Question

Find​ P(A or B or​ C) for the given probabilities. Upper P left parenthesis Upper A right parenthesis equals 0.34 comma Upper P left parenthesis Upper B right parenthesis equals 0.23 comma Upper P left parenthesis Upper C right parenthesis equals 0.12 Upper P left parenthesis Upper A and Upper B right parenthesis equals 0.11 comma Upper P left parenthesis Upper A and Upper C right parenthesis equals 0.04 comma Upper P left parenthesis Upper B and Upper C right parenthesis equals 0.07 Upper P left parenthesis Upper A and Upper B and Upper C right parenthesis equals 0.01 Upper P left parenthesis Upper A or Upper B or Upper C right parenthesis equals nothing

Answer 1

Answer:

$P(A \cup B \cup C)=0.48$

Step-by-step explanation:

Given the following probabilities for events A and B

$P(A)=0.34\\P (B )= 0.23\\ P (C )= 0.12\\ P(A \cap B )= 0.11\\ P (A \cap C )= 0.04\\ P (B \cap C )= 0.07\\P(A \cap B \cap C)=0.01$

We want to find P(A ∪ B ∪ C).

Using the  inclusion/exclusion formula for the union of three events:

P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(A ∩ B) − P(A ∩ C)−P(B∩C)+P(A∩B∩C).

$P(A \cup B \cup C)=0.34+0.23+0.12-0.11-0.04-0.07+0.01\\P(A \cup B \cup C)=0.48$

Therefore, P(A or B or​ C) = 0.48