Answer to Question #224445 in Statistics and Probability for Qasim

Question

From a box containing 5 dimes and 3 nickels, 4 coins are selected at random without replacement.

Find the probability distribution for the total T of the 4 coins. Express the probability distribution graphically as a probability histogram. Find the Expected Value and Standard Deviation of the number of coins.

Answer 1

For the purposes of this question, the expected probability outcomes are calculated as follows:

Let:

$D=Dimes$

$N=Nickels$

$1$ $dime=10cents$

$1$ $nickel=5cents$

$​$ $P(4D \& 0N)=\dfrac{\dbinom{5}{4}\dbinom{3}{0}}{\dbinom{5+3}{4}}=\dfrac{1}{14}$

$P(3D \& 1N)=\dfrac{\dbinom{5}{3}\dbinom{3}{1}}{\dbinom{5+3}{4}}=\dfrac{3}{7}$

$P(2D \& 2N)=\dfrac{\dbinom{5}{2}\dbinom{3}{2}}{\dbinom{5+3}{4}}=\dfrac{3}{7}$

$P(1D \& 3N)=\dfrac{\dbinom{5}{1}\dbinom{3}{3}}{\dbinom{5+3}{4}}=\dfrac{1}{14}$

The probability distribution table is presented as follows:

$E\left(T\right)=\frac{1}{14}\left(0.40\right)+\frac{3}{7}\left(0.35\right)+\frac{3}{7}\left(0.30\right)+\frac{1}{14}\left(0.25\right)$

$E\left(T\right)=0.325$

$E\left(T^2\right)=\frac{1}{14}\left(0.40\right)^2+\frac{3}{7}\left(0.35\right)^2+\frac{3}{7}\left(0.30\right)^2+\frac{1}{14}\left(0.25\right)^2$

$E\left(T^2\right)=\frac{1.7945}{14}$

$V\:ar\:T=\sigma ^2=E\left(T^2\right)-\left(E\left(T\right)\right)^2$

$V\:ar\:T=\sigma \:^2=\frac{1.4975}{14}-\left(0.325\right)^2\approx 0.001339\:\:$

$\sigma =\sqrt{\sigma ^2}=\sqrt{0.001339}\approx 0.0366$