4.9 Suppose a bowl has 9 chips; one chip is labeled \"1\", three chips are label

Question

4.9 Suppose a bowl has 9 chips; one chip is labeled "1", three chips are labeled "3", and five chips are labeled "5". Suppose two chips are selected at random with replacement. Let the random variable X equal the absolute difference between the two draws (e.g. if the first draw is a 1 (11) and the second draw is a 5 (52), then the absolute difference is 1-5-4) (a) Find the probability distribution of X (b) Find the probability that both draws are the same (c) Find the probability that both draws are not the same. (d) Given that both draws are not the same, determine the probability that the absolute difference is equal to 2, i.e. find P(X -2X > 0) (e) On average, what is X equal to? 4.10 Repeat question 4.9(a) assuming the chips are drawn without replacement.

Explanation / Answer

Q.4.9 There are one chip labeled '1' , three chips labeled '3' , five chips labeled '5'

selected at random with replacement. SO there are 9 x 9 = 81 ways to select the chips

X = absolute difference between two draws

so X can have values = 0, 2, 4

so X would be 0 when we get same type of draws in each case. (1,1) ,(3,3) ,(5,5)

f(x) = 1/9 * 1/9 + 3/9 * 3/9 + 5/9 * 5/9 = 35/81 for X = 0

X would be two when draws are (1,3) , (3,1) , (3,5), (5,3)

f(x) = 2 * (1/9) * (3/9) + 2 * (3/9) * (5/9) = 36/81

X would be 4 when draws are (1,5) and (5,1)

f(x) = 2 * (1/9) * (5/9) = 10/81

so f(x) = 35/81 ; x = 0

= 36/81; x =2

= 10/81 ; x = 4

(b) Probability that both draws are same = P(X = 0) = 35/81

(c) Probaiblity that both draws are different = 1 - P(X = o) = 1 - 35/81 = 46/81

(d) P(X =2 l X >0) = (36/81) / (46/81) = 18/23

(e) E(X) = 0 * 35/81 + 2 * 36/81 + 4 * 10/81 = 112/81 = 1.38

Question 4.10

Q.4.9 There are one chip labeled '1' , three chips labeled '3' , five chips labeled '5'

selected at random without replacement so number of ways are 9 * 8 = 72

X = absolute difference between two draws

so X can have values = 0, 2, 4

so X would be 0 when we get same type of draws in each case. (3,3) ,(5,5) . HEre case (1,1) is not there as there is only one chip of 1.

f(x) = 3/9 * 2/8 + 5/9 * 4/8 = 26/72 for X = 0

X would be two when draws are (1,3) , (3,1) , (3,5), (5,3)

f(x) = 1/9 * 3/8 + 3/9 * 1/8 + 3/9 * 5/8 + 5/9 * 3/8 = 36/72

X would be 4 when draws are (1,5) and (5,1)

f(x) = 1/9 * 5/8 + 5/9 * 1/8 = 10/72

so f(x) = 26/72 ; x = 0

= 36/72; x =2

= 10/72 ; x = 4

(b) Probability that both draws are same = P(X = 0) = 26/72

(c) Probaiblity that both draws are different = 1 - P(X = o) = 1 - 26/72 = 46/72

(d) P(X =2 l X >0) = (36/72) / (46/72) = 18/23

(e) E(X) = 0 * 26/72 + 2 * 36/72 + 4 * 10/72 =  112/72 = 14/9 = 1.556

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