1. Tardigrade
  2. Question
  3. Mathematics
  4. Let A, B, C be three events such that P (A) = 0.3 P(B) = 0.4, P(C) = 0.8, P(A ∩ B) = 0.08, P(A ∩ C) = 0.28, P(A ∩ B ∩ C) = 0.09. If P(A ∪ B ∪ C) ge 0.75, then

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. Let A, B, C be three events such that P (A) = 0.3 P(B) = 0.4, P(C) = 0.8, $P(A \cap B) = 0.08, P(A \cap C) = 0.28, P(A \cap B \cap C) = 0.09$. If $P(A \cup B \cup C) \ge 0.75$, then

Probability

Solution:

Since $ P\left(A \cup B \cup C\right)\ge0.75 , $ therefore
$0.75 \le P\left(A\cup B\cup C\right) \le1$
$ \Rightarrow 0.75 \le P\left(A\right) + P\left(B\right) + P\left(C\right) - P\left(A \cap B\right) - P\left(B \cap C\right) - P\left(A \cap C\right) + P\left(A\cap B\cap C\right) \le1$
$ \Rightarrow 0.75 \le 0.3 + 0.4 + 0.8 - 0.08 - P\left(B\cap C\right) - 0.28 + 0.09 \le 1 $
$\Rightarrow 0.75 \le1.23 - P\left(B \cap C\right) \le1 $
$\Rightarrow -0.48 \le- P\left(B \cap C\right) \le - 0.23$
$ \Rightarrow 0.23 \le P\left(B\cap C \right)\le 0.48$