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) =  \[\frac{^{2}{}{C}_2 \times^{2}{}{C}_0 \times ^{1}{}{C}_0}{^{5}{}{C}_2}\]

\[= \frac{1 \times 1 \times 1}{10}\]
\[ = \frac{1}{10}\]

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

P(11) = \[\frac{^{2}{}{C}_2 \times ^{2}{}{C}_0 \times ^{1}{}{C}_0}{^{5}{}{C}_2}\]

\[= \frac{1 \times 1 \times 1}{10}\]
\[ = \frac{1}{10}\]

. "1" and "2" on them

⇒p(12) =`(2 c_1 xx ^2 c_1 xx ^1 c_0)/(5 c _2)`

 

	"1's"	"2's"	"3's"	Total
Available	2	2	1	5
To Choose	1	1	0	2
Choices	2C1	2C1	1C0	5C2

=`(2/1xx2/1xx1)/10`

=`(2xx2xx1)/10`

=`4/10`

=`2/5`

"1" and "3" on them 

⇒p(13) =`(2 c_1 xx ^2 c_ 0xx ^1 c_1)/(5 c _2)`

	"1's"	"2's"	"3's"	Total
Available	2	2	1	5
To Choose	1	0	1	2
Choices	2C1	2C0	1C1	5C2

`=(2/1xx1xx1/1)/10`

`=(2xx1xx1)/10`

=`2/10`

`=1/5`

"2" and "2" on them 

⇒p(22) =`(2 c_1 xx ^2 c_2 xx ^1 c_1)/(5 c _2)`

	"1's"	"2's"	"3's"	Total
Available	2	2	1	5
To Choose	0	2	0	2
Choices	2C0	2C2	1C0	5C2

 

`=(1xx1xx1)/10`

`=1/10`

"2" and "3" on them 

⇒p(23) =`(2 c_0 xx ^2 c_1 xx ^1 c_1)/(5 c _2)`

	"1's"	"2's"	"3's"	Total
Available	2	2	1	5
To Choose	0	1	1	2
Choices	2C0	2C1	1C1	5C2

`=(1xx2/1xx1)/10`

`= (1xx2xx1)/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 u 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	x2	px2
		[x × P (X = x)]		[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		=13.6

The Expected Value of the sum

⇒ Expectation of "x"

⇒E (x) = ∑ px

          = 3.6

Variance of the sum of the numbers on the cards

⇒ var (x) = E (x²) - (E (X))² 

⇒ var (x) = ∑ px² - (∑ 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	\[\frac{50}{10}\]
		`∑`pixi =\[\frac{36}{10} = 3 . 6\]	`∑`pixi 2=\[\frac{138}{10} = 13 . 8\]

\[\text{ Mean }  = \sum p_i x_i = 3 . 6\]
\[\text{ Variance }  = \sum p_i {x_i}2^{}_{} - \left( \text{ Mean } \right)^2 = 13 . 8 - 12 . 96 = 0 . 84\]

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
		\[\sum\nolimits_{}^{}\] pixi = 2.3	\[\sum\nolimits_{}^{}\] pixi2 = 5.7

 

\[\text{ Mean } = \sum p_i y_i = 2 . 3\]
\[\text{ Variance } = \sum p_i {y_i}2^{}_{} - \left( \text{ Mean } \right)^2 = 5 . 7 - 5 . 29 = 0 . 41\]