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The probabilities that A and b can solve...

The probabilities that A and b can solve a problem independently are `1/3` and `1/4` respectively. If both try to solve the problem independently, find the probability that:
(i) problem will be solved

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To solve the problem, we need to find the probability that at least one of A or B solves the problem. We can denote the probabilities as follows: - Let \( P(A) \) be the probability that A solves the problem, which is \( \frac{1}{3} \). - Let \( P(B) \) be the probability that B solves the problem, which is \( \frac{1}{4} \). ### Step 1: Understand the problem We need to find the probability that the problem is solved, which can be expressed as: \[ P(A \cup B) \] This represents the probability that either A solves the problem, or B solves the problem, or both. ### Step 2: Use the formula for the union of two events The formula for the probability of the union of two independent events is: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] ### Step 3: Calculate \( P(A \cap B) \) Since A and B are independent, we can calculate \( P(A \cap B) \) as: \[ P(A \cap B) = P(A) \times P(B) \] Substituting the values: \[ P(A \cap B) = \frac{1}{3} \times \frac{1}{4} = \frac{1}{12} \] ### Step 4: Substitute values into the union formula Now we can substitute \( P(A) \), \( P(B) \), and \( P(A \cap B) \) into the union formula: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] \[ P(A \cup B) = \frac{1}{3} + \frac{1}{4} - \frac{1}{12} \] ### Step 5: Find a common denominator To add these fractions, we need a common denominator. The least common multiple (LCM) of 3, 4, and 12 is 12. We can rewrite the fractions: - \( \frac{1}{3} = \frac{4}{12} \) - \( \frac{1}{4} = \frac{3}{12} \) Now substituting these values: \[ P(A \cup B) = \frac{4}{12} + \frac{3}{12} - \frac{1}{12} \] ### Step 6: Perform the addition and subtraction Now we can perform the arithmetic: \[ P(A \cup B) = \frac{4 + 3 - 1}{12} = \frac{6}{12} \] ### Step 7: Simplify the fraction Finally, we simplify \( \frac{6}{12} \): \[ P(A \cup B) = \frac{1}{2} \] ### Final Answer The probability that the problem will be solved is \( \frac{1}{2} \). ---

To solve the problem, we need to find the probability that at least one of A or B solves the problem. We can denote the probabilities as follows: - Let \( P(A) \) be the probability that A solves the problem, which is \( \frac{1}{3} \). - Let \( P(B) \) be the probability that B solves the problem, which is \( \frac{1}{4} \). ### Step 1: Understand the problem We need to find the probability that the problem is solved, which can be expressed as: \[ P(A \cup B) \] ...
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NAGEEN PRAKASHAN ENGLISH-PROBABIILITY-Exercise 13 B
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  2. Given that the events A and B are such that P(A)=1/2, P(AuuB)=3/5a n d...

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  3. If A and B are independent events such that P(A)=0.3 and P(B)=0.5, the...

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  4. If A and B are two events such that P(A) = (1)/(4), P(B) = (1)/(2) and...

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  5. If A and B are two independent events, then the probability of occur...

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  6. A and B are two events such that P(A)=3/5, P(B)=3/10 and P(AuuB)=1/2 ...

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  7. There are 10 black and 8 white balls in a bag. Two balls are drawn wit...

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  8. A card is drawn from a well shuffled pack of 52 cards . In which of th...

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  9. The probabilities that A and b can solve a problem independently are 1...

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  10. Three cards are drawn one by one without replacement from a well shuf...

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  11. In three throws of a dice, find the probability of getting odd number ...

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  12. In a hostel 60% of the students read Hindi news paper, 40% read Englis...

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  13. A speaks truth in 75% and B in 80% of the cases. In what percentage of...

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  14. The odds in favour of A whose age is 45 years will alive upto 60 years...

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  15. In a company there are two vacancies. A man and his wife come for inte...

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  16. A bag contains 50 tickets numbered 1, 2, 3, .., 50 of which five are ...

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  17. In a company there are two vacancies. A man and his wife come for inte...

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