Statistics And Probability Question 114
Question: If A, B, C are 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 find the range of $ x=P(B\cap C) $ lies in the interval
Options:
A) $ 0.23\le x\le 0.48 $
B) $ 0.23\le x\le 0.47 $
C) $ 0.22\le x\le 0.48 $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
- [a] Since $ 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) $ Or $ P(A\cup B\cup C)=0.3+0.4+0.8-(0.08+0.28) $ $ +P(B\cap C)+0.09 $ $ F=1.23-P(B\cap C) $ Or $ P(B\cap C)=1.23-P(A\cup B\cup C) $ But we know that $ 0\le P(A\cup B\cup C)1 $ Hence $ 0.23\le P(B\cap C)\le 0.48 $
BETA
