Question

Five bulbs of which three are defective are to be tried in two bulb points in a dark room

Answer 1

Answer:

D 7

Step-by-step explanation:

Total no. of bulb= 5

no. of defected bulb= 3

no. of not defected bulb=2

Total no. of bulb combination = 5C2

=5!/2!(5-2)!

= 5!/2!3!

= 5×4×3×2×1/2×1×3×2×1

=120/12

=10

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

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

3!/2!(3-2)!

= 3!/2! 1!

= 3×2×1/2×1×1

= 6/2

=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 which is option d