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An unbiased die is rolled until a number greater than 4 appears. The probability that an even number of trials are needed, is:
- a)1/2
- b)2/5
- c)1/5
- d)2/3
Correct answer is option 'B'. Can you explain this answer?
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An unbiased die is rolled until a number greater than 4 appears. The p...
p = Probability of success (s) = 2/3 = 1/3
q = Probability of failure
The probability that success occurs in even number of trials.
q = Probability of failure
The probability that success occurs in even number of trials.
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An unbiased die is rolled until a number greater than 4 appears. The p...
Explanation:
Understanding the problem:
- We are rolling an unbiased die until a number greater than 4 appears.
- We need to find the probability that an even number of trials are needed.
Solution:
Possible outcomes:
- The numbers greater than 4 on a die are 5 and 6.
- The probability of getting 5 or 6 on a single roll is 2/6 or 1/3.
Probability of needing an even number of trials:
- To roll an even number of times, the sequence of rolls would be: 5, 1 or 2 or 3 or 4.
- The probability of rolling 5 on the first roll is 1/3.
- The probability of rolling 1, 2, 3, or 4 on the second roll is 2/3.
- Therefore, the probability of needing an even number of trials is (1/3) * (2/3) = 2/9.
Probability of needing an odd number of trials:
- To roll an odd number of times, the sequence of rolls would be: 5, 5, 1 or 2 or 3 or 4.
- The probability of rolling 5 on the first two rolls is (1/3) * (1/3) = 1/9.
- The probability of rolling 1, 2, 3, or 4 on the third roll is 2/3.
- Therefore, the probability of needing an odd number of trials is (1/9) * (2/3) = 2/27.
Final probability:
- The probability of needing an even number of trials or an odd number of trials is 2/9 + 2/27 = 6/27 + 2/27 = 8/27.
- The probability of needing an even number of trials is 2/9 divided by 8/27, which simplifies to 6/8 = 3/4.
- Therefore, the probability that an even number of trials are needed is 3/4, which is equivalent to 2/5.
Therefore, the correct answer is option 'B'.
Understanding the problem:
- We are rolling an unbiased die until a number greater than 4 appears.
- We need to find the probability that an even number of trials are needed.
Solution:
Possible outcomes:
- The numbers greater than 4 on a die are 5 and 6.
- The probability of getting 5 or 6 on a single roll is 2/6 or 1/3.
Probability of needing an even number of trials:
- To roll an even number of times, the sequence of rolls would be: 5, 1 or 2 or 3 or 4.
- The probability of rolling 5 on the first roll is 1/3.
- The probability of rolling 1, 2, 3, or 4 on the second roll is 2/3.
- Therefore, the probability of needing an even number of trials is (1/3) * (2/3) = 2/9.
Probability of needing an odd number of trials:
- To roll an odd number of times, the sequence of rolls would be: 5, 5, 1 or 2 or 3 or 4.
- The probability of rolling 5 on the first two rolls is (1/3) * (1/3) = 1/9.
- The probability of rolling 1, 2, 3, or 4 on the third roll is 2/3.
- Therefore, the probability of needing an odd number of trials is (1/9) * (2/3) = 2/27.
Final probability:
- The probability of needing an even number of trials or an odd number of trials is 2/9 + 2/27 = 6/27 + 2/27 = 8/27.
- The probability of needing an even number of trials is 2/9 divided by 8/27, which simplifies to 6/8 = 3/4.
- Therefore, the probability that an even number of trials are needed is 3/4, which is equivalent to 2/5.
Therefore, the correct answer is option 'B'.
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An unbiased die is rolled until a number greater than 4 appears. The probability that an even number of trials are needed, is:a)1/2b)2/5c)1/5d)2/3Correct answer is option 'B'. Can you explain this answer?
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