Comprehension

Rohit, Jaspreet, and Alia appeared for an interview for three vacancies in the same post. The probability of Rohit's selection is \( \frac{1}{5} \), Jaspreet's selection is \( \frac{1}{3} \), and Alia's selection is \( \frac{1}{4} \). The events of selection are independent of each other.

Based on the above information, answer the following questions:

Question: 1

What is the probability that at least one of them is selected?

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To find the probability of "at least one" event happening, use the complement rule: \( P({At least one}) = 1 - P({None}) \).
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Solution and Explanation

Step 1: Probability of no one being selected
The probability that none of them are selected is: \[ P({No one selected}) = \left( 1 - \frac{1}{5} \right) \cdot \left( 1 - \frac{1}{3} \right) \cdot \left( 1 - \frac{1}{4} \right) = \frac{4}{5} \cdot \frac{2}{3} \cdot \frac{3}{4} = \frac{2}{5}. \] Step 2: Probability of at least one being selected
The probability that at least one of them is selected is: \[ P({At least one selected}) = 1 - P({No one selected}) = 1 - \frac{2}{5} = \frac{3}{5}. \] Final Result: The probability that at least one of them is selected is \( \frac{3}{5} \).
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Question: 2

Find \( P(G \cap \overline{H}) \), where \( G \) is the event of Jaspreet's selection and \( \overline{H} \) denotes the event that Rohit is not selected.

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For independent events, the probability of their intersection is the product of their individual probabilities.
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Solution and Explanation

Step 1: Compute the probability
The probability \( P(G \cap \overline{H}) \) is given by: \[ P(G \cap \overline{H}) = P(G) \cdot P(\overline{H}), \] where \( P(G) = \frac{1}{3} \) and \( P(\overline{H}) = 1 - P(H) = 1 - \frac{1}{5} = \frac{4}{5} \). \[ P(G \cap \overline{H}) = \frac{1}{3} \cdot \frac{4}{5} = \frac{4}{15}. \] Final Result: The probability \( P(G \cap \overline{H}) \) is \( \frac{4}{15} \).
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Question: 3

Find the probability that exactly one of them is selected.

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To calculate "exactly one" selection, consider all cases where one succeeds, and the others fail, then sum the probabilities.
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Solution and Explanation

Step 1: Compute the probability of exactly one being selected
The probability of exactly one being selected is: \[ P({Exactly one selected}) = P(R) \cdot P(\overline{J}) \cdot P(\overline{A}) + P(\overline{R}) \cdot P(J) \cdot P(\overline{A}) + P(\overline{R}) \cdot P(\overline{J}) \cdot P(A), \] where: \[ P(\overline{J}) = 1 - P(J) = \frac{2}{3}, \quad P(\overline{A}) = 1 - P(A) = \frac{3}{4}, \quad P(\overline{R}) = 1 - P(R) = \frac{4}{5}. \] Substitute the values: \[ P({Exactly one selected}) = \frac{1}{5} \cdot \frac{2}{3} \cdot \frac{3}{4} + \frac{4}{5} \cdot \frac{1}{3} \cdot \frac{3}{4} + \frac{4}{5} \cdot \frac{2}{3} \cdot \frac{1}{4}. \] Simplify each term: \[ P({Exactly one selected}) = \frac{6}{60} + \frac{12}{60} + \frac{8}{60} = \frac{26}{60} = \frac{13}{30}. \] Final Result: The probability that exactly one of them is selected is \( \frac{13}{30} \).
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Question: 4

Find the probability that exactly two of them are selected.

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For "exactly two" events, consider all pairs of selections and one failure, and sum their probabilities.
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Solution and Explanation

Step 1: Compute the probability of exactly two being selected
The probability of exactly two being selected is: \[ P({Exactly two selected}) = P(R) \cdot P(J) \cdot P(\overline{A}) + P(R) \cdot P(\overline{J}) \cdot P(A) + P(\overline{R}) \cdot P(J) \cdot P(A). \] Substitute the values: \[ P({Exactly two selected}) = \frac{1}{5} \cdot \frac{1}{3} \cdot \frac{3}{4} + \frac{1}{5} \cdot \frac{2}{3} \cdot \frac{1}{4} + \frac{4}{5} \cdot \frac{1}{3} \cdot \frac{1}{4}. \] Simplify each term: \[ P({Exactly two selected}) = \frac{3}{60} + \frac{2}{60} + \frac{4}{60} = \frac{9}{60} = \frac{3}{20}. \] Final Result: The probability that exactly two of them are selected is \( \frac{3}{20} \).
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