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As a generalization of Example 5.3(figure), consider a test of n circuits such that each circuit is acceptable with probability p, independent of the outcome of any other test. Show that the joint PMF of X, the number of acceptable circuits, and Y, the number of acceptable circuits found before observing the first reject, is PX,Y(x,y) =  ((n-y-1)C(x-y))*p^(x)*(1-p)^(n-x) For 0 ≤ y ≤ x < n p^(n) For x=y=n 0 otherwise Hint: For 0 ≤ y ≤ x < n, show that {X = x, Y = y} = A ∩ B ∩ C, where A:   The first y tests are acceptable. B:   Test y + 1 is a rejection. C:   The remaining n − y − 1 tests yield x − y acceptable circuits

As a generalization of Example 5.3(figure), consider a test of n circuits such that each circuit is acceptable with probability p, independent of the outcome of any other test. Show that the joint PMF of X, the number of acceptable circuits, and Y, the number of acceptable circuits found before observing the first reject, is PX,Y(x,y) =  ((n-y-1)C(x-y))*p^(x)*(1-p)^(n-x) For 0 ≤ y ≤ x < n p^(n) For x=y=n 0 otherwise Hint: For 0 ≤ y ≤ x < n, show that {X = x, Y = y} = A ∩ B ∩ C, where A:   The first y tests are acceptable. B:   Test y + 1 is a rejection. C:   The remaining n − y − 1 tests yield x − y acceptable circuits

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
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As a generalization of Example 5.3(figure), consider a test of n circuits such that each circuit is acceptable with probability p, independent of the outcome of any other test. Show that the joint PMF of X, the number of acceptable circuits, and Y, the number of acceptable circuits found before observing the first reject, is

PX,Y(x,y) = 

  • ((n-y-1)C(x-y))*p^(x)*(1-p)^(n-x) For 0 ≤ y ≤ x < n
  • p^(n) For x=y=n
  • 0 otherwise

Hint: For 0 ≤ y ≤ x < n, show that {X = x, Y = y} = A ∩ B ∩ C,

where

A:   The first y tests are acceptable.

B:   Test y + 1 is a rejection.

C:   The remaining n − y − 1 tests yield x − y acceptable circuits

Example 5.3 Test two integrated circuits one after the other. On each test, the possible outcomes are a (accept) and r (reject). Assume that all circuits are acceptable with probability o.9 and that the outcomes of successive tests are independent. Count the number of acceptable circuits X and count the number of successful tests Y before you observe the first reject. (If both tests are successful, let Y = 2.) Draw a tree diagram for the experiment and find the joint PMF Px,Y(x, u). 0.9 aaa X=2,Y=2 0.9 O rear X=1,Y=1 0.9 G era X-1,Y-0 *err X=0,Y -0 P ļaa] = 0.81, P[ra] = 0.09, The experiment has the tree diagram shown to the left. The sample space of the experiment is S = {aa, ar, ra, rr} . (5.7) Observing the tree diagram, we compute P[ar] = 0.09, (5.8) P[rr] = 0.01. (5.9) Each outcome specifies a pair of values X and Y. Let g(s) be the function that transforms each out- come s in the sample space S into the pair of random variables (X, Y). Then g(aa) = (2, 2), g(ar) = (1, 1), 9(ra) = (1,0), g(rr) = (0,0). (5.10) For each pair of values x, y, Px.y(x, y) is the sum of the probabilities of the outcomes for which X = x and Y = y. For example, Px,y(1, 1) = P[ar].
Transcribed Image Text:Example 5.3 Test two integrated circuits one after the other. On each test, the possible outcomes are a (accept) and r (reject). Assume that all circuits are acceptable with probability o.9 and that the outcomes of successive tests are independent. Count the number of acceptable circuits X and count the number of successful tests Y before you observe the first reject. (If both tests are successful, let Y = 2.) Draw a tree diagram for the experiment and find the joint PMF Px,Y(x, u). 0.9 aaa X=2,Y=2 0.9 O rear X=1,Y=1 0.9 G era X-1,Y-0 *err X=0,Y -0 P ļaa] = 0.81, P[ra] = 0.09, The experiment has the tree diagram shown to the left. The sample space of the experiment is S = {aa, ar, ra, rr} . (5.7) Observing the tree diagram, we compute P[ar] = 0.09, (5.8) P[rr] = 0.01. (5.9) Each outcome specifies a pair of values X and Y. Let g(s) be the function that transforms each out- come s in the sample space S into the pair of random variables (X, Y). Then g(aa) = (2, 2), g(ar) = (1, 1), 9(ra) = (1,0), g(rr) = (0,0). (5.10) For each pair of values x, y, Px.y(x, y) is the sum of the probabilities of the outcomes for which X = x and Y = y. For example, Px,y(1, 1) = P[ar].
Px y(x, y)| y = 0 y =1 y=2 I = 0 I = 1 I = 2 0.01 0.09 0.09 0.81 .09 1 .01 .09 The joint PMF can be represented by the table on left, or, as shown below, as a set of labeled points in the x, y plane where each point is a possible value (probability > 0) of the pair (x, y), or as a simple list: (0.81 z= 2, y = 2, 0.09 z= 1, y = 1, Pxy (1, y) = {0.09 r= 1, y = 0, 0.01 z= 0, y = 0. 0. otherwise Note that all of the probabilities add up to 1. This reflects the second axiom of probability (Section 1.3) that states P[S] = 1. Using the notation of random variables, we write this as ΣΣ .y (π, 9) = 1 PESX VESY (5.11) As defined in Chapter 3, the range Sxis the set of all values of X with nonzero probability and simil- arly for Sy. It is easy to see the role of the first axiom of probability in the PMF: Px.y(x, u) z o for all pairs x, y. The third axiom, which has to do with the union of mutually exclusive events, takes us to another important property of the joint PMF. We represent an event B as a region in the X, Y plane. Figure 5.2 shows two examples of events. We would like to find the probability that the pair of random variables (X, Y) is in the set B. When (X, Y) E B, we say the event B occurs. Moreover, we write P[B]as a shorthand for P[(X, Y) E B]. The next theorem says that we can find P[B] by adding the probabilities of all points (x, y) that are in B.
Transcribed Image Text:Px y(x, y)| y = 0 y =1 y=2 I = 0 I = 1 I = 2 0.01 0.09 0.09 0.81 .09 1 .01 .09 The joint PMF can be represented by the table on left, or, as shown below, as a set of labeled points in the x, y plane where each point is a possible value (probability > 0) of the pair (x, y), or as a simple list: (0.81 z= 2, y = 2, 0.09 z= 1, y = 1, Pxy (1, y) = {0.09 r= 1, y = 0, 0.01 z= 0, y = 0. 0. otherwise Note that all of the probabilities add up to 1. This reflects the second axiom of probability (Section 1.3) that states P[S] = 1. Using the notation of random variables, we write this as ΣΣ .y (π, 9) = 1 PESX VESY (5.11) As defined in Chapter 3, the range Sxis the set of all values of X with nonzero probability and simil- arly for Sy. It is easy to see the role of the first axiom of probability in the PMF: Px.y(x, u) z o for all pairs x, y. The third axiom, which has to do with the union of mutually exclusive events, takes us to another important property of the joint PMF. We represent an event B as a region in the X, Y plane. Figure 5.2 shows two examples of events. We would like to find the probability that the pair of random variables (X, Y) is in the set B. When (X, Y) E B, we say the event B occurs. Moreover, we write P[B]as a shorthand for P[(X, Y) E B]. The next theorem says that we can find P[B] by adding the probabilities of all points (x, y) that are in B.
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