Question

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 1

Answer:

here the correct is 2 trials

Answer 2

if one blub is correct so room lighted or both blub correct room lighted

only room not lighted when both bulb are defective

so we find room not lighted event A and it substracted from 1 so we get room lighting ans...

³c2 = 3×2/2×1 = 3 way room not light

5c2 = 5×4/2×1 = 10 total number of way

room lighted ans.. = 10-3 = 7

plese like ans..