A box contains 24 transistors, 4 of which are defective. If 4 are sold at random, find the following probabilities. [4*5 marks] A box contains 24 transistors, 4 of which are defective. If 4 are sold at random, find the following probabilities.

a) Exactly 2 are defectives.

b) None is defective.

c) All are defective.

d) At least 1 is defective.

a) 10.73%

b) 45.60%

c) 0.0094%

d) 54.4%

Step-by-step explanation:

The probability of one transistor being defective is equal the number of defective transistors over the total number of transistors

a)

The probability of the first one being defective is 4/24, and then the probability of the second one being defective is 3/23, because we have one less defective transistor. The probability of the third one being not defective is 20/22, and the fourth one not being defective is 19/21. We also multiply the final probability by a combination of 4 choose 2, because the defective transistors can be any 2 of the group of 4:

C (4,2) = 4! / (2! * 2!) = 6

P = 6 * (4/24) * (3/23) * (20/22) * (19/21) = 0.1073 = 10.73%

b)

Similar to the letter a), we have:

First one not defective: P = 20/24

Second one not defective: P = 19/23

Third one not defective: P = 18/22

Fourth one not defective: P = 17/21

P = (20/24) * (19/23) * (18/22) * (17/21) = 0.4560 = 45.60%

c)

First one defective: P = 4/24

Second one defective: P = 3/23

Third one defective: P = 2/22

Fourth one defective: P = 1/21

P = (4/24) * (3/23) * (2/22) * (1/21) = 0.000094 = 0.0094%

d)

If we want at least 1 defective, we just need to subtract 100% from the case where all 4 are not defective:

P = 1 - 0.4560 = 0.5440 = 54.4%