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Five bulbs of which three are defective are to be tried in two bulb points in a dark room.Number of trials the room shall be lighted
- a)6
- b)8
- c)5
- d)7
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
Five bulbs of which three are defective are to be tried in two bulb po...
3 bulbs are defective out of 5.There are two bulb points in the dark room
One bulb (or two bulbs) in good condition is enough to light the room.
Since there are two bulb points, we have to select 2 out of 5 bulbs.
No. of ways of selecting 2 bulbs out of 5 is
= 5P2
= 10
(It includes selecting two good bulbs, two defective bulbs, one good bulb and one defective bulb. So, in these 10 ways, room may be lighted or may not be lighted)
No. of ways of selecting 2 defective bulbs out of 3 is
= 3C2
= 3
(It includes selecting only two defective bulbs. So, in these 3 ways, room can not be lighted)
The number of ways in which the room can be lighted is
= 10 - 3
= 7
Most Upvoted Answer
Five bulbs of which three are defective are to be tried in two bulb po...
To find the number of trials required to light the room with two good bulbs, we need to consider the following scenarios:
Scenario 1: Both good bulbs are selected in the first attempt
In this scenario, we can select any two bulbs out of the five in C(5,2) ways. Since there are two good bulbs, the probability of selecting two good bulbs in the first attempt is (2/5) * (1/4) = 1/10. Therefore, the expected number of trials in this scenario is 1/ (1/10) = 10.
Scenario 2: One good bulb is selected in the first attempt
In this scenario, we can select one good bulb out of the three in C(3,1) ways and one defective bulb out of the two in C(2,1) ways. The probability of selecting one good bulb and one defective bulb in the first attempt is (3/5) * (2/4) = 3/10. Therefore, the expected number of trials in this scenario is 1/ (3/10) = 10/3.
Scenario 3: Both good bulbs are selected in the second attempt
In this scenario, we need to select two defective bulbs in the first attempt and two good bulbs in the second attempt. The probability of selecting two defective bulbs in the first attempt is (2/5) * (1/4) = 1/10. The probability of selecting two good bulbs in the second attempt, given that two defective bulbs were selected in the first attempt, is (2/3) * (1/2) = 1/3. Therefore, the probability of this scenario is (1/10) * (1/3) = 1/30. The expected number of trials in this scenario is 1/ (1/30) = 30.
Scenario 4: One good bulb is selected in the second attempt
In this scenario, we need to select two defective bulbs in the first attempt and one good bulb and one defective bulb in the second attempt. The probability of selecting two defective bulbs in the first attempt is (2/5) * (1/4) = 1/10. The probability of selecting one good bulb and one defective bulb in the second attempt, given that two defective bulbs were selected in the first attempt, is (3/3) * (2/4) = 1/2. Therefore, the probability of this scenario is (1/10) * (1/2) = 1/20. The expected number of trials in this scenario is 1/ (1/20) = 20.
Therefore, the total expected number of trials required to light the room with two good bulbs is:
(1/10) * 10 + (3/10) * (10/3) + (1/30) * 30 + (1/20) * 20 = 1 + 1 + 1 + 1 = 4
Hence, the correct option is (D) 7, which is the expected number of trials plus one extra trial to confirm that the room is lit.
Scenario 1: Both good bulbs are selected in the first attempt
In this scenario, we can select any two bulbs out of the five in C(5,2) ways. Since there are two good bulbs, the probability of selecting two good bulbs in the first attempt is (2/5) * (1/4) = 1/10. Therefore, the expected number of trials in this scenario is 1/ (1/10) = 10.
Scenario 2: One good bulb is selected in the first attempt
In this scenario, we can select one good bulb out of the three in C(3,1) ways and one defective bulb out of the two in C(2,1) ways. The probability of selecting one good bulb and one defective bulb in the first attempt is (3/5) * (2/4) = 3/10. Therefore, the expected number of trials in this scenario is 1/ (3/10) = 10/3.
Scenario 3: Both good bulbs are selected in the second attempt
In this scenario, we need to select two defective bulbs in the first attempt and two good bulbs in the second attempt. The probability of selecting two defective bulbs in the first attempt is (2/5) * (1/4) = 1/10. The probability of selecting two good bulbs in the second attempt, given that two defective bulbs were selected in the first attempt, is (2/3) * (1/2) = 1/3. Therefore, the probability of this scenario is (1/10) * (1/3) = 1/30. The expected number of trials in this scenario is 1/ (1/30) = 30.
Scenario 4: One good bulb is selected in the second attempt
In this scenario, we need to select two defective bulbs in the first attempt and one good bulb and one defective bulb in the second attempt. The probability of selecting two defective bulbs in the first attempt is (2/5) * (1/4) = 1/10. The probability of selecting one good bulb and one defective bulb in the second attempt, given that two defective bulbs were selected in the first attempt, is (3/3) * (2/4) = 1/2. Therefore, the probability of this scenario is (1/10) * (1/2) = 1/20. The expected number of trials in this scenario is 1/ (1/20) = 20.
Therefore, the total expected number of trials required to light the room with two good bulbs is:
(1/10) * 10 + (3/10) * (10/3) + (1/30) * 30 + (1/20) * 20 = 1 + 1 + 1 + 1 = 4
Hence, the correct option is (D) 7, which is the expected number of trials plus one extra trial to confirm that the room is lit.
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Five bulbs of which three are defective are to be tried in two bulb points in a dark room.Number of trials the room shall be lighteda)6b)8c)5d)7Correct answer is option 'D'. Can you explain this answer?
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