Question

In: Statistics and Probability

The maintenance department of a building must replace 8 defective lighting tubes. This service has 40...

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

Solutions

Expert Solution

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         

orchestra answered 1 year ago
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