A stockroom currently has 30 components of a certain type, of which 9 were provided by supplier 1, 7 by supplier 2, and 14 by supplier 3. Six of these are to be randomly selected for a particular assembly. Let X = the number of supplier 1's components selected, Y = the number of supplier 2's components selected, and p(x, y) denote the joint pmf of X and Y. (a) What is p(2, 1)? [Hint: Each sample of size 6 is equally likely to be selected. Therefore, p(2, 1) = (number of outcomes with X = 2 and Y = 1)/(total number of outcomes). Now use the product rule for counting to obtain the numerator and denominator.] (Round your answer to five decimal places.) P(2, 1) = | (b) Using the logic of part (a), obtain p(x, y). (This can be thought of as a multivariate hypergeometric distribution-sampling without replacement from a finite population consisting of more than two categories. Round your answers to five decimal places.) P(x, y) 1 2 4 6 4 6 3. 3.
A stockroom currently has 30 components of a certain type, of which 9 were provided by supplier 1, 7 by supplier 2, and 14 by supplier 3. Six of these are to be randomly selected for a particular assembly. Let X = the number of supplier 1's components selected, Y = the number of supplier 2's components selected, and p(x, y) denote the joint pmf of X and Y. (a) What is p(2, 1)? [Hint: Each sample of size 6 is equally likely to be selected. Therefore, p(2, 1) = (number of outcomes with X = 2 and Y = 1)/(total number of outcomes). Now use the product rule for counting to obtain the numerator and denominator.] (Round your answer to five decimal places.) P(2, 1) = | (b) Using the logic of part (a), obtain p(x, y). (This can be thought of as a multivariate hypergeometric distribution-sampling without replacement from a finite population consisting of more than two categories. Round your answers to five decimal places.) P(x, y) 1 2 4 6 4 6 3. 3.
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
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Chapter1: Combinatorial Analysis
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Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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Question
![A stockroom currently has 30 components of a certain type, of which 9 were provided by supplier 1, 7 by supplier 2, and 14 by supplier 3.
Six of these are to be randomly selected for a particular assembly. Let X = the number of supplier 1's components selected, Y = the number
of supplier 2's components selected, and p(x, y) denote the joint pmf of X and Y.
(a) What is p(2, 1)? [Hint: Each sample of size 6 is equally likely to be selected. Therefore,
p(2, 1) = (number of outcomes with X = 2 and Y = 1)/(total number of outcomes). Now use the product rule for counting to obtain the
numerator and denominator.] (Round your answer to five decimal places.)
Р(2, 1) %3D
(b) Using the logic of part (a), obtain p(x, y). (This can be thought of as a multivariate hypergeometric distribution-sampling without
replacement from a finite population consisting of more than two categories. Round your answers to five decimal places.)
p(x, y)
2.
3
4
1
2
Y 3
4](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff8536f70-6afb-49b2-94d8-7d5f9820999c%2F636a9754-6256-49e0-816d-1299cef699fc%2Fs3tb08_processed.png&w=3840&q=75)
Transcribed Image Text:A stockroom currently has 30 components of a certain type, of which 9 were provided by supplier 1, 7 by supplier 2, and 14 by supplier 3.
Six of these are to be randomly selected for a particular assembly. Let X = the number of supplier 1's components selected, Y = the number
of supplier 2's components selected, and p(x, y) denote the joint pmf of X and Y.
(a) What is p(2, 1)? [Hint: Each sample of size 6 is equally likely to be selected. Therefore,
p(2, 1) = (number of outcomes with X = 2 and Y = 1)/(total number of outcomes). Now use the product rule for counting to obtain the
numerator and denominator.] (Round your answer to five decimal places.)
Р(2, 1) %3D
(b) Using the logic of part (a), obtain p(x, y). (This can be thought of as a multivariate hypergeometric distribution-sampling without
replacement from a finite population consisting of more than two categories. Round your answers to five decimal places.)
p(x, y)
2.
3
4
1
2
Y 3
4
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