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A bag contains tickets numbered 1 to 30. Three tickets are chosen at random. Find the probability that all the tickets have two digit numbers.
- a)11/58
- b)19/58
- c)21/30
- d)2/3
- e)5/8
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
A bag contains tickets numbered 1 to 30. Three tickets are chosen at r...
Calculation
- Total number of ways to choose 3 tickets out of 30 = 30C3
- Total number of ways to choose 3 two-digit tickets = 10C3 (since there are 10 two-digit numbers from 10 to 30)
- Probability = (Number of favorable outcomes) / (Total number of outcomes) = 10C3 / 30C3
Solution
- Calculate the combinations:
- 30C3 = 30! / (3! * (30-3)!) = 30 * 29 * 28 / 6 = 4060
- 10C3 = 10! / (3! * (10-3)!) = 10 * 9 * 8 / 6 = 120
- Substitute into the probability formula:
- Probability = 120 / 4060 = 19 / 58
Therefore, the probability that all three tickets chosen have two-digit numbers is 19/58, which corresponds to option 'b'.
- Total number of ways to choose 3 tickets out of 30 = 30C3
- Total number of ways to choose 3 two-digit tickets = 10C3 (since there are 10 two-digit numbers from 10 to 30)
- Probability = (Number of favorable outcomes) / (Total number of outcomes) = 10C3 / 30C3
Solution
- Calculate the combinations:
- 30C3 = 30! / (3! * (30-3)!) = 30 * 29 * 28 / 6 = 4060
- 10C3 = 10! / (3! * (10-3)!) = 10 * 9 * 8 / 6 = 120
- Substitute into the probability formula:
- Probability = 120 / 4060 = 19 / 58
Therefore, the probability that all three tickets chosen have two-digit numbers is 19/58, which corresponds to option 'b'.
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A bag contains tickets numbered 1 to 30. Three tickets are chosen at random. Find the probability that all the tickets have two digit numbers.a)11/58b)19/58c)21/30d)2/3e)5/8Correct answer is option 'B'. Can you explain this answer?
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