Suppose events [tex]\( A, B, \)[/tex] and [tex]\( C \)[/tex] are independent and
[tex]\[ P(A) = \frac{1}{2} \quad P(B) = \frac{1}{5} \quad P(C) = \frac{1}{3} \][/tex]

Find the probability. (Enter the probability as a fraction.)
[tex]\[ P(\overline{A \cap B \cap C}) = \square \][/tex]



Answer :

To solve this problem, we need to find the probability of the complement of the intersection of three independent events [tex]\( A \)[/tex], [tex]\( B \)[/tex], and [tex]\( C \)[/tex].

Given:
[tex]\[ P(A) = \frac{1}{2}, \quad P(B) = \frac{1}{5}, \quad P(C) = \frac{1}{3} \][/tex]

1. Calculate the probability of the intersection of [tex]\( A \)[/tex], [tex]\( B \)[/tex], and [tex]\( C \)[/tex] ([tex]\( A \cap B \cap C \)[/tex]).

Since the events are independent, the probability of the intersection of [tex]\( A \)[/tex], [tex]\( B \)[/tex], and [tex]\( C \)[/tex] is:
[tex]\[ P(A \cap B \cap C) = P(A) \times P(B) \times P(C) \][/tex]
Substitute the values:
[tex]\[ P(A \cap B \cap C) = \frac{1}{2} \times \frac{1}{5} \times \frac{1}{3} = \frac{1}{30} \][/tex]

2. Calculate the probability of the complement of [tex]\( A \cap B \cap C \)[/tex].

The complement of [tex]\( A \cap B \cap C \)[/tex] is denoted as [tex]\( \overline{A \cap B \cap C} \)[/tex]. The probability of the complement of an event is [tex]\( 1 \)[/tex] minus the probability of the event itself.
[tex]\[ P(\overline{A \cap B \cap C}) = 1 - P(A \cap B \cap C) \][/tex]
Substitute the value we found for [tex]\( P(A \cap B \cap C) \)[/tex]:
[tex]\[ P(\overline{A \cap B \cap C}) = 1 - \frac{1}{30} = \frac{30}{30} - \frac{1}{30} = \frac{29}{30} \][/tex]

Therefore, the probability [tex]\( P(\overline{A \cap B \cap C}) \)[/tex] is:
[tex]\[ \boxed{\frac{29}{30}} \][/tex]