Question

A room has 3 lamps. From a collection of 10 light bulbs of which 6 are not good, a person selects 3 at random and pits them in a socket. The probability that he will have light, is

• A. 5/6
• B. 1/2
• C. 1/6
• D. None of these

Answer 1

Answer: (A)

5/6

There are 3 possibilities :
Either we get 1 correct bulb, or 2 or 3.
Hence total chances of having a light $=\left[3{(}^6{C}_2{\times }^4{C}_1\right)+3{(}^6{C}_1{\times }^4{C}_2\left){+}^4{C}_3\right]{/}^{10}{C}_3=\frac{5}{6}$