ECE313 Summer Problem Set 4. Reading: RVs, mean, variance, and coniditional probability

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1 ECE Summer 0 Problem Set Reading: RVs, mean, variance, and coniditional probability Quiz Date: This Friday Note: It is very important that you solve the problems first and check the solutions afterwards. Problem Consider a random variable X with the following pmf: p X i c p i, for i,, Find the value of c such that p X i is a valid pmf. Furthermore, a Find EX and Var [X. b If it is known that E [X + 9, what is p? c If it is known that Var [X +, what is p? For the pmf to be valid we need p X i. So where we have used the fact that p X i c p i c p j i0 c j0 p c p x i x for x <. This implies that we must have c p. Before obtaining the mean and the variance, let us obtain two useful formulas. Take the derivative with respect to x on both sides of i0 xi x to obtain Differentiation for a second time yields ix i x i i x i x i

2 ECE Summer 0 UIUC a For the mean we have E [X ip X i i p i p p To obtain the variance, first we find E [ X : E [ X i p X i i p i p p p i i + i p i p i i p i p + i p i p i p p i i p i + p i p p p + p p p The variance is Var [X p p p p p. b E [X + E [X + 9 E [X p /. c Var [X + Var [X Var [X 6 p p 6. The last equality is satisfied if 6p + p 0. The solutions to this equation are { /, /}. Only / is acceptable why?. Problem Example.. of the text. Problem Suppose that you and nine of your friends are planning on going to a concert, four of your friends are also taking ECE, and five are not. Unfortunately, your car only fits five people so you randomly choose four friends to go with you all friends are equally likely to be chosen, and the rest will miss the concert. When you get there, it turns out that those who are taking ECE get in for free, and everyone else pays $0. Let X be the total amount of money you and your friends end up paying to get into the concert. a What values can X take? b Find the pmf of X. c Find the expected value of X. d If instead of getting in for free, the ECE students have to pay z dollars, what is the expected value of the money that you have to pay?

3 ECE Summer 0 UIUC a Let Y be the number of non-ece students that go with you, then Y {0,,,, } and X 0Y. Therefore, X {0, 0, 0, 0, 0}. 9 b You choose four of your nine friends to ride with you, so that Ω 6. Let us find the size of the event {Y i} for i {0,,,, }: There are i ways to choose i non-ece students from a set of non-ece students, and there are i ways to choose the remaining students from the ECE friends. So {Y i} i i and thus or equivalently, p X k P {Y k/0} k/0 k/0 6 p X 0 P {Y 0} p X 0 P {Y } p X 0 P {Y } 0 p X 0 P {Y }, for k {0, 0, 0, 0, 0} , 0 6, 60 6, 0 6, p X 0 P {Y } 6 6. c E[X u i u i p X u i d Let ˆX be a random variable denoting the new amount to be paid. Then, ˆX 0Y + z Y X z + z. 0 Therefore, E[ ˆX E[X z 0 + z 00 9 Problem z 0 + z. Let C be a real-valued constant, and let n be an integer. The pmf for a discrete-type random variable X is given by p X i Ci for integers i n and zero else. Some helpful summation identities can be found in Appendix 6. of the course notes. a Find the constant C such that p X is a valid pmf. b Find E[X. c Find E [ + X.

4 ECE Summer 0 UIUC a The pmf has to sum up to one, therefore and so C nn+. b By definition, E[X u i u i p X u i u i p X u i nn + Ci C i c We have E [ + X + E [ X. Using LOTUS, Hence E [ + X n+ n+. Problem nn + i C, [ E ui p X u i X ui nn + nn + nn + n + nn + 6 i i nn + n n + n + To solve this problem, use the general form of the rule P AB P A P B A that was discussed in class. The general form is P A A A n P A P A A P A A A P A n A A A n. An ordinary deck of playing cards is randomly divided into piles of cards each. What is the probability that each pile has exactly ace? Define events in E i, i,,, as follows: E {the ace of spades is in any of the piles}, E {the ace of spades and the ace of hearts are in different piles}, E {the ace of spades, the ace of hearts, and the ace of diamonds are in different piles}, E {all four aces are in different piles}.

5 ECE Summer 0 UIUC The desired probability is P E P E E E E since E E E E E. By the given rule P E E E E P E P E E P E E E P E E E E. Now, P E. Since the pile containing the ace of spades receives of the remaining cards, the ace of hearts must be one of the other 9 cards. We have P E E 9 Similarly, the piles containing the aces of spades and hearts will receive of the remaining 0 cards. So the ace of diamonds must be one of the other 6 cards: P E E E 6 0 Finally, P E E E E 9 Hence, the probability that each hands has exactly one ace is Problem We randomly choose cards out of a deck of cards. a What is the probability of full house given that the first card and the second card have different numbers? b What is the probability of full house given that the first card and the second card have the same number? c Should the sum of the answers in the previous parts be one? a It does not matter what the actual numbers are so for definiteness we assume the first card is C and the second card is C. That is, P full house st and nd cards have different numbers P full house stc, ndc. Now, we have two cases. Number appears twice and appears three times or vice versa. So P full house stc, ndc b Similar to the previous part, 0 P full house st and nd cards have same number P full house stc, ndd. Again we have two cases. Number appears twice or number appears three times. If appears twice, there are ways to choose the number that appears three times and ways to choose which suits appear. If appears three times, there are ways to choose the suit of the third, ways to choose the number that appear twice, and ways to choose the suits of the number that appears twice. So P full house stc, ndd c No. We have P A B + P A c B. The probabilities in this problem are P A B and P A B c so the sum of the probabilities is not necessarily. 0