KCET · Maths · Probability
Probability of occurrence of an event A is \(\dfrac{1}{2}\) and that of B is \(\dfrac{3}{10}\). If A and B are mutually exclusive, then the probability of occurrence of neither A nor B is
- A \(\dfrac{4}{5}\)
- B \(\dfrac{3}{5}\)
- C \(\dfrac{2}{5}\)
- D \(\dfrac{1}{5}\)
Answer & Solution
Correct Answer
(D) \(\dfrac{1}{5}\)
Step-by-step Solution
Detailed explanation
Given \(P(A) = \dfrac{1}{2}\) and \(P(B) = \dfrac{3}{10}\).
Since A and B are mutually exclusive events, \(P(A \cap B) = 0\).
The probability of occurrence of either A or B is given by \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\).
Substituting the values, \(P(A \cup B) = \dfrac{1}{2} + \dfrac{3}{10} - 0 = \dfrac{5 + 3}{10} = \dfrac{8}{10} = \dfrac{4}{5}\).
The probability of occurrence of neither A nor B is \(P(A' \cap B') = 1 - P(A \cup B)\).
\(P(A' \cap B') = 1 - \dfrac{4}{5} = \dfrac{1}{5}\).
Answer: \(\dfrac{1}{5}\)
Since A and B are mutually exclusive events, \(P(A \cap B) = 0\).
The probability of occurrence of either A or B is given by \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\).
Substituting the values, \(P(A \cup B) = \dfrac{1}{2} + \dfrac{3}{10} - 0 = \dfrac{5 + 3}{10} = \dfrac{8}{10} = \dfrac{4}{5}\).
The probability of occurrence of neither A nor B is \(P(A' \cap B') = 1 - P(A \cup B)\).
\(P(A' \cap B') = 1 - \dfrac{4}{5} = \dfrac{1}{5}\).
Answer: \(\dfrac{1}{5}\)
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