Question

A room has 3 sockets for lamps. From a collection of 10 light bulbs 6 are defective. A person selects 3 at random and puts them in every socket. What is the probability that the room will be lit?

Answer 1

Answer:

98/125

Step-by-step explanation:

Hi,

Given that 6 out of 10 bulbs are defective,

hence probability of a randomly selected bulb to be defective is,

P(D) = 6/10 = 3/5

The room is not lit only if all the 3 randomly selected bulbs are

defective.

Probability for all 3 bulbs being defective = 3/5*3/5*3/5 = 27/125

So, Probability for room being lit will be 1 - Probability for all 3

bulbs being defective = 1 - 27/125

= 98/125(which includes all the cases where at least

one of the bulb is non-defective).

Hence the probability that room is lit is 98/125.

Hope, it helps !