Question

Given that A,B and C are events such that $ P(A) = P(B) = P(C) = \dfrac{1}{5} $ , $ P(A \cap B) = P(B \cap C) = 0 $ and $ P(A \cap C) = \dfrac{1}{{10}} $ . The probability that at least one of the events $ A,B \;or\; C $ occurs is
A) $ \dfrac{1}{2} $
B) $ \dfrac{2}{3} $
C) $ \dfrac{2}{5} $
D) $ \dfrac{3}{5} $

Answer 1

Hint: We first calculate the value of probability of intersection of A, B and C. Using this value we further find the probability of occurrence of at least one of the events. Basic sets formulas are also being used in this question.

Complete step by step solution:
It is given in the question that
 $ P(A) = P(B) = P(C) = \dfrac{1}{5} $ ,
 $ P(A \cap B) = P(B \cap C) = 0 $
 $ P(A \cap C) = \dfrac{1}{{10}} $
We know that ,
 $
  P(A \cap B \cap C) \leqslant P(A \cap B) \\
  P(A \cap B \cap C) \leqslant 0 - (i) \;
  $
We also know that the value of $ P(A \cap B \cap C) $ will be a positive integer values
 $ P(A \cap B \cap C) \geqslant 0 - (ii) $
Comparing the above two equations we get, $ P(A \cap B \cap C) = 0 $
Now, for calculating the probability we use this formula,
P(At least one A,B or C)= $ P(A \cup B \cup C) $
 $
  P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) + P(A \cap B \cap C) \\
\Rightarrow P(A \cup B \cup C) = \dfrac{1}{5} + \dfrac{1}{5} + \dfrac{1}{5} - 0 - \dfrac{1}{{10}} - 0 + 0 \\
\Rightarrow P(A \cup B \cup C) = \dfrac{3}{5} - \dfrac{1}{{10}} \\
\Rightarrow P(A \cup B \cup C) = \dfrac{{6 - 1}}{{10}} \\
\Rightarrow P(A \cup B \cup C) = \dfrac{5}{{10}} = \dfrac{1}{2} \;
  $
So, the correct answer is “Option A”.

Note: When the intersection of A,B and C is zero it implies that the event is exhaustive, that means the events are completely independent and none of the sets is dependent on the other.