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  • There are 5 cards numbered 1 to 5, one number on one card. Two cards are drawn at random without replacement. Let X denote the sum of the numbers on two cards drawn. Find the mean and variance of X.

There are 5 cards numbered 1 to 5, one number on one card. Two cards are drawn at random without replacement. Let X denote the sum of the numbers on two cards drawn. Find the mean and variance of X.

Answers (1)

Solution

The sample space is

S = { (1,2),(1,3),(1,4),(1,5),
(2,1),(2,3),(2,4),(2,5),
(3,1),(3,2),(3,4),(3,5),
(4,1),(4,2),(4,3),(4,5)
(5,1),(5,2),(5,3),(5,4)}

Total Sample Space, n(S) = 20

Let the random variable be X which denotes the sum of the numbers on the cards drawn.
∴ X = 3, 4, 5, 6, 7, 8, 9
At X = 3
The cards whose sum is 3 are (1,2), (2,1)

P(X)=\frac{2}{20}=\frac{1}{10}
At  x=4  
The cards whose sum is 4 are (1,3),(3,1)
P(X)=\frac{2}{20}=\frac{1}{10}
At  X=5 
The cards whose sum is 5 are (1,4),(2,3),(3,2),(4,1)
P(X)=\frac{4}{20}=\frac{1}{5}
At  X=6
The cards whose sum is 6 are (1,5),(2,4),(4,2),(5,1)
P(X)=\frac{4}{20}=\frac{1}{5}
At  x=7 

The cards whose sum is 7 are (2,5),(3,4),(4,3),(5,2)
P(X)=\frac{4}{20}=\frac{1}{5}
At X = 8
The cards whose sum is 8 are (3,5), (5,3)
P(X)=\frac{2}{20}=\frac{1}{10}

At X = 9
The cards whose sum is 9 are (4,5), (5,4)

\begin{aligned} &\begin{array}{l} P(X)=\frac{2}{20}=\frac{1}{10} \\\\ \therefore \text { Mean, } E(X)=\Sigma \times P(X) \\\\ =3 \times \frac{1}{10}+4 \times \frac{1}{10}+5 \times \frac{1}{5}+6 \times \frac{1}{5}+7 \times \frac{1}{5}+8 \times \frac{1}{10}+9 \times \frac{1}{10} \\\\ =\frac{3}{10}+\frac{2}{5}+1+\frac{6}{5}+\frac{7}{5}+\frac{4}{5}+\frac{9}{10} \\\\ =\frac{3+4+10+12+14+8+9}{10} \\\\ =\frac{60}{10} \\ =6 \end{array}\\ &\text { And, }\\ &\Sigma X^{2} P(X)=3^{2} \times \frac{1}{10}+4^{2} \times \frac{1}{10}+5^{2} \times \frac{1}{5}+6^{2} \times \frac{1}{5}+7^{2} \times \frac{1}{5}+8^{2} \times \frac{1}{10}\\ &+9^{2} \times \frac{1}{10} \end{aligned}

\begin{aligned} &=\frac{9}{10}+\frac{16}{10}+5+\frac{36}{5}+\frac{49}{5}+\frac{64}{10}+\frac{81}{10}\\ &=\frac{9+16+50+72+98+64+81}{10}\\ &=\frac{390}{10}\\ &=39\\ &\text { Therefore, }\\ &\operatorname{Var} \mathrm{X}=\Sigma \mathrm{X}^{2} \mathrm{P}(\mathrm{X})-[\Sigma \mathrm{XP}(\mathrm{X})]^{2}\\ &=39-36\\ &=3 \end{aligned}

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