Question:
Two students π΄ and π΅ are assigned to solve a problem separately. The (conditional) probability that π΄ can solve the problem given that π΅ cannot solve it, is 1/5. The (conditional) probability that π΅ can solve the problem given that π΄ can solve the problem is 3/5 . The probability that π΄ can solve the problem is 1/10.Then, the probability that π΅ can solve the problem is ______ (rounded off to one decimal place).
Two students π΄ and π΅ are assigned to solve a problem separately. The (conditional) probability that π΄ can solve the problem given that π΅ cannot solve it, is 1/5. The (conditional) probability that π΅ can solve the problem given that π΄ can solve the problem is 3/5 . The probability that π΄ can solve the problem is 1/10.Then, the probability that π΅ can solve the problem is ______ (rounded off to one decimal place).
Updated On: Feb 10, 2025
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Correct Answer: 0.8
Solution and Explanation
\[ P(A) = \frac{1}{10}, \quad P(B|A) = \frac{3}{5}, \quad P(A|B^c) = \frac{1}{5} \] Using the law of total probability: \[ P(B) = P(B|A) P(A) + P(B|A^c) P(A^c) \] Since, \[ P(A^c) = 1 - P(A) = 1 - \frac{1}{10} = \frac{9}{10} \] And given: \[ P(B|A^c) = 1 - P(A|B^c) = 1 - \frac{1}{5} = \frac{4}{5} \] Substituting the values: \[ P(B) = \frac{3}{5} \times \frac{1}{10} + \frac{4}{5} \times \frac{9}{10} \] \[ = \frac{3}{50} + \frac{36}{50} = \frac{39}{50} = 0.78 \]
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