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University Level Answer.

Question 6: A shipment of 10 cameras includes 3 that are defectives. a. 4 cameras from the shipment were randomly selected without replacement and tested. Let X be the number of defectives selected. Find P(X-2). b. Let Y be the number of cameras from the shipment were randomly selected without replacement and tested until all the detectives were found. Find P(Y = y).

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Answer #1

(a)

Here x can take values 0, 1, 2, 3. Number of ways of selecting 4 cameras out of 10 is

101 210 (10.4) = 4!(10-4)!

Number of ways of selecting x defective out of 3 and 4-x non defective out of 10-3=7 non defectives is

331 7! 3! 7! ca, r)C(7,4-x) = 30240

So

144 C(10, 4)

(b)

Here y can take values 3,4,...10. When Y=3 then all defective are tested at first three selection so

3 2 13-2.1 10 9 8 10.9.8

When Y=4 then two defectives are in first three and one at 4th selections so

3.2.1 C(3,2) 10.9.8 10 9 8 7

Likewise

3.2.1 10.9.8 10 9 8 7 6

3 2 1 7 6 5 10 9 8 7 6 5 3-2.1 C(5,2) 10-9.8

Continuing in this manner

3.21 CỦy_ 1.2) . . У- 3, 4. 5, ...10 10.9.8 120

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