A fancy fair die of six faces is made as follows: two faces have exactly 1 dot e

Question

A fancy fair die of six faces is made as follows: two faces have exactly 1 dot each, two faces have exactly 2 dots each, and the two remaining faces have exactly 3 dots each. This fancy die is tossed three times and each time the face up is recorded. Let X denote the difference between the sum of the number of dots recorded on the first two tosses and the number of dots recorded on the third toss. (a) Find and plot the PMF of X. (b) Find the mean and variance of X. (c) Find the probability that |X| greaterthanorequalto k for all possible k.

Explanation / Answer

now X= (Sum of number of dots at first two tosses) - (number of dots on third toss)

sample space of all the three tosses=

123,122,121,213,212,211,113,112,111,223,222,221,131,132,133,311,312,313,331,332,333,321,322,323,231,232,233

sample space of X= 0,1,2,1,1,2,-,0,1,1,2,3,3,2,1,3,2,1,5,4,3,4,3,2,4,3,2 (we get following samples after adding first two tosses and subtracting third toss)

(a) PMF will be :

(b) mean = 1*6/26 + 2* 7/26 + 3* 6/26 + 4* 3/26 + 5* 1/26 =2.03

variance = 1*1*6/26 + 2*2*7/26 + 3*3* 6/26 + 4*4* 3/26 + 5*5*1/26 = 5.85

(please consider * as multiply sign)

(c) for mode of X ,sample size will be  0,1,2,1,1,2,1,0,1,1,2,3,3,2,1,3,2,1,5,4,3,4,3,2,4,3,2

P(|x|<=k)

P(|x|>=k)=1 - P(|x|<=k) =

PMF probability of getting exactly one dot probability of getting two dots probability of getting three dots 1/3 1/3 1/3

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