SampleTest2SolW14
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University of Winnipeg**We aren't endorsed by this school
Course
MATH 1202
Subject
Mathematics
Date
Jan 12, 2025
Pages
5
Uploaded by renesansi
SAMPLE TEST #2: SOLUTIONSTotal Marks: 28Instructions:- Nonprogrammable and nongraphical calculators are allowed.- This is a closed book test: no books, notes or formula sheets are allowed.- Start each question on a new page.- Show as much detail as you can, i.e. show ALL your work.- Venn diagrams may be used to obtain numerical answers but may only act as aids in other problems.- Should you give more than one solution to a single question, none will be graded.Question 1: Consider the random experiment of tossing a coin three times and observing the result (Heador Tail) for each toss. LetYdenote the difference between the number of Heads obtained and the numberof Tails obtained (i.e., number of Heads minus number of Tails).(a) By constructing a table showing the value ofYfor each outcome in the sample spaceS, find the rangeofY,RY.(b) Suppose that the probability of Head on any given toss is 0.2. Find the PMF ofY.(4 marks: 2-2)Solution:(a) Here the sample space isS={HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}. We can identifythe value ofYfor each outcomeωinS:ωYHHH3HHT1HTH1THH1HTT-1THT-1TTH-1TTT-3Thus, it is clear thatRY={-3,-1,1,3}.(b) We find the PMF ofYusing the above table:ωYProbabilityHHH30.008HHT10.032HTH10.032THH10.032HTT-10.128THT-10.128TTH-10.128TTT-30.5121
So,Pr(Y=-3)=0.512,Pr(Y=-1)=3(0.128)=0.384,Pr(Y=1)=3(0.032)=0.096 andPr(Y=3)=0.008,i.e.,pY(y) =0.512ify=-3,0.384ify=-1,0.096ify= 1,0.008ify= 3,0otherwise.Question 2: Prove theLack-of-Memoryproperty of the geometric distribution, i.e., provePr(X=n+k|X > n) =Pr(X=k)fork, n,∈ N.(4 marks)Solution:LHS=Pr(X=n+k|X > n)RHS=Pr(X=k)=Pr(X=n+k, X > n)Pr(X > n)=p(1-p)k-1=Pr(X=n+k)Pr(X > n)=p(1-p)n+k-1(1-p)n=p(1-p)k-1=RHS2Question 3: From an animal population of 2000 individuals, 200 animals have been caught, tagged, andreleased to mix into the population. After these individuals have had an opportunity to mix with the restof the population, a random sample of 50 animals is taken. LetXdenote the number of tagged animals inthe second sample.(a) Give the exact expression forPr(X= 5).(b) Find a good numerical approximation for the previous probability.(4 marks: 2-2)Solution:(a) HereX∼H(2000,50,0.10), i.e.Xhas a hypergeometric distribution with parametersN=2000,n=50andp=0.10. And so its pmf is:pX(x) =(200x)(180050-x)(200050)forx= 0,1, . . . ,50,0otherwise.ThusPr(X=5)=pX(5) =(2005)(180045)(200050).(b) SinceNp-x=200-50=150≥0, we can approximate the probability using the Binomial distributionwith parametersn=50 andp=0.10 and soPr(X=5).=(505)(0.10)5(1-0.10)50-5.= 0.1849.2
Question 4: A service station has both self-service and full-service gas pumps. There is one gas pump foreach type of service and each gas pump has two hoses. LetXdenote the number of hoses being used at theself-service gas pump at a particular time andYdenote the number of hoses being used at the full-servicegas pump. The joint PMF ofXandYappears below:ypX,Y(x, y)01200.100.040.02x10.080.200.0620.060.140.30Then:(a) What isPr(X= 1, Y= 1)?(b) ComputePr(X≤1, Y≤1).(c) Obtain the marginal PMF ofXand using it, computePr(X≤1).(6 marks: 1-2-3)Solution:(a)Pr(X=1 andY=2)=pX,Y(1,2)=0.20(b) By FPF,Pr(X≤1 andY≤1)=∑(x,y):x≤1,y≤1pX,Y(x, y)=0.10+0.04+0.08+0.20=0.42.(c) We find the marginal PMF ofXby summing across the columns:pX(x)=∑2y=0pX,Y(x, y). Therefore:pX(0)=∑2y=0pX,Y(0, y)=0.10+0.04+0.02=0.16pX(1)=∑2y=0pX,Y(1, y)=0.08+0.20+0.06=0.34pX(2)=∑2y=0pX,Y(2, y)=0.06+0.14+0.30=0.50That is:pX(x) =0.16ifx= 0,0.34ifx= 1,0.50ifx= 2,0otherwise.We find the marginal pmf ofYby summing across the rows:pY(y)=∑2x=0pX,Y(x, y). Therefore:pY(0)=∑2x=0pX,Y(x,0)=0.10+0.08+0.06=0.20pY(1)=∑2x=0pX,Y(x,1)=0.04+0.20+0.14=0.38pY(2)=∑2x=0pX,Y(x,2)=0.02+0.06+0.30=0.38That is:pY(y) =0.24ify= 0,0.38ify= 1,0.38ify= 2,0otherwise.Now,Pr(X≤1)=Pr(X=0 or 1)=Pr(X=0)+Pr(X=1)=pX(0)+pX(1)=0.16+0.34=0.50.3
Question 5: A bin of five transistors is known to contain two that are defective. The transistors are to betested one at a time until the defective ones are found. LetXdenote the number of tests made until thefirst defective andYthe number of additional tests until the second defective is found.(a) Find the joint PMF ofXandY.(b) Obtain the conditional PMF ofYgiven thatX=3.(6 marks: 3-3)Solution:(a) Let us consider this a permutation problem, i.e. the transistors are in the bin in a certain order andwe have five positions for the five transistors. First we note thatPr(X=x, Y=y)=Pr(a defective inthexth position andY=y) means the other defective is in the (x+y)th position. Then, since thereare(52)= 10 possible ways to pick which position the defectives are in, and the eventX=x, Y=yinvokes one of these positions (orderings), it has probability 1/10. That is:pX,Y(x, y) =(110x, y= 1,2,3,4, x+y≤5,0otherwise.(b) First we notePr(X=3)=∑y∈RYpX,Y(3, y) =pX,Y(3,1) +pX,Y(3,2) = 2/10 = 0.20. Now, sincepY|X=3(Y|X= 3) =Pr(Y=y|X= 3) =Pr(Y=y, X= 3)Pr(X= 3),we havePr(Y= 1|X= 3) =Pr(Y= 1, X= 3)Pr(X= 3)=0.100.20= 0.50Pr(Y= 2|X= 3) =Pr(Y= 2, X= 3)Pr(X= 3)=0.100.20= 0.50That is:pY|X=3(Y|X= 3) =0.50ify= 1,2,0otherwise.Question 6: The probability mass function of a random variableXis given by:x01pX(x)2/31/3Another random variableY, independent ofX, has the following probability mass function:y012pY(y)4/72/71/7Find the probability mass function ofZ=X+|Y-1|.(4 marks)4
Solution: First we note that sinceXandYare independent, thenpX,Y(x, y) =pX(x)pY(y) forx∈RXandy∈RY. Now, we see that the possible values ofZto be 0, 1 and 2. So, by summing over the appropriatepairs (x, y), we have:Pr(Z=0)=Pr(X=0,Y=1)=(2/3)(2/7)=4/21.Pr(Z=1)=Pr(X=0,Y=2 orX=1,Y=1 orX=0,Y=0)=(2/3)(1/7)+(1/3)(2/7)+(2/3)(4/7)=12/21.Pr(Z=2)=Pr(X=1,Y=0 orX=1,Y=2)=(1/3)(4/7)+(1/3)(1/7)=5/21.Now the PMF ofZcan be written as:pZ(z) =4/21ifz= 0,12/21ifz= 1,5/21ifz= 2,0otherwise.5