Question

A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective? Suppose the bulb drawn in previous case is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?

Answer 1

Given : 

4 out of 20 bulbs are defective (i.e.) 

Number of defective bulbs = 4 

Number of non-defective bulbs = 20 – 4 = 16 

If a bulb is drawn at random, the total outcomes are = 20 

Number of outcomes favourable to ‘defective bulb’ = 4 

∴ Probability of getting a defective bulb

= \(\frac{No.\,of\,favourable\,outcomes}{Total\,no.\,of\,outcomes}\)

\(=\frac{4}{20}=\frac{1}{5}\)

Suppose a non-defective bulb is drawn and not replaced, then the bulbs remaining are = 19 

∴ Total outcomes in drawing a bulb from the rest = 19

Number of favourable outcomes in drawing nondefective bulb from the rest = 16 – 1 = 15 

∴ Probability of getting a non-defective bulb in the second draw

= \(\frac{No.\,of\,favourable\,outcomes}{Total\,no.\,of\,outcomes}\)

= 15/19