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If E(1) and E(2) be two events such that...

If `E_(1)` and `E_(2)` be two events such that `P(E_(1))=0.3, P(E_(2))=0.2` and `P(E_(1)nnE_(2))=0.1,` then find:
(i) `P(barE_(1)nnE_(2))` (ii) `P(E_(1)nnbarE_(2))`

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The correct Answer is:
To solve the problem, we need to find two probabilities based on the given information. Let's break it down step by step. Given: - \( P(E_1) = 0.3 \) - \( P(E_2) = 0.2 \) - \( P(E_1 \cap E_2) = 0.1 \) ### Part (i): Find \( P(\bar{E_1} \cap E_2) \) 1. **Understand the formula**: The probability of the intersection of the complement of \( E_1 \) and \( E_2 \) can be expressed as: \[ P(\bar{E_1} \cap E_2) = P(E_2) - P(E_1 \cap E_2) \] 2. **Substitute the known values**: - We know \( P(E_2) = 0.2 \) and \( P(E_1 \cap E_2) = 0.1 \). \[ P(\bar{E_1} \cap E_2) = 0.2 - 0.1 \] 3. **Calculate the result**: \[ P(\bar{E_1} \cap E_2) = 0.1 \] ### Part (ii): Find \( P(E_1 \cap \bar{E_2}) \) 1. **Understand the formula**: The probability of the intersection of \( E_1 \) and the complement of \( E_2 \) can be expressed as: \[ P(E_1 \cap \bar{E_2}) = P(E_1) - P(E_1 \cap E_2) \] 2. **Substitute the known values**: - We know \( P(E_1) = 0.3 \) and \( P(E_1 \cap E_2) = 0.1 \). \[ P(E_1 \cap \bar{E_2}) = 0.3 - 0.1 \] 3. **Calculate the result**: \[ P(E_1 \cap \bar{E_2}) = 0.2 \] ### Final Answers: - (i) \( P(\bar{E_1} \cap E_2) = 0.1 \) - (ii) \( P(E_1 \cap \bar{E_2}) = 0.2 \)
To solve the problem, we need to find two probabilities based on the given information. Let's break it down step by step. Given: - \( P(E_1) = 0.3 \) - \( P(E_2) = 0.2 \) - \( P(E_1 \cap E_2) = 0.1 \) ### Part (i): Find \( P(\bar{E_1} \cap E_2) \) ...
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