Question

Please explain

A box contains 42 light bulbs of which eight are defective. If one person selects 20 bulbs from the box in a random manner, and a second person then takes the remaining 22 bulbs, what is the probability that all eight defective bulbs will be obtained by the same person?

Answer 1

Solution :

Total lightbulbs = N = 42

defective lightbulbs = 8

First person has selected 20 bulbs and second person has remaining 22 lightbulbs.

P(First person with all eight defective bulbs) = (COMBIN(8,8)*COMBIN(34,12))/COMBIN(42,20) = 0.0011

P(Second person with all eight defective bulbs) = (COMBIN(8,8)*COMBIN(34,14))/COMBIN(42,22) = 0.0027

P(all eight defective bulbs will be obtained by the same person) = P(First person with all eight defective bulbs) + P(Second person with all eight defective bulbs)

= 0.0011+0.0027 = 0.0038

Answer : 0.0038