there three urns. the first urn contains 3 green balls and 2 white balls, the second contain 5 green balls and 6 white balls and the third urn contain 2 green balls and 4 white. one ball is drawn from each urn. find the mean variance of the probability distribution of the discrete random variable "number of white balls drawn".
Answers
Probability distribution of White Balls :
White balls Probability
0 1/11
1 58/165
2 68/165
3 8/55
Mean number of white Balls = 266/165
Step-by-step explanation:
Urn1 : 3 Green 2 white 5 Total
Urn2 : 5 Green 6 white 11 Total
Urn3 : 2 Green 4 white 6 Total
White Balls
0 White balls - no white balls from any Urn
0 (3/5)(5/11)(2/6) = 1/11
1 White Ball 3 cases - 1 from Urn1 , Urn2 or Urn 3
1 (2/5)(5/11)(2/6) + (6/11)(3/5)(2/6) + (4/6)(3/5)(5/11)
2/33 + 6/55 + 2/11
(10 + 18 + 30)/165 = 58/165
2 White Balls Urn1 , Urn2 , Urn1Urn3 , Urn2 , Urn3
2 (2/5)(6/11)(2/6) + (2/5)(4/6)(5/11) + (6/11)(4/6)(3/5)
4/55 + 4/33 + 12/55
(12 + 20 + 36)/165 = 68/165
3 White Balls 1 from each urn
3 (2/5)(6/11)(4/6) = 8/55
Probability distribution :
White balls Probability
0 1/11
1 58/165
2 68/165
3 8/55
Mean = 0(1/11) + 1(58/165) + 2(68/165) + 3*(8/55)
= 0 + 58/165 + 136/55 + 24/55
= ( 58 + 136 + 72)/165
= 266/165
Mean number of white Balls = 266/165
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Answer:
Step-by-step explanation: