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
Box contains five cards $1, 1, 2, 2, 3$.
Here, $X$ denotes the sum of the two number on cards drawn.
$Y$ denotes the maximum of the two number.
So, $X = 2, 3, 4, 5 Y = 1, 2, 3 P(X = 2) = P(1)P(1) =\frac{2}{5}\times\frac{1}{4}$
$=0.1 P(X = 3) = P(1)P(2) + P(2)P(1) =\frac{2}{5}\times\frac{2}{4}+\frac{2}{5}\times\frac{2}{4}$
$=0.4 P(X = 4) = P(2)P(2) + P(1)P(3) + P(3)P(1) =\frac{2}{5}\times\frac{1}{4}+\frac{2}{5}\times\frac{1}{4}+\frac{1}{5}\times\frac{2}{4}$
$=0.3 P(X = 5) = P(2)P(3) + P(3)P(2) =\frac{2}{5}\times\frac{2}{4}+\frac{1}{5}\times\frac{2}{4}$
$=0.2 Probability distribution for X$
Now,
Mean $=\sum\text{xp}$ Mean $= 3.6$ Variance $=\sum\text{x}^2\text{p}-(\text{mean})^2 =13.8-(3.6)^2 =13.8-12.96$ Variance $= 0.84 P(Y = 1) = P(1)P(1) =\frac{2}{5}\times\frac{1}{4} =\frac{2}{20} =0.1 P(Y = 2) = P(1)P(2) + P(2)P(1) + P(2)P(2)$
$ =\frac{2}{5}\times\frac{2}{4}+\frac{2}{5}\times\frac{2}{4}+\frac{2}{5}\times\frac{1}{4}$
$=0.5 P(Y = 3) = P(1)P(3) + P(2)P(3) + P(3)P(1) + P(3)P(2)$
$=\frac{2}{5}\times\frac{1}{4}+\frac{2}{5}\times\frac{1}{4}+\frac{1}{5}\times\frac{2}{4}+\frac{1}{5}\times\frac{2}{4}$
$=0.4$ Probability distribution for $Y$ is
Mean $=\sum\text{xp}=2.3$ Variance $=\sum\text{x}^2\text{p}-(\text{mean})^2 =5.1-(2.3)^2$ Variance $= 0.41$
Here, $X$ denotes the sum of the two number on cards drawn.
$Y$ denotes the maximum of the two number.
So, $X = 2, 3, 4, 5 Y = 1, 2, 3 P(X = 2) = P(1)P(1) =\frac{2}{5}\times\frac{1}{4}$
$=0.1 P(X = 3) = P(1)P(2) + P(2)P(1) =\frac{2}{5}\times\frac{2}{4}+\frac{2}{5}\times\frac{2}{4}$
$=0.4 P(X = 4) = P(2)P(2) + P(1)P(3) + P(3)P(1) =\frac{2}{5}\times\frac{1}{4}+\frac{2}{5}\times\frac{1}{4}+\frac{1}{5}\times\frac{2}{4}$
$=0.3 P(X = 5) = P(2)P(3) + P(3)P(2) =\frac{2}{5}\times\frac{2}{4}+\frac{1}{5}\times\frac{2}{4}$
$=0.2 Probability distribution for X$
| $x:$ | $2$ | $3$ | $4$ | $5$ |
| $P(x):$ | $0.1$ | $0.4$ | $0.3$ | $0.2$ |
| $x_i$ | $p_i$ | $x_ip_i$ | $x_i^2p_i$ |
| $2$ | $0.1$ | $0.1$ | $0.4$ |
| $3$ | $0.4$ | $1.2$ | $3.6$ |
| $4$ | $0.3$ | $1.2$ | $4.8$ |
| $5$ | $0.2$ | $1.0$ | $5.0$ |
| $ $ |
$ $ |
$\sum\text{xp}=3.6$ | $\sum\text{x}^2\text{p}=13.8$ |
$ =\frac{2}{5}\times\frac{2}{4}+\frac{2}{5}\times\frac{2}{4}+\frac{2}{5}\times\frac{1}{4}$
$=0.5 P(Y = 3) = P(1)P(3) + P(2)P(3) + P(3)P(1) + P(3)P(2)$
$=\frac{2}{5}\times\frac{1}{4}+\frac{2}{5}\times\frac{1}{4}+\frac{1}{5}\times\frac{2}{4}+\frac{1}{5}\times\frac{2}{4}$
$=0.4$ Probability distribution for $Y$ is
| $x:$ | $1$ | $2$ | $3$ |
| $p(x):$ | $0.1$ | $0.5$ | $0.4$ |
| $y_i$ | $p_i$ | $y_ip_i$ | $y_i^2p_i$ |
| $1$ | $0.1$ | $0.1$ | $0.1$ |
| $2$ | $0.5$ | $1.0$ | $2.0$ |
| $3$ | $0.4$ | $1.2$ | $3.6$ |
| $ $ |
$ $ |
$\sum\text{xp}=2.3$ | $\sum\text{x}^2\text{p}=5.7$ |
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