In: Statistics and Probability
The maintenance department of a building must replace 8 defective lighting tubes. This service has 40 tubes in stock with 5 broken tubes. The person who has to replace the tubes decides to choose 10 at random from the stock. What is the probability that the ten tubes are enough to replace the 8 that are defective?
Answer: 0.910712332
Non-defective tubes = 35
Defective tubes= 5
Total tubes = 40
Number of tubes selected = 10
Let X = Number of Non-defectives:
To find P(X8):
Case1: X = 8
Number of ways of selecting 10 tubes from 40 tubes = 40C10 = 8.47660528 X 108
Number of ways of selecting 8 non-defectives from 35 non-defectives = 35C8 = 2.353582 X 107
Number of ways of selecting 2 defectives from 5 defectives = 5C2 = 10
So,
P(X = 8) = (2.353582 X 107 X 10/ 8.47660528 X 108 )
= 0.2776561987 (1)
Case 2: X = 9
Number of ways of selecting 10 tubes from 40 tubes = 40C10 = 8.47660528 X 108
Number of ways of selecting 9 non-defectives from 35 non-defectives = 35C9 = 7.060746 X 107
Number of ways of selecting 1 defectives from 5 defectives = 5C21 = 5
So,
P(X = 9) = ( 7.060746 X 107 X 5/ 8.47660528 X 108 )
= 0.416484298 (2)
Case 3: X = 10
Number of ways of selecting 10 tubes from 40 tubes = 40C10 = 8.47660528 X 108
Number of ways of selecting 10 non-defectives from 35 non-defectives = 35C10 = 1.83579396 X 108
So,
P(X = 10) = ( 1.83579396X 108 / 8.47660528 X 108 )
= 0.21657183 (3)
Adding the probability values of P(X = 8) ,P(X = 9) and P(X =
10), given by (1), , (2) & (3), we get the required reslt as
follows:
P(X8)=0.910712332
So,
Answer is:
0.21657183