Tuesday

January 27, 2015

January 27, 2015

Posted by **Randy** on Tuesday, March 12, 2013 at 1:06pm.

(a) The outcome of interest is the number on each of the two balls we select. List the complete sample space of outcomes.

(b) What is the probability of selecting two even-numbered balls?

(c) What is the probability that the number on the second selected ball is greater than the number on the first selected ball?

(d) Let X be the absolute difference of the two numbers on the selected balls |X1 - X2|, where X1 is the number on the first selected ball and X2 is the second selected ball. Find the probability distribution of X.

(e) What is the probability that X = 2 if the number 4 is not selected?

(f) Find the expected value of X.

(g) Find the variance of X.

- Statistics -
**bobpursley**, Tuesday, March 12, 2013 at 1:11pmI am not certain what your question is here. Do you want your work checked?

- Statistics -
**Randy**, Tuesday, March 12, 2013 at 3:23pma) SS= {12,13,14,21,22,23,24,34,43}

b) Selecting two even numbers

SS= {22,24,42}

P(getting 22) = (2/5)(1/4)= 0.1

P(getting 24) = (2/5)(1/4)= 0.1

P(getting 42) = (1/5)(2/4)= 0.1

P(getting 2 even number balls) = P(22)+P(24)+P(42)= 0.1+0.1+0.1= 0.3

c) The number on the second selected ball is greater than that of the first selected ball

SS={12,13,14,23,24,34}

P(getting 12)= (1/5)(2/4)= 0.1

P(getting 13)= (1/5)(1/4)= 0.05

P(getting 14)= (1/5)(1/4)= 0.05

P(getting 23)= (2/5)(1/4)= 0.1

P(getting 24)= (2/5)(1/4)= 0.1

P(getting 34)= (1/5)(1/4)= 0.05

P(number on the 2nd selected is greater than 1st selected ball)

= P(12)+P(13)+P(14)+P(23)+P(24)+P(34) 0.1+0.05+0.05+0.1+0.1+0.05=0.9

this is what I have so far... I'm having issues with question d, my probabilities aren't adding up to 1

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