Question

An urn contains five black marbles and one orange marbles. Four marbles are drawn out one at a time. For each marble, if it is black the marble is set aside, but if it is orange it is returned to the urn before the next marble is drawn. Let X be the number of black marbles drawn from the urn. Find the probability distribution for X and find the expectation value and variance of X

Answer 1

Number of black marble = 5.

Number of orange marble = 1.

Total number of marbles = 6.

We have to draw four marble one each at a time .

We draw orange marble with replacement ( i.e Orange marble replaced back to the urn before the next draw.)

and we draw black marbles without replacement ( i.e. Black marble does not replaced back to the urn before the next draw.

Consider the outcomes

O : Orange and B : Black.

The Possible outcomes ( sample space) is

S = { OOOO, OOOB, OOBO, OBOO, BOOO, OOBB, OBOB, OBBO, BOOB, BOBO, BBOO, OBBB, BOBB, BBOB, BBBO, BBBB }

Outcomes	Probability	Number of Black Marbles
OOOO	(1/6)*(1/6)*(1/6)*(1/6)= 1/ 1296	0
OOOB	(1/6)*(1/6)*(1/6)*(5/6) = 5 /1296	1
OOBO	(1/6)*(1/6)*(5/6)*(1/5) = 1/216	1
OBOO	(1/6) *(5/6) * (1/5) * (1/5) =1/180	1
BOOO	(5/6) * (1/5) * (1/5) * (1/5) = 1/150	1
OOBB	(1/6)*(1/6) *(5/6) *(4/5)=4/216	2
OBOB	(1/6)*(5/6)*(1/5)*(4/5) = 4 / 180	2
OBBO	(1/6) * (5/6) * (4/5) * (1/4) =1/36	2
BOBO	(5/6) * (1/5) * (4/5) * (1/4)= 1/30	2
BOOB	(5/6) * ( 1/5) * (1/5) * (4/5) = 4/150	2
BBOO	(5/6) * ( 4/5) * (1/4) * (1/4) =1/24	2
OBBB	(1/6) * (5/6) * (4/5) * (3/4) = 3/36	3
BOBB	(5/6) * (1/5) * (4/5) * (3/4) = 3/30	3
BBOB	(5/6)*(4/5)*(1/4) * (3/4) = 3/24	3
BBBO	(5/6) * (4/5) * (3/4) * (1/3) = 1/6	3
BBBB	(5/6) * (4/5) * (3/4) * ( 2/3) = 1/3	4

Let X denotes the number of black marbles

X takes values 0,1, 2, 3 ,4.

P ( X =0) = 1/1296 = 0.000772.

P(X=1 ) = (5/1296) + ( 1/216) + ( 1/180) + ( 1/150) = 0.02071

P(X=2) = (4 / 216)+ ( 4 / 180) + (1/36) + ( 1/30 ) + (4/150) + ( 1/ 24)=0.170185

P(X=3) = ( 3/36 ) + ( 3/30 ) + ( 3/24 ) + ( 1/3) = 0.475000

P(X=4) = 1/3 = 0.333333

Hence the probability distribution of X is

x	Probability	x*p	x^2*p
0	0.0008	0.0000	0.0000
1	0.0207	0.0207	0.0207
2	0.1702	0.3404	0.6807
3	0.4750	1.4250	4.2750
4	0.3333	1.3333	5.3333
Total	1.000000	3.119412	10.309778

Expected value of X =3.1194

Variance of X

Variance of X = 0.579122