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# A box contains 6 red balls and 2 black balls. Two balls are drawn at random from it without replacement. If x denotes the number of red balls drawn, then find $E\left( x \right)$

Last updated date: 17th Jun 2024
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Hint: We can take the possible value for x and its probability. Then we can find $E\left( x \right)$ by taking the summation of the product of the random variable and its corresponding probability.

We have a box containing 2 black balls and 6 red balls. We are taking 2 balls from the box without replacement. We are given that ${\text{x}}$ is the number of red balls drawn from the box. As the box has 6 red balls and we are taking only 2 balls, x can take values 0, 1, and 2.
When $x = 0$, we have no red balls. So, its probability is given by the number of ways of selecting 2 black balls from 2 black balls divided by number of ways of selecting 2 balls from 8 balls.
$P\left( 0 \right) = \dfrac{{{}^2{C_2}}}{{{}^8{C_2}}} = \dfrac{{\dfrac{{2!}}{{2!\left( 0 \right)!}}}}{{\dfrac{{8!}}{{2!\left( 6 \right)!}}}} = \dfrac{{1 \times 2}}{{8 \times 7}} = \dfrac{1}{{28}}$
When $x = 1$, we have 1 red ball and 1 black ball. So, its probability is given by the number of ways of selecting 1 black ball from 2 black balls and 1 red ball from 6 red balls divided by the number of ways of selecting 2 balls from 8 balls.
$P\left( 1 \right) = \dfrac{{{}^2{C_1} \times {}^6{C_1}}}{{{}^8{C_2}}} = \dfrac{{\dfrac{{2!}}{{1!\left( 1 \right)!}} \times \dfrac{{6!}}{{1!\left( 5 \right)!}}}}{{\dfrac{{8!}}{{2!\left( 6 \right)!}}}} = \dfrac{{2 \times 6 \times 2}}{{8 \times 7}} = \dfrac{3}{7}$
When $x = 2$, we have 2 red balls. So, its probability is given by the number of ways of selecting 2 red balls from 6 red balls divided by the number of ways of selecting 2 balls from 8 balls.
$P\left( 2 \right) = \dfrac{{{}^6{C_2}}}{{{}^8{C_2}}} = \dfrac{{\dfrac{{6!}}{{2!\left( 4 \right)!}}}}{{\dfrac{{8!}}{{2!\left( 6 \right)!}}}} = \dfrac{{6 \times 5}}{{8 \times 7}} = \dfrac{{15}}{{28}}$
The expected value for a random variable x is given by
$E\left( x \right) = \sum _{i = 1}^n{x_i} \times P\left( {{x_i}} \right) \\ = 0 \times P\left( 0 \right) + 1 \times P\left( 1 \right) + 2 \times P\left( 2 \right) \\ = 0 \times \dfrac{1}{{28}} + 1 \times \dfrac{3}{7} + 2 \times \dfrac{{15}}{{28}} \\ = \dfrac{6}{{14}} + \dfrac{{15}}{{14}} \\ = \dfrac{{21}}{{14}} \\ = \dfrac{3}{2} \\$
Thus, the required expected value is $\dfrac{3}{2}$.

Note: The concepts of random variables, probability and expected value are used in this problem. $E\left( x \right)$of a discrete random variable is given by $E\left( x \right) = \sum _{i = 1}^n{x_i} \times P\left( {{x_i}} \right)$ and $E\left( x \right)$of a continuous random variable is given by $E\left( x \right) = \int\limits_{ - \infty }^\infty {xP\left( x \right)}$.