Question #223684

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.


1
Expert's answer
2021-08-08T17:00:49-0400
P(4D&0N)=(54)(30)(5+34)=114P(4D \& 0N)=\dfrac{\dbinom{5}{4}\dbinom{3}{0}}{\dbinom{5+3}{4}}=\dfrac{1}{14}

P(3D&1N)=(53)(31)(5+34)=37P(3D \& 1N)=\dfrac{\dbinom{5}{3}\dbinom{3}{1}}{\dbinom{5+3}{4}}=\dfrac{3}{7}

P(2D&2N)=(52)(32)(5+34)=37P(2D \& 2N)=\dfrac{\dbinom{5}{2}\dbinom{3}{2}}{\dbinom{5+3}{4}}=\dfrac{3}{7}

P(1D&3N)=(51)(33)(5+34)=114P(1D \& 3N)=\dfrac{\dbinom{5}{1}\dbinom{3}{3}}{\dbinom{5+3}{4}}=\dfrac{1}{14}


T0.400.350.300.25p(T)1143737114\begin{matrix} T & 0.40 & 0.35 & 0.30 & 0.25\\ \\ p(T) & \dfrac{1}{14} & \dfrac{3}{7} & \dfrac{3}{7}& \dfrac{1}{14} \end{matrix}

E(T)=114(0.40)+37(0.35)+37(0.30)+114(0.25)E(T)=\dfrac{1}{14}(0.40)+\dfrac{3}{7}(0.35)+\dfrac{3}{7}(0.30)+\dfrac{1}{14}(0.25)

=0.325=0.325

E(T2)=114(0.40)2+37(0.35)2+37(0.30)2E(T^2)=\dfrac{1}{14}(0.40)^2+\dfrac{3}{7}(0.35)^2+\dfrac{3}{7}(0.30)^2

+114(0.25)2=1.497514+\dfrac{1}{14}(0.25)^2=\dfrac{1.4975}{14}

Var(T)=σ2=E(T2)(E(T))2Var(T)=\sigma^2=E(T^2)-(E(T))^2

=1.497514(0.325)20.001339=\dfrac{1.4975}{14}-(0.325)^2\approx0.001339

σ=σ20.0013390.0366\sigma=\sqrt{\sigma^2}\approx\sqrt{0.001339}\approx0.0366

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