Question

5 bulbs of which 3 are defective are to be tried in two bulb points in a dark room number of trials the room shall be lighted is

Answer 1

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.

Answer 2

Answer:

7

Explanation:

Total no. of bulb= 5

no. of defected bulb= 3

no. of not defected bulb=2

Total no. of bulb combination = 5C2

=10

( since a room can be lighted with one bulb also)

total no. of bulb combination when room shall not light = 3C2

=3

Now,

Total no. of trial when room shall light

=10-3

=7

Hence, number of trial when the room shall be lighted is 7