Question

SHORT ANSWER ( You must show all work to receive partial credit. All answers must be rigorously justified using concepts and tools from class. In the event that an answer is logically obvious to you without any mathematical justification, you must urite a justification in words.): 13. Suppose the random variables X and Y have the joint pmf given below, where the probabil- ities pi 20, p2 2 0, p3 2 0, and p 20 are such that Pi + p2 + p3 + p4 = 1. 0 1 0 P2 Рз P4 (a) Write the marginal distributions in the table in terms of pr. p2. P3, and p4. Label which marginal distribution is for X and which marginal distribution is for Y (b) Find E[E[YIX]] in terms of p.P2. P3, and p (c) Find E[X4 Y = 1] in terms of p1; P2. Pa, and РА- 14. All electronic circuit board failures can be classified into one of the following mutually ex- clusive and collectively exhaustive categories: assembly defects (40%), electrical component defects (35%), and mechanical defects (25%). Suppose ten boards fail independently, and let the random variables X, Y, and Z denote the number of each type of defect among the boards, respectively. (That is, X for assembly, Y for electrical, and Z for mechanical.) What is P(X 8,Z=1 | Y = 1}?

Answer 1

8444) Pyly ofer pv PitP2 TIP3 | P4 P3 + P4 P(x) PitP3 P2 + Py Px(x)2 Py ly) - Marginal pdf of x Marginal pdf of 4 6) ELELYIx] Now, E[Yx)= EY Pyxlyke Posle lyle) = PLAY

• Far x=0, Y=0 i 841x Cyla)= P10g) = Ploo) - Pi Px(0) P&P3 Pxco) For x=0,421 (ارہ)P . Pyx (41x)= Pxto) Pitp3 i. Pysy to can be written as follows Y=4 L P1 P4/xly/x-) 14

E[4x] = {y Pylx lyx) = + 1x 0X PL PitP3 P3 PitP3 P3 pidP3 & El € [4/x]] = PitP3 c) E[x+/y=1] = {xt Px/y (x1y=1) P21 Pyly (214-13 = Play) = P1,1) Py(1) Pyll) For Vz x=0 Pxly lxly21) = Plou PYLU P3t Ry

Far xal Pxly Caly=1) - PUI) = Pyll) P₃ + P4 50) Pxly can be written as x = x Pxly Calyas) V4 per canyons P3 3+ Py Pzt Py >> E [ x4/4=1] = { x* Px1y (4191) *x P Potpu + (х ч PztPy PatPy