Question

Two cards are selected at random from a box which contains five cards numbered 1, 1, 2, 2, and 3. Let X denote the sum and Y the maximum of the two numbers drawn. Find the probability distribution, mean and variance of X and Y.

Answer 1

There are 5 cards numbered 1, 1, 2, 2 and 3.

Let:
X = Sum of two numbers on cards = 2, 3, 4, 5

Y = Maximum of two numbers = 1, 2, 3


Thus, the probability distribution of X is given by

X	P(X)
2	$\frac{1}{10}$
3	$\frac{4}{10}$
4	$\frac{3}{10}$
5	$\frac{2}{10}$

Total number of cards = 2 cards with "1" on them + 2 cards with "2" on them + 1 card with "3" on it
= 5

Total number of possible choices (in drawing the two cards)= Number of ways in which the two cards can be drawn from the total 5
= ⁵C₂
=$\frac{5!}{2!3!}$
= 10

Probabilty that the two cards drawn would be having the numbers 1 and 1 on them is P(11).

P(11) = $\frac{{}^{2}C_2\times {}^{2}C_0\times {}^{1}C_0}{{}^{5}C_2}$
$$\begin{gathered} =\frac{1\times 1\times 1}{10} \\ =\frac{1}{10} \end{gathered}$$

Probabilty that the two cards drawn would be having the numbers 1 and 2 on them is P(11).

P(11) = $\frac{{}^{2}C_2\times {}^{2}C_0\times {}^{1}C_0}{{}^{5}C_2}$
$$\begin{gathered} =\frac{1\times 1\times 1}{10} \\ =\frac{1}{10} \end{gathered}$$

"1" and "2" on them

⇒ P(12)	=	2C1 × 2C1 × 1C0 5C2
		"1's" "2's" "3's" Total Available 2 2 1 5 To Choose 1 1 0 2 Choices 2C1 2C1 1C0 5C2
	=	2 1 × 2 1 × 1 10
	=	2 × 2 × 1 10
	=	4 10
	=	2 5

"1" and "3" on them

⇒ P(13)	=	2C1 × 2C0 × 1C1 5C2
		"1's" "2's" "3's" Total Available 2 2 1 5 To Choose 1 0 1 2 Choices 2C1 2C0 1C1 5C2
	=	2 1 × 1 × 1 1 10
	=	2 × 1 × 1 10
	=	2 10
	=	1 5

"2" and "2" on them

⇒ P(22)	=	2C0 × 2C2 × 1C1 5C2
		"1's" "2's" "3's" Total Available 2 2 1 5 To Choose 0 2 0 2 Choices 2C0 2C2 1C0 5C2
	=	1 × 1 × 1 10
	=	1 10

"2" and "3" on them

⇒ P(23)	=	2C0 × 2C1 × 1C1 5C2
		"1's" "2's" "3's" Total Available 2 2 1 5 To Choose 0 1 1 2 Choices 2C0 2C1 1C1 5C2
	=	1 × 2 1 × 1 10
	=	1 × 2 × 1 10
	=	2 10
	=	1 5

Probability for the sum of the numbers on the cards drawn to be

	2 ⇒ P(X = 2)	=	P(11)
		=	1 10
	3 ⇒ P(X = 3)	=	P(12)
		=	4 10
	4 ⇒ P(X = 4)	=	P(13 or 22) i.e. P(13 ∪ 22)
		=	P(13) + P(22)
		=	2 10 + 1 10
		=	3 10
	5 ⇒ P(X = 2)	=	P(23)
		=	2 10

The probabilty distribution of "x" would be

x	2	3	4	5
P(X = x)	1 10	4 10	3 10	2 10

Calculations for Mean and Standard Deviations

	x	P (X = x)	px [x × P (X = x)]	x2	px2 [x2 × P (X = x)]
	2	1 10	2 10	4	4 10
	3	4 10	12 10	9	36 10
	4	3 10	12 10	16	48 10
	5	2 10	10 10	25	50 10
Total		1	36 10		138 10
			= 3.6		= 13.8

The Expected Value of the sum

⇒ Expectation of "x"

⇒ E (x)	=	Σ px
	=	3.6

Variance of the sum of the numbers on the cards

⇒ var (x)	=	E (x2) − (E(x))2
⇒ var (x)	=	Σ px2 − (Σ px)2
	=	13.8 − (3.6)2
	=	13.8 − 12.96
	=	0.84

Standard Deviation of the sum of the numbers on the cards

⇒ SD (x)	=	+ √ Var (x)
	=	+ √ 0.84
	=	+ 0.917

Computation of mean and variance

xi	pi	pixi	pixi2
2	$\frac{1}{10}$	$\frac{2}{10}$	$\frac{4}{10}$
3	$\frac{4}{10}$	$\frac{12}{10}$	$\frac{36}{10}$
4	$\frac{3}{10}$	$\frac{12}{10}$	$\frac{48}{10}$
5	$\frac{2}{10}$	1	$\frac{50}{10}$
		${\sum }_{}^{}$pixi = $\frac{36}{10}=3.6$	${\sum }_{}^{}$pixi2 = $\frac{138}{10}=13.8$

$$\begin{gathered} \mathrm{Mean}=\sum p_i{x}_i=3.6 \\ \mathrm{Variance}=\underset{}{\overset{}{\sum p_i{x_i}^2}}-{\left(\mathrm{Mean}\right)}^2=13.8-12.96=0.84 \end{gathered}$$


Thus, the probability distribution of Y is given by

Y	P(Y)
1	0.1
2	0.5
3	0.4

Computation of mean and variance

yi	pi	piyi	piyi2
1	0.1	0.1	0.1
2	0.5	1	2
3	0.4	1.2	3.6
		${\sum }_{}^{}$pixi = 2.3	${\sum }_{}^{}$pixi2 = 5.7

$$\begin{gathered} \mathrm{Mean}=\sum p_i{y}_i=2.3 \\ \mathrm{Variance}=\underset{}{\overset{}{\sum p_i{y_i}^2}}-{\left(\mathrm{Mean}\right)}^2=5.7-5.29=0.41 \end{gathered}$$