Answer to Question #257976 in Statistics and Probability for Mohammed

Question

1.    10 fair die are rolled. Find the expected value of the sum of the faces showing.

2.    An urn contains 2 red chips and 3 white chips.

a.    Two are drawn out at random without replacement. Let X denote the number of red chips in the sample. Calculate Var(X).

b.    Calculate Var(X) if the sampling is done with replacement.

Answer 1

1. For onefair dice

$$E(X)=\dfrac{1}{6}(1)+\dfrac{1}{6}(2)+\dfrac{1}{6}(3)+\dfrac{1}{6}(1)+\dfrac{1}{6}(2)+\dfrac{1}{6}(3)$$

$$=3.5$$

Expected value is linear, meaning that the expected value of the sum of random variables is the sum of their expected values. This means that if you roll $n$  dice, the expected value of the sum of the faces is $n$ times the expected value of a single face, or $3.5n.$

For 10 fair die

$$E(Y)=10(3.5)=35$$

The expected value of the sum of the faces showing is $35.$

2.

a.  Without replacement.

$$P(X=0)=\dfrac{\dbinom{2}{0}\dbinom{3}{2}}{\dbinom{2+3}{2}}=\dfrac{1(3)}{10}=\dfrac{3}{10}$$

$$P(X=1)=\dfrac{\dbinom{2}{1}\dbinom{3}{1}}{\dbinom{2+3}{2}}=\dfrac{2(3)}{10}=\dfrac{6}{10}$$

$$P(X=2)=\dfrac{\dbinom{2}{2}\dbinom{3}{0}}{\dbinom{2+3}{2}}=\dfrac{1(1)}{10}=\dfrac{1}{10}$$

$$E(X)=\dfrac{3}{10}(0)+\dfrac{6}{10}(1)+\dfrac{1}{10}(2)=\dfrac{8}{10}$$

$$E(X^2)=\dfrac{3}{10}(0)^2+\dfrac{6}{10}(1)^2+\dfrac{1}{10}(2)^2=\dfrac{10}{10}$$

$$Var(X)=E(X^2)-(E(X))^2=\dfrac{10}{10}-(\dfrac{8}{10})^2=0.36$$

b.  With replacement.

$$P(X=0)=\dfrac{3}{5}(\dfrac{3}{5})=\dfrac{9}{25}$$

$$P(X=1)=\dfrac{2}{5}(\dfrac{3}{5})+\dfrac{3}{5}(\dfrac{2}{5})=\dfrac{12}{25}$$

$$P(X=2)=\dfrac{2}{5}(\dfrac{2}{5})=\dfrac{4}{25}$$

$$E(X)=\dfrac{9}{25}(0)+\dfrac{12}{25}(1)+\dfrac{4}{25}(2)=\dfrac{20}{25}=\dfrac{4}{5}$$

$$E(X^2)=\dfrac{9}{25}(0)^2+\dfrac{12}{25}(1)^2+\dfrac{4}{25}(2)^2=\dfrac{28}{25}$$

$$Var(X)=E(X^2)-(E(X))^2=\dfrac{28}{25}-(\dfrac{4}{5})^2=0.48$$