Question

There is a box of 20 marbles. Of these marbles, 6 are red, 8 are green and 6 are blue. 6 marbles are randomly selected from the box without replacement. Let X be the number of marbles that are red or blue, and let Y be the number of marbles that are blue.

a. What is the probability the first and second marbles are red, the third and fourth are blue and the fifth and sixth are green?

b. What is the probability we obtain 2 red, 2 blue and 2 green?

c. Find E (X).
d. Find Var(Y).
e. Find P (X = Y ).

Answer 1

The total number of ways in which 6 marbles can be selected out of 20 without replacement is .

(2) - 38760

a. The number of ways in which the given sequence of balls come out can be calculated using the successive multiplication rule of probability.

There are 6 red balls so the probability of selecting a red ball out of 20 is (6/20). Now there are 5 red and 19 total balls left. The probability of selecting another red ball is (5/19). The remaining probabilities can be calculated similarly.

	Total balls	Red	Blue	Green	Ball selected	Probability
	20	6	6	8	RED	6/20
	19	5	6	8	RED	5/19
	18	4	6	8	BLUE	6/18
	17	4	5	8	BLUE	5/17
	16	4	4	8	GREEN	8/16
	15	4	4	7	GREEN	7/15

Hence, the required probability = (6/20)*(5/19)*(6/18)*(5/17)*(8/16)*(7/15) = 0.0307

b. In this case, there is no particular order in which the balls has to come out. Thus the problem boils down to selecting the given number of balls of each color.

The number of ways in which the given sequence of marbles come out is :

(2) (2) (3) 6500

Hence, the required probability = (no. of ways in which the given event can occur)/(total no. of ways in which events can occur)

$P=\frac{6300}{38760}=0.163$

c. Let RB be a fictitious color so that each one of the red and blue marbles belong to it. Thus there are 12 RB marbles and 8 Green marbles. Thus X is the number of RB marbles out of the 6 selected marbles.

   (ALL GREEN)

28 P(X =0) = 38760

(5 GREEN AND 1 RB)

672 P(X = 1) = 38760

(4 GREEN AND 2 RB)

P(X =2) = 4620 38760

$P(X=3)=\frac{\binom{12}{3}\binom{8}{3}}{\binom{20}{6}} = \frac{12320}{38760}$ (3 GREEN AND 3 RB)

$P(X=4)=\frac{\binom{12}{4}\binom{8}{2}}{\binom{20}{6}} = \frac{13860}{38760}$ (2 GREEN AND 4 RB)

$P(X=5)=\frac{\binom{12}{5}\binom{8}{1}}{\binom{20}{6}} = \frac{6336}{38760}$ (1 GREEN AND 5 RB)

$P(X=6)=\frac{\binom{12}{6}\binom{8}{0}}{\binom{20}{6}} = \frac{924}{38760}$ (ALL RB)

Now that we have the discrete probability distribution, we can easily calculated the Expectation using

$E(X) = \sum_{x=0}^{6}P(X=x) \times x$

Putting all the values,

$E(X) = 0 \times \frac{28}{38760} +1 \times \frac{672}{38760}+ 2\times \frac{4620}{38760}+3 \times \frac{12320}{38760}+4 \times \frac{13860}{38760}+5 \times \frac{6336}{38760}+6 \times \frac{924}{38760}$

E(X) = 0 +672 +9240 + 36960 + 55440 + 31680 + 5544 38760 = 138816/38760

Hence, E(X) = 3.58.

d.

Similar to the last problem, the other two balls can be combined as RG. Thus there are 6 blue balls and 14 RG balls. Again, using the same procedure to find out the probabilities, we get the following results.

Y (no. of Blue balls)	Number of RG balls	P(Y = y)	Y^2
0	6	$\frac{\binom{6}{0}\binom{14}{6}}{\binom{20}{6}} = \frac{3003}{38760}$	0
1	5	$\frac{\binom{6}{1}\binom{14}{5}}{\binom{20}{6}} = \frac{12012}{38760}$	1
2	4	$\frac{\binom{6}{2}\binom{14}{4}}{\binom{20}{6}} = \frac{15015}{38760}$	4
3	3	$\frac{\binom{6}{3}\binom{14}{3}}{\binom{20}{6}} = \frac{7280}{38760}$	9
4	2	$\frac{\binom{6}{4}\binom{14}{2}}{\binom{20}{6}} = \frac{1365}{38760}$	16
5	1	$\frac{\binom{6}{5}\binom{14}{1}}{\binom{20}{6}} = \frac{84}{38760}$	25
6	0	$\frac{\binom{6}{6}\binom{14}{0}}{\binom{20}{6}} = \frac{1}{38760}$	36

Using the formula for Expectation, we get

$E(Y) = 0 \times \frac{3003}{38760} +1 \times \frac{12012}{38760}+ 2\times \frac{15015}{38760}+3 \times \frac{7280}{38760}+4 \times \frac{1365}{38760}+5 \times \frac{84}{38760}+6 \times \frac{1}{38760}$

$E(Y) = \frac{0 \times 3003 +1 \times 12012+ 2 \times 15015+3 \times 7280+4 \times 1365+5 \times 84+6 \times 1}{38760}$

Hence, $E(Y) = \frac{69768}{38760} = 1.8$

In order to calculate the variance, we will need E(Y^2).

$E(Y^2) = \sum_{y=0}^{6}P(Y=y) \times y^2$

$E(Y^2) = \frac{0 \times 3003 +1 \times 12012+ 4 \times 15015+9 \times 7280+16 \times 1365+25 \times 84+36 \times 1}{38760}$

As we know, var(Y) = E(Y^2) - (E(Y))^2, we get