Question #221083
A box contains 24 transistors, 4 of which are defective. If 4 are sold at random, find the
following probabilities.
a. Exactly 2 are defective. (5 marks)
c. All are defective. (5 marks)
b. None is defective. (5 marks)
d. At least 1 is defective. (5 marks)
1
Expert's answer
2021-07-29T20:37:03-0400

a. Exactly 2 are defective:

P(2)=C(2;4)×(1/6)2×(5/6)2=6×(1/6)2×(5/6)2=0.1157.P(2) = C(2;4)×(1/6)^2×(5/6)^2 = 6×(1/6)^2×(5/6)^2 = 0.1157.

c. All are defective:

P(4)=(1/6)4=0.00077.P(4) = (1/6)^4 = 0.00077.

b. None is defective:

P(0)=(5/6)4=0.4823.P(0) = (5/6)^4 = 0.4823.

d. At least 1 is defective:

P(>=1) = 1 - P(0) = 1 - 0.4823 = 0.5177.


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