Q. $2$ dice are thrown. Suppose a random variable $X$ is assigned a value $2k$ , if the sum on the dice is equal to $k$ , then the expected value of $X$ is
NTA AbhyasNTA Abhyas 2020Probability
Solution:
$k$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $11$ $12$ $P\left(k\right)$ $\frac{1}{36}$ $\frac{2}{36}$ $\frac{3}{36}$ $\frac{4}{36}$ $\frac{5}{36}$ $\frac{6}{36}$ $\frac{5}{36}$ $\frac{4}{36}$ $\frac{3}{36}$ $\frac{2}{36}$ $\frac{1}{36}$
$k=$ sum on dice
Expected value $=\frac{1}{36}\times 4+\frac{2}{36}\times 6+\frac{3}{36}\times 8+\frac{4}{36}\times 10$
$+\frac{5}{36}\times 12+\frac{6}{36}\times 14+\frac{5}{36}\times 16+\frac{4}{36}\times 18$
$+\frac{3}{36}\times 20+\frac{2}{36}\times 22+\frac{1}{36}\times 24$
$=\frac{504}{36}=14$
| $k$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ | $9$ | $10$ | $11$ | $12$ |
|---|---|---|---|---|---|---|---|---|---|---|---|
| $P\left(k\right)$ | $\frac{1}{36}$ | $\frac{2}{36}$ | $\frac{3}{36}$ | $\frac{4}{36}$ | $\frac{5}{36}$ | $\frac{6}{36}$ | $\frac{5}{36}$ | $\frac{4}{36}$ | $\frac{3}{36}$ | $\frac{2}{36}$ | $\frac{1}{36}$ |