Question

A bin of 5 transistors is known to contain 2 that are defective. The transistors are to be tested, one at a time, until the defective ones are identified. Denote by N1 the number of tests made until the first defective is identified and by N2 the number of additional tests until the second defective is identified. Find the joint probability mass function of N1 and N2.

Short answer

(a)

(b)$$\begin{gathered} E\left[X\right]=2.5 \\ E\left[Y\right]=2.55 \end{gathered}$$

(c)$$\begin{gathered} Var\left(X\right)=1.05 \\ Var\left(Y\right)=1.0275 \end{gathered}$$

Step-by-step solution

Introduction

N1 is the number of tests performed until the first defect is discovered, and N2 represents the number of additional tests performed until the second problem is discovered.

Step 2:Given Information

X and Y's combined distribution

$$\begin{gathered} Formulaused: \\ P(X=x)=\sum _{y}P(X=x,Y=y) \\ Calculation:ProbabilitymassfunctionofX \\ P(X=1)=P(X=1,Y\&=1)+P(X=1,Y=2)+P(X=1,Y=3) \\ \&+P(X=1,Y=4) \\ P(X=1)=0.08+0.06\&+0.04+0.02=0.20 \\ P(X=2)=P(X=2,Y\&=1)+P(X=2,Y=2)+P(X=2,Y=3) \\ \&+P(X=2,Y=4) \\ P(X=2)=0.06+0.12\&+0.08+0.04=0.30 \end{gathered}$$

$$\begin{gathered} Findingforothertwovalues \\ P(X=3)=P(X=3,Y=1)+P(X=3,Y=2)+P(X=3,Y=3) \\ +P(X=3,Y=4) \\ P(X=3)=0.03+0.09\&+0.12+0.06=0.3 \\ P(X=4)=P(X=4,Y=1)+P(X=4,Y=2)+P(X=4,Y=3) \\ +P(X=4,Y=4) \\ P(X=4)=0.01+0.03+0.07+0.09=0.20 \\ ProbabilitymassfunctionofY \\ P(Y=1)=P(X=1,Y=1)\&+P(X=2,Y=1)+P(X=3,Y=1) \\ +P(X=4,Y=1) \\ P(Y=1)=0.08+0.06\&+0.03+0.01=0.18 \\ P(Y=2)=P(X=1,Y=2)+P(X=2,Y=2)+P(X=3,Y=2) \\ +P(X=4,Y=2) \\ P(Y=2)=0.06+0.12\&+0.09+0.03=0.30 \\ Findingforothertwovalues \\ P(Y=3)=P(X=1,Y=3)+P(X=2,Y=3)+P(X=3,Y=3) \\ +P(X=4,Y=3) \\ P(Y=3)=0.04+0.08\&+0.12+0.07=0.31 \\ P(Y=4)=P(X=1,Y=4)+P(X=2,Y=4)+P(X=3,Y=4) \\ +P(X=4,Y=4) \\ P(Y=4)=0.02+0.04\&+0.06+0.09=0.21 \end{gathered}$$

Step 3:  Finding E[X] and E[y]

$$\begin{gathered} Formulaused: \\ E\left[X\right]=\sum _{x}x\times P(X=x) \\ E\left[Y\right]=\sum _{y}y\times P(Y=y) \\ Calculation:U\sin gthedatafromsubparta \\ E\left[X\right]=1\times 0.20+2\times 0.30+3\times 0.30+4\times 0.20=2.5 \\ E\left[Y\right]=1\times 0.18+2\times 0.30+3\times 0.31+4\times 0.21=2.55 \end{gathered}$$

 var(X) and var(Y)

Formula used:

$$\begin{gathered} Var\left(X\right)=E\left(X^2\right)-\left(E\right[X]{)}^2 \\ E\left(X^2\right)=\sum _x{x}^2\times P(X=x) \\ Calculation: \\ VarianceofX \\ E\left(X^2\right)=1^2\times 0.20+2^2\times 0.30+3^2\times 0.30+4^2\times 0.20 \\ E\left(X^2\right)=7.3 \\ Var(X)=7.3-2.5^2=1.05 \\ VarianceofY \\ E\left(Y^2\right)=1^2\times 0.18+2^2\times 0.30+3^2\times 0.31+4^2\times 0.21 \\ E\left(Y^2\right)=7.53 \\ Var(Y)=7.53-2.{55}^2=1.0275 \end{gathered}$$