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A bag contains 15 tickets numbered 1 to 15. A ticket is drawn and replaced. Then one more ticket is drawn and replaced. The probability that first number drawn is even and second is odd is
- a)56/225
- b)26/578
- c)57/289
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A bag contains 15 tickets numbered 1 to 15. A ticket is drawn and repl...
In the first draw, we have 7 even tickets out of 15 and in the second we have 8 odd tickets out of 15.
Thus, (7/15) x (8/15) = 56/225.
Thus, (7/15) x (8/15) = 56/225.
Most Upvoted Answer
A bag contains 15 tickets numbered 1 to 15. A ticket is drawn and repl...
Solution:
To find the probability that the first number drawn is even and the second number drawn is odd, we need to consider two scenarios:
Scenario 1: First number drawn is even
In this scenario, there are 7 even numbers (2, 4, 6, 8, 10, 12, 14) out of the total 15 numbers in the bag. So the probability of drawing an even number on the first draw is 7/15.
Scenario 2: Second number drawn is odd
In this scenario, there are 8 odd numbers (1, 3, 5, 7, 9, 11, 13, 15) out of the total 15 numbers in the bag. So the probability of drawing an odd number on the second draw is 8/15.
Since both draws involve replacing the ticket back into the bag, the probability of each draw is independent of the other.
To find the overall probability of both scenarios happening, we multiply the probabilities of each scenario together:
P(First number drawn is even and second number drawn is odd) = P(First number drawn is even) * P(Second number drawn is odd)
= (7/15) * (8/15)
= 56/225
Therefore, the correct answer is option A) 56/225.
To find the probability that the first number drawn is even and the second number drawn is odd, we need to consider two scenarios:
Scenario 1: First number drawn is even
In this scenario, there are 7 even numbers (2, 4, 6, 8, 10, 12, 14) out of the total 15 numbers in the bag. So the probability of drawing an even number on the first draw is 7/15.
Scenario 2: Second number drawn is odd
In this scenario, there are 8 odd numbers (1, 3, 5, 7, 9, 11, 13, 15) out of the total 15 numbers in the bag. So the probability of drawing an odd number on the second draw is 8/15.
Since both draws involve replacing the ticket back into the bag, the probability of each draw is independent of the other.
To find the overall probability of both scenarios happening, we multiply the probabilities of each scenario together:
P(First number drawn is even and second number drawn is odd) = P(First number drawn is even) * P(Second number drawn is odd)
= (7/15) * (8/15)
= 56/225
Therefore, the correct answer is option A) 56/225.
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