STAT 302: Assignment 1
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1 STAT 302: Assignment 1 Due date: Feb 4, Hand to Mailbox besides LSK An imaginary small university has 3 programs A, B and C. If an applicant is a girl, her probability of being admitted to these programs (A, B, and C are 20%, 50% and 60% respectively. If an applicant is a boy, his probability of being admitted to these programs are 15%, 45% and 40%. (a If a girl were equally liely to apply to one of these 3 programs, what would the probability that he will be admitted to this university? If a boy were equally liely to apply to one of these 3 programs, what would the probability that she will be admitted to this university? (b Suppose instead, a girl applies to programs A, B or C with probabilities 50%, 30% and 20%. What is probability that a girl applicant will be admitted? Similarly, a boy applies to programs A, B or C with probabilities 10%, 50% and 40%. What is probability that a boy applicant will be admitted? (c If a girl applicant is nown to be admitted, what is the probability that this applicant applied to program A, assume (b is the real world situation? (a Let A g be the event that the girl is admitted. Let E 1, E 2 and E 3 be the events that she applies to programs A, B and C, respectively. You may employ the formula for total probability: P(A g =P(A E 1 P(E 1 +P(A E 2 P(E 2 +P(A E 3 P(E 3 =( /3=13/30. If it were a boy applicant, the computation is the same but the conditional probabilities are different. In obvious notation, we have P(A b =P(A E 1 P(E 1 +P(A E 2 P(E 2 +P(A E 3 P(E 3 =( /3= 10/30. Note that this average admission rate for boys is lower in this case. (b Employing the same notation. For a girl applicant, we have P(E 1 = 0.5, P(E 2 =0.3 and P(E 3 =0.2. Thus, P(A g = P(A E 1 P(E 1 +P(A E 2 P(E 2 +P(A E 3 P(E 3 = ( =0.37. For a boy applicant, we have P(E 1 =0.1, P(E 2 =0.5 and P(E 3 =0.4. Thus, P(A b = P(A E 1 P(E 1 +P(A E 2 P(E 2 +P(A E 3 P(E 3 = ( =
2 (c In terms of the notation introduced earlier, we are computing compute P(E 1 A g. Straightforwardly, we have P(E 1 A g =P(A g E 1 P(E 1 /P(A g =( /0.37=1/37. Remar: The phenomenon revealed in (a and (b is called Simpson s Paradox: one student can beat another student in every category under consideration yet with lower overall average. 2. The probability for a passenger not showing up is 5%. For this reason, AirCanada sells 205 ticets for a 200-seat airplane at $300cdn each. If the 201th passenger shows up, she will be refunded 300 plus a $500 compensation. The same for the 202nd and so on. AirCanada pocets all money from seat-sale if 200 or fewer passengers show up. Assume individual passengers will act independently. (a Give a reason why the independence assumption might be wrong? Nevertheless, please compute in (b and (c under this assumption. (b Let X be net income of the AirCanada from the above described random experiment. Compute the probability function of X. (c Compute the mean and variance of X. : (a If a couple are traveling, they will act concordantly in any realistic world. (b If Y is the number of passengers exceeding 200. Then Y can tae values of 0, 1, 2,..., 5. The incomes X= Y= Y. The probability for Y=1, 2,..., 5 can be computed based on binomial probability mass function: ( 205 P(Y== ( ( for =1, 2,..., 10 and P(Y=0=1 5 P(Y== =1 The other probabilities and some useful numbers are P(Y = This table translates into the probability function of X as follows: 2
3 x P(X = x xp(x = x x 2 P(X=x (c Based on these numbers, we get E(X= and E(X 2 = This gives Var(X= Remar: On average, AirCanada maes extra $1474cdn per flight if this is the case. 3. There are 10 blac marbles and 20 white marbles in an urn. (a Suppose n= marbles are randomly chosen without replacement, and X of them are blac. Find the probability mass function of X. (b Suppose n= marbles are randomly chosen with replacement, and Y of them are blac. Find the probability mass function of Y. Compute E{exp(Y}. (c Suppose the marbles are taen out one by one. Let Z be the number of blac balls drawn before the first white ball. Derive the probability mass function of Z. (a In this experiment, there are possible outcomes, when marbles are regarded as non-identifiable. This is, the sample space has this size. The random variable X can at most tae values 0, 1, 2,...,. The size of event X=xhas size ( ( x x for x=0, 1,...,. Thus, the probability mass function of X is given by with x=0, 1,...,. P(X=x= 20 x( x (b When the marbles are sampled with replacement, X has binomial distribution with p=1/3. Its pmf is ( P(X=x= (1/3 x (2/3 x x for x=0, 1,...,. 3
4 We also have E{exp(X}= x=0 ( (1/3 x (2/3 x exp(x=(1/3 {2+e}. x (c It is simpler to wor on P(Z. When the first balls drawn are all blac, then Z. Thus, and P(Z 0=1. P(Z Hence, the pmf of Z is given by with =0, 1,..., 10. = 10 9 ( (31 = 10!(30! (10!(30! = P(Z==P(Z P(Z +1=, Let N be a random variable with Poisson distribution and parameterµ=3. (a Compute the probability that N is even. Given that N is even, what is the probability of N= 6? (b Compute E{N(N 1(N 2} and E{exp(tN} for any real number t. (c Compute E(N 3 and Var{exp(N}. (a It is nown that P(N=2= 32 exp( 3. Hence, (2! P(N is even= P(N= 2= 1 2 =0 For the second half, we have P(N=6 N is even= 1+exp( 6 exp( 3{exp( 3+exp(3}=. 2 N=6 P(N is even = 1 40(e 3 + e 3. (b E{N(N 1(N 2}= =0 ( 1( 2 3! exp( 3=33 = 27. 4
5 (c or E{exp(tN} = =0 exp(t 3! exp( 3=exp(3et 3. E(N 3 = d{exp(3et 3} dt = 57. t=0 E(N 3 =E{N+3N(N 1+N(N 1(N 2}=57. var(exp(n=e{exp(2n} {E exp(n} 2 = exp(3e 2 3 exp(6e (a Let A, B be two independent events. Show that A c and B c are also independent. (b Let A, B, and C be three events such that P(C>0. Show that P(A B C= P(A C + P(B C P(AB C. (c Let A, B, and C be three events such that P(B > 0, P(C > 0, and that B and C are mutually exclusion. Prove to disprove that Proof: (a It is seen that P(A c B c P(A B+P(A C=P(A B C. = 1 P(A B=1 {P(A+P(B P(AB} = 1 P(A P(B+P(AP(B A and B are independent = {1 P(A}{1 P(B}=P(A c P(B c. By definition of independence, we have shown that A c and B c are independent. (b We find as required. P(A B C = P(AC BC P(C = P(AC+P(BC P(ABC P(C = P(AC + P(BC P(C P(C P(ABC P(C = P(A C+P(B C P(AB C (c This is false. If A=S, the sample space, we get but They are apparently unequal. P(A B+P(A C=2 P(A B C=1. 5
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