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An urn contains six balls of each of the three colors: red, blue, green. Find the expected number of different colors obtained when three balls are drawn:

a. with replacement;

b. without replacement.

The correct answers are:

a: 19/9

b: 2.1912

I just don't know how to get to get to the answers.

Thanks for any help

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We can also use recurrences, where A,B,C are the number of balls of three colors:

a) \begin{align*} f(a,b,c) = \left\{\begin{matrix} 3-\left(\lfloor\frac{a}{A}\rfloor + \lfloor\frac{b}{B}\rfloor+\lfloor\frac{c}{C}\rfloor\right)& \text{if }A+B+C-(a+b+c) = 3\\ \dfrac{a\cdot f(a-1,b,c)+b\cdot f(a,b-1,c)+c\cdot f(a,b,c-1)}{a+b+c} & \text{otherwise} \end{matrix}\right. \end{align*}

b) \begin{align*} f(a,b,c) = \left\{\begin{matrix} 3-\left(\lfloor\frac{a}{A}\rfloor + \lfloor\frac{b}{B}\rfloor+\lfloor\frac{c}{C}\rfloor\right)& \text{if }A+B+C-(a+b+c) = 3\\ \dfrac{A\cdot f(a-1,b,c)+B\cdot f(a,b-1,c)+C\cdot f(a,b,c-1)}{A+B+C} & \text{otherwise} \end{matrix}\right. \end{align*}

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The count, $X,Y$, of different colours drawn with and without replacement (respectively) will have the support: $\{1, 2, 3\}$.   Find the probability of each, for the two experiments, then use:

$$\begin{align}\mathsf E(X) & = \sum_{k=1}^3 k\;\mathsf P(X{=}k) \\[1ex] & = \mathsf P(X{=}1)+2\,\mathsf P(X{=}2)+3\,\mathsf P(X{=}3) \\[1ex] & = 2 -\mathsf P(X=1)+\mathsf P(X=3) \\[2ex]\mathsf E(Y) & = 2 -\mathsf P(Y=1)+\mathsf P(Y=3) \end{align}$$

$\mathsf P(X=1), \mathsf P(Y=1)$ are the probabilities that all three balls are the same colour.

$\mathsf P(X=3), \mathsf P(Y=3)$ are the probabilities that all three balls are of distinct colours.

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Let us take a card deck that contains only three colors : spade, diamond and heart, and suits from 1 to 6.

1) Define a poker hand as a hand of three cards. There are several types of hands, full color, pair and tricolor. Let count the hands without replacement :

tricolor : $ {6 \choose 1} {6 \choose 1} {6 \choose 1} = 216 $

pair : $ {3 \choose 1} {2 \choose 1} {6 \choose 1} {6 \choose 2} = 540 $

full color : $ {3 \choose 1} {6 \choose 3}= 60 $

their sum is $ {18 \choose 3} = 816 $ , the average # of colors is ${ 3×216 + 2×540 + 1×60 \over 816 }\approx 2.1912 $

2) When replacing, the G.F. is $(x+y+z)^3$

The needed coefficients satisfy 6 + 18 + 3 = 27;

The average # of colors is ${ 3×6 + 2×18 + 1×3 \over 27 } = {19 \over 9} $

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