1. We have a box containing three cards: a) one card has the number 1 on it (b)

Question

1. We have a box containing three cards: a) one card has the number 1 on it (b) one card has the number 2 on it, and (c) one card has the number 3 on it You randomly sample two cards from the box one-by-one with replacement. Define a randonm variable X as the sum of values on the two cards (a) Make a table showing the probability distribution for X. Your table should have the same hcadings as the table in question 3, where column 1 heading is X and column 2 heading is p(X), and there should be one row for each possible value of X A good way to start for working out any discrete distribution is to first find the support (set of possible values of X). Then use LTP to figure out the probability for each value in the support. Make a table of these results, and presto you have the probability distribution I suggest you check that the probabilities in your distribution sum to 1. (b) Find the expected value of X. That is find E(x) E(X) = X,Pr(X) where X1, . . . , Xc are the values in the support (c) By just looking at the distribution table you make in part (a), you could have figured out the value of E(X) without using the formula given by equation (1). Say what specifically it is about this distribution that allows you to know E(X) without using the formula. (d) Find the variance of X

Explanation / Answer

a) The combinations possible are

1,1 X=2

1,2 X=3

1,3 X=4

2,1 X=3

2,2 X=4

2,3 X=5

3,1 X= 4

3,2 X=5

3,3 X=6

P(x)

X P(x)

2 1/9

3 2/9

4 3/9

5 2/9

6 1/9

b) Expectation of X = 1/9* (2) +2/9*(3)+3/9*4+2/9*5+1/9*6= 4

c) distribution is symmentrical, So E(x)= Median value =4

d) Var (x) = 1/9* (2-4)^2 +2/9*(3-4)^2+3/9*(4-4)^2+2/9*(5-4)^2+1/9*(6-4)^2 = 2* [1/9*4 +2/9]=2*2/3=4/3

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