Suppose that for Intel’s Core 2 E8200 processor, 75% operate properly only at 2
ID: 3171556 • Letter: S
Question
Suppose that for Intel’s Core 2 E8200 processor, 75% operate properly only at 2.8 GHz, while 20% are capable of operating properly at 3.2 GHz. The remaining 5% are defective.
a.What is the probability that if 20 of the processors are randomly sampled, 15 operate only at 2.8 GHz, 4 are capable of operating at 3.2 GHz, and the remaining processor is defective?
b.What is the probability that if 20 of the processors are randomly sampled at least one will be defective?
c.What are the expected value and variance for the number of processors that must be selected until exactly 10 have been found that are capable of operating properly at 3.2GHz?
d.What is the probability that the 20 th processor selected is the 5 th capable of operating properly at 3.2GHz?
e.Suppose that in a lot of 16 processors, 10 operate only at 2.8 GHz, 4 are capable of operating at 3.2 GHz, and two are defective. If six processors are selected from the lot, what is the probability exactly four operate only at 2.8 GHz?
Explanation / Answer
Solution
For convenience in explaining and presenting, let
A represent the event that the processor operates properly at 2.8 GHz,
B represent the event that the processor operates properly at 3.2 GHz, and
D represent the event that the processor is defective.
Then, given statements,
“75% operate properly only at 2.8 GHz, while 20% are capable of operating properly at 3.2 GHz. The remaining 5% are defective.” =>
P(A) = 0.75 ……………………………………………………………………..(1)
P(B) = 0.20 ……………………………………………………………………..(2)
P(D) = 0.05 ……………………………………………………………………..(3)
Part (a)
To find probability that out of the 20 processors randomly sampled, 15 operate only at 2.8 GHz, 4 are capable of operating at 3.2 GHz, and the remaining processor is defective, we need P(A = 15, B = 4 and C = 1).
P(A = 15, B = 4 and C = 1) = P(A = 15) x P(B = 4) x P(C = 1).
Now, any 15 out of 20 can be A in 20C15 (= 15504) ways and probability of all 15 being A = 0.7515. Out of the remaining 5, 4 can be B in 5C4 (= 5) ways and probability of all 4 being B = 0.204 and P(remaining one is D) = 0.05.
Thus, P(A = 15, B = 4 and C = 1) = 15504 x 5 x 0.7515 x 0.204 x 0.05
= 0.0829 ANSWER
Part (b)
P(at least 1 defective) = 1 – P(no defective)
= 1 – 0.9520[because probability of a defective = 0.05 ]
= 1 – 0.3585 = 0.6415 ANSWER
Part (c)
probability that the 20 th processor selected is the 5 thcapable of operating properly at 3.2GHz => out of first 19, 4 are B, remaining 15 are not B and 20th is B.
So, probability = 19C4 x 0.24 x 0.815 x 0.2 = 0.0436 ANSWER
Part (d)
In a lot of 16 processors, 10 operate only at 2.8 GHz, 4 are capable of operating at 3.2 GHz, and two are defective. =>
P(A) = 10/16, P(B) = 4/16 and P(D) = 2/16.
If six processors are selected from the lot, exactly four operate only at 2.8 GHz => (4A & 2B) or (4A & 2D) or (4A, 1B & 1D).
So, P(out of six processors selected from the lot, exactly four operate only at 2.8 GHz)
= P(4A & 2B) + P(4A & 2D) + P(4A, 1B & 1D)
= 6C4(10/16)4(4/16)2 + 6C4(10/16)4(2/16)2 + 6C4(10/16)4 (2C1)(4/16)(2/16)
= 0.3219 ANSWER
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