying their respective probabilities. So, \(P(C_1^c \cap C_2^c) = P(C_1^c) \cdot P(C_2^c) = 0.75 \cdot 0.75 = 0.5625\).

Calculating the Union Probability

On analyzing the intersection with \(C_3\), it can be deduced that since \(C_3\) is independent of the other two events, the probability of the union of \(C_1^c \cap C_2^c\) and \(C_3\) will be equal to the sum of their individual probabilities, according to union of independent events formula. Hence, \(P[(C_1^c \cap C_2^c) \cup C_3] = P(C_1^c \cap C_2^c) + P(C_3) = 0.5625 + 0.25 = 0.8125\). We can therefore conclude that the probability of the union of the complements of \(C_1\) and \(C_2\) with \(C_3\) is 0.8125.