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A box contains 12 electric lamps of which 5 are defectives. A man selects three lamps at random. What is the expected number of defective lamps in his selection?
- a)1.25
- b)2.5
- c)1.05
- d)2.03
Correct answer is option 'A'. Can you explain this answer?
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A box contains 12 electric lamps of which 5 are defectives. A man sele...
Solution:
Given,
Total number of lamps = 12
Number of defective lamps = 5
We need to find the expected number of defective lamps in the selection of 3 lamps at random.
To solve this problem, we can use the concept of probability.
Probability of selecting a defective lamp in the first draw = 5/12
Probability of selecting a non-defective lamp in the first draw = 7/12
After the first draw, there will be 11 lamps left, out of which 4 are defective and 7 are non-defective.
Probability of selecting a defective lamp in the second draw, given that the first lamp was defective = 4/11
Probability of selecting a non-defective lamp in the second draw, given that the first lamp was defective = 7/11
Similarly, after the second draw, there will be 10 lamps left, out of which 3 are defective and 7 are non-defective.
Probability of selecting a defective lamp in the third draw, given that the first two lamps were defective = 3/10
Probability of selecting a non-defective lamp in the third draw, given that the first two lamps were defective = 7/10
Using the above probabilities, we can find the probability of selecting different combinations of defective and non-defective lamps in the selection of 3 lamps at random.
Number of defective lamps in the selection of 3 lamps at random can be 0, 1, 2, or 3.
Probability of selecting 0 defective lamps = (7/12) x (6/11) x (5/10) = 35/264
Probability of selecting 1 defective lamp = [(5/12) x (7/11) x (4/10)] + [(7/12) x (5/11) x (4/10)] + [(7/12) x (6/11) x (3/10)] = 35/66
Probability of selecting 2 defective lamps = [(5/12) x (4/11) x (7/10)] + [(5/12) x (7/11) x (3/10)] + [(7/12) x (5/11) x (3/10)] = 25/66
Probability of selecting 3 defective lamps = (5/12) x (4/11) x (3/10) = 3/88
Expected number of defective lamps in the selection of 3 lamps at random can be calculated as follows:
Expected number of defective lamps = (0 x 35/264) + (1 x 35/66) + (2 x 25/66) + (3 x 3/88) = 5/4 = 1.25
Therefore, the expected number of defective lamps in the selection of 3 lamps at random is 1.25. Hence, option A is the correct answer.
Given,
Total number of lamps = 12
Number of defective lamps = 5
We need to find the expected number of defective lamps in the selection of 3 lamps at random.
To solve this problem, we can use the concept of probability.
Probability of selecting a defective lamp in the first draw = 5/12
Probability of selecting a non-defective lamp in the first draw = 7/12
After the first draw, there will be 11 lamps left, out of which 4 are defective and 7 are non-defective.
Probability of selecting a defective lamp in the second draw, given that the first lamp was defective = 4/11
Probability of selecting a non-defective lamp in the second draw, given that the first lamp was defective = 7/11
Similarly, after the second draw, there will be 10 lamps left, out of which 3 are defective and 7 are non-defective.
Probability of selecting a defective lamp in the third draw, given that the first two lamps were defective = 3/10
Probability of selecting a non-defective lamp in the third draw, given that the first two lamps were defective = 7/10
Using the above probabilities, we can find the probability of selecting different combinations of defective and non-defective lamps in the selection of 3 lamps at random.
Number of defective lamps in the selection of 3 lamps at random can be 0, 1, 2, or 3.
Probability of selecting 0 defective lamps = (7/12) x (6/11) x (5/10) = 35/264
Probability of selecting 1 defective lamp = [(5/12) x (7/11) x (4/10)] + [(7/12) x (5/11) x (4/10)] + [(7/12) x (6/11) x (3/10)] = 35/66
Probability of selecting 2 defective lamps = [(5/12) x (4/11) x (7/10)] + [(5/12) x (7/11) x (3/10)] + [(7/12) x (5/11) x (3/10)] = 25/66
Probability of selecting 3 defective lamps = (5/12) x (4/11) x (3/10) = 3/88
Expected number of defective lamps in the selection of 3 lamps at random can be calculated as follows:
Expected number of defective lamps = (0 x 35/264) + (1 x 35/66) + (2 x 25/66) + (3 x 3/88) = 5/4 = 1.25
Therefore, the expected number of defective lamps in the selection of 3 lamps at random is 1.25. Hence, option A is the correct answer.
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A box contains 12 electric lamps of which 5 are defectives. A man selects three lamps at random. What is the expected number of defective lamps in his selection?a)1.25b)2.5c)1.05d)2.03Correct answer is option 'A'. Can you explain this answer?
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