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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 is
- 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...
5 bulbs
3 defective
2 good
2 bulbs placed in a room.
how many trials required to get the room to light.
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you only need 1 good bulb to light the room.
the number of possible combinations would be C(5,2) which is equal to 10 possible combinations.
the number of possible combinations of 2 that would involve only defective bulbs would be C(3,2) which is equal to 3.
10 - 3 leaves 7 combinations where the room will be lighted.
let's see if this works out.
let d1, d2, d3 represent the defective bulbs.
let g1, g2 represent the good bulbs.
the total possible combinations are:
d1, d2 = room is dark *****
d1, d3 = room is dark *****
d1, g1 = room is light
d1, g2 = room is light
d2, d3 = room is dark *****
d2, g1 = room is light
d2, g2 = room is light
d3, g1 = room is light
d3, g2 = room is light
g1, g2 = room is light
there are 3 possible situations where the room will still be dark.
there are 7 possible situations where the room will be light.
Most Upvoted Answer
Five bulbs of which three are defective are to be tried in two bulb po...
Solution:
To find the number of trials the room shall be lighted, we need to consider the following cases:
Case 1: Both bulbs are good
In this case, we can choose any two bulbs out of the five in 5C2 ways. The probability of selecting two good bulbs is (2/5) * (1/4) = 1/10. Therefore, the number of trials the room shall be lighted in this case is 5C2 * (1/10) = 10 * (1/10) = 1.
Case 2: One bulb is good and one bulb is defective
In this case, we can choose one good bulb out of the two good bulbs in 2C1 ways and one defective bulb out of the three defective bulbs in 3C1 ways. The probability of selecting one good and one defective bulb is (2/5) * (3/4) = 3/10. Therefore, the number of trials the room shall be lighted in this case is 2C1 * 3C1 * (3/10) = 2 * 3 * (3/10) = 18/10 = 9/5.
Case 3: Both bulbs are defective
In this case, we can choose any two defective bulbs out of the three in 3C2 ways. The probability of selecting two defective bulbs is (3/5) * (2/4) = 3/10. Therefore, the number of trials the room shall be lighted in this case is 3C2 * (3/10) = 3 * (3/10) = 9/10.
Total number of trials the room shall be lighted is the sum of the number of trials in all the cases. Therefore, the total number of trials is 1 + 9/5 + 9/10 = 7.
Hence, the correct option is (D) 7.
To find the number of trials the room shall be lighted, we need to consider the following cases:
Case 1: Both bulbs are good
In this case, we can choose any two bulbs out of the five in 5C2 ways. The probability of selecting two good bulbs is (2/5) * (1/4) = 1/10. Therefore, the number of trials the room shall be lighted in this case is 5C2 * (1/10) = 10 * (1/10) = 1.
Case 2: One bulb is good and one bulb is defective
In this case, we can choose one good bulb out of the two good bulbs in 2C1 ways and one defective bulb out of the three defective bulbs in 3C1 ways. The probability of selecting one good and one defective bulb is (2/5) * (3/4) = 3/10. Therefore, the number of trials the room shall be lighted in this case is 2C1 * 3C1 * (3/10) = 2 * 3 * (3/10) = 18/10 = 9/5.
Case 3: Both bulbs are defective
In this case, we can choose any two defective bulbs out of the three in 3C2 ways. The probability of selecting two defective bulbs is (3/5) * (2/4) = 3/10. Therefore, the number of trials the room shall be lighted in this case is 3C2 * (3/10) = 3 * (3/10) = 9/10.
Total number of trials the room shall be lighted is the sum of the number of trials in all the cases. Therefore, the total number of trials is 1 + 9/5 + 9/10 = 7.
Hence, the correct option is (D) 7.
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Community Answer
Five bulbs of which three are defective are to be tried in two bulb po...
Possibilities of room shall be lighted is....
3c3
3c2
3c1
then
3c3+3c2+3c1
1+3+3..
=7
3c3
3c2
3c1
then
3c3+3c2+3c1
1+3+3..
=7
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