KCET · Maths · Probability
Probability of at least one of the events A and B occur is \(0.6\). If A and B occur simultaneously with probability \(0.2\), then \(P(\bar{A}) + P(\bar{B})\) is
- A \(1\)
- B \(0.8\)
- C \(0.6\)
- D \(1.2\)
Answer & Solution
Correct Answer
(D) \(1.2\)
Step-by-step Solution
Detailed explanation
Given \(P(A \cup B) = 0.6\) and \(P(A \cap B) = 0.2\).
Using the addition theorem of probability:
\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
\(0.6 = P(A) + P(B) - 0.2\)
\(P(A) + P(B) = 0.8\)
We need to find \(P(\bar{A}) + P(\bar{B})\):
\(P(\bar{A}) + P(\bar{B}) = (1 - P(A)) + (1 - P(B))\)
\(P(\bar{A}) + P(\bar{B}) = 2 - (P(A) + P(B))\)
\(P(\bar{A}) + P(\bar{B}) = 2 - 0.8 = 1.2\)
Answer: \(1.2\)
Using the addition theorem of probability:
\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
\(0.6 = P(A) + P(B) - 0.2\)
\(P(A) + P(B) = 0.8\)
We need to find \(P(\bar{A}) + P(\bar{B})\):
\(P(\bar{A}) + P(\bar{B}) = (1 - P(A)) + (1 - P(B))\)
\(P(\bar{A}) + P(\bar{B}) = 2 - (P(A) + P(B))\)
\(P(\bar{A}) + P(\bar{B}) = 2 - 0.8 = 1.2\)
Answer: \(1.2\)
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