SOLUTION: 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? answer is 7 pls solve this thank u

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Question 658824: 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? answer is 7 pls solve this thank u
Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
5 bulbs
3 defective
2 good
2 bulbs placed in a room.
how many trials required to get the room to light.
-----
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.


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