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A collection of tickets comes in four colors: red, blue, white, and green. There are twice as many reds as blues, equal numbers of blues and whites, and three times as many green as whites. I choose 5 tickets at random with replacement. Let $X$ be the number of different colors that appear.

a) Find a numerical expression for $P(X\ge 4)$.

b) Find a numercial expression for $E(X)$.

My understanding of this problem is that $P(X\ge 4)$ is the same as $P(X=4)$ since there are only four colors to choose so that X cannot be 5. Then it becomes a hypergeometric distribution. But the answer on the solution manual is pretty scary which is $$\dfrac{5!}{1!1!1!2!}[2(1/7)^3(2/7)(3/7)+(1/7)^2(2/7)^2(3/7)+(1/7)^2(2/7)(3/7)^2] $$

and $E(X)$ is $$ 2[1-(6/7)^5]+1-(5/7)^5 + 1- (4/7)^5$$ Could someone explain a little bit for me? I am so confused.

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2 Answers 2

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Here’s an extended explanation for the first part.

Let $n$ be the number of blue tickets; then there are $n$ white tickets, $2n$ red tickets, and $3n$ green tickets, for a total of $7n$ tickets. Thus, $1/7$ of the tickets are blue, $1/7$ are white, $2/7$ are red, and $3/7$ are green. This means that the probability of drawing a blue ticket is $1/7$, as is the probability of drawing a white ticket, while the probability of drawing a red ticket is $2/7$, and that of drawing a green ticket is $3/7$.

As you say, the probability of getting at least $4$ colors is the same as the probability of getting exactly $4$ colors. Since we’re drawing $5$ tickets, this happens only when we draw two tickets of one color and one of each of the other three colors. This can happen in four ways, one for each of the colors that can be doubled. Let’s look at just one of these ways to start with: what’s the probability of drawing two red tickets and one of each of the other three colors?

The tickets could come out in lots of different orders; some of the possibilities are RWRBG, WBRRG, and GWRBR. It’s not hard to calculate the probability of any one of these results: for instance, the probability of the drawing RWRBG is $$\left(\frac27\right)\left(\frac17\right)\left(\frac27\right)\left(\frac17\right)\left(\frac37\right)=\left(\frac17\right)^2\left(\frac27\right)^2\left(\frac37\right).$$ In fact, every order of draws that results in two reds and one of each of the other colors has this same probability, $(1/7)^2(2/7)^2(3/7)$. To find the total probability of getting two reds and one of each of the other colors, we need to know in how many different orders such tickets can be drawn.

In the string of $5$ draws there are $\binom52$ different places in which the two reds could appear. Once we know where they are, the blue can appear in any of the $3$ remaining places, the white in either of the $2$ that remain after that, and at that point there’s no choice for the green. This gives us a total of $$\binom52 \cdot 3\cdot 2 = \frac{5!}{2!3!}\cdot 3\cdot 2 = \frac{5!}{2}=\frac{5!}{1!1!1!2!}$$ orders, and hence a total probability of $$\frac{5!}{1!1!1!2!}\left(\frac17\right)^2\left(\frac27\right)^2\left(\frac37\right)$$ of drawing two reds and one of each of the other colors. This is the middle term of that scary answer (after you multiply through to get rid of the square brackets).

The other two terms are found similarly. The third term corresponds to drawing two greens (hence the $(3/7)^2$ and one of each of the other three colors. The first term actually combines two terms: the probability of drawing two blues and one of each of the other colors is $(1/7)^2(1/7)(2/7)(3/7)=(1/7)^3(2/7)(3/7)$, and so is the probability of drawing two whites and one of each of the other two colors, so they’ve been combined into a single term with an extra factor of $2$. The number of possible orders is the same in each case, which is why the factor of ${5!\choose 1!1!1!2!}$ can be pulled out of all the terms.

For the second part, notice that the probability of not drawing a blue ticket is $6/7$. Thus, $(6/7)^5$ is the probability of drawing no blue tickets in $5$ draws, and $1-(6/7)^5$ is the probability of drawing at least one blue ticket in $5$ draws. $1-(6/7)^5$ is also the probability of drawing at least one white ticket in $5$ draws. By similar reasoning, $1-(5/7)^5$ is the probability of drawing at least one red ticket in $5$ draws, and $1-(4/7)^5$ is the probability of drawing at least one green ticket in $5$ draws. Let $D_b$ be the event of drawing at least one blue ticket, $D_w$ that of drawing at least one white ticket, $D_r$ that of drawing at least one red ticket, and $D_g$ that of drawing at least one green ticket; we’ve just calculated the probabilities of these four events.

The probability of $D_b$ is $1-(6/7)^5$, so in the long run $D_b$ occurs in that fraction of all $5$-ticket draws. Similarly, $D_r$ occurs in the fraction $1-(5/7)^5$ of all $5$-ticket draws. If I make $n$ $5$-ticket draws, I expect $D_b$ and $D_w$ each to occur about $(1-(6/7)^5)n$ times, $D_r$ to occur about $(1-(5/7)^5)n$ times, and $D_g$ to occur about $(1-(4/7)^5)n$ times. And the total number of occurrences of any of these four events in the whole set of $n$ $5$-ticket draws will be about $$\begin{align*} &\left(1-\left(\frac67\right)^5\right)n+\left(1-\left(\frac67\right)^5\right)n+\left(1-\left(\frac57\right)^5\right)n+\left(1-\left(\frac47\right)^5\right)n\\ &=n\left(1-\left(\frac67\right)^5+1-\left(\frac67\right)^5+1-\left(\frac57\right)^5+1-\left(\frac47\right)^5\right). \end{align*}$$

Now $X$, the number of different colors drawn, is simply the number of these four events that actually occur: if three of them occur, we’ve drawn tickets of three different colors, and so on. If on average we get $$n\left(1-\left(\frac67\right)^5+1-\left(\frac67\right)^5+1-\left(\frac57\right)^5+1-\left(\frac47\right)^5\right)$$ of them in $n$ $5$-ticket draws, on average we get $(1/n)$-th of this, or $$1-\left(\frac67\right)^5+1-\left(\frac67\right)^5+1-\left(\frac57\right)^5+1-\left(\frac47\right)^5$$ of them, in one $5$-ticket draw. In other words, $E(X)$, the expected number of colors, is $$1-\left(\frac67\right)^5+1-\left(\frac67\right)^5+1-\left(\frac57\right)^5+1-\left(\frac47\right)^5,$$ which (after combining the first two terms) is the expression given in the solution manual.

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The formula for $\mathrm E(X)$ is explained here.

The formula for $\mathrm P(X=4)$ goes as follows. One wants to draw a color $i$ twice and each of the three colors $j\ne i$ once. The prefactor with the factorials is the number of ways to choose the positions of the two draws which produced color $i$ and of each of the three unique draws which produced each color $j\ne i$. For each given set of positions, the probability to get exactly the desired result is $p_i^2$ times the product of the $p_j$ for $j\ne i$. It happens that two terms are equal in the resulting sum inside the brackets because two probabilities $p_i$ and $p_j$ are equal but in general the formula is $$ \mathrm P(X=4)={5\choose 2,1,1,1}\sum\limits_{i=1}^4p_iq,\qquad\text{where}\ q=\prod\limits_{i=1}^4p_i. $$ In your case, the probabilities $p_i$ are $\frac17$, $\frac17$, $\frac27$, and $\frac37$.

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