0
$\begingroup$

An urn contains $10$ white balls, marked by the numbers $ 1, 2, . . . , 10$, and $3$ blue balls, marked by $11, 12, 13$. $5$ balls are drawn randomly

a) With replacement.

b) Without replacement.

Find the distribution of the minimum of the numbers marked on the balls.

I started a) by marking $\{x\}_{i=1}^5$ the result received at stage $i$.

As $ x= \min{\{x\}_{i=1}^5}$ and $k$ is the minimum number drawn, we could say that:

$P(x\le k) = P(\min{\{x\}_{i=1}^5} \ge k) =* $

  • Because $ {\{x\}_{i=1}^5} \ge \min{\{x\}_{i=1}^5} \ge k $ then $ {\{x\}_{i=1}^5} \ge k $

= $ P((x_1\ge k) \cap (x_2\ge k) \cap...\cap (x_5\ge k)) = *$

  • The Balls are independent

= $ \prod_1^5P(x_i \ge k) = * $

  • $ P(x_i \ge k)= \frac{13-k}{13}$

$ (1- \frac{k}{13})^5. $

Am I on the right path?

Thanks in advance.

2 Answers 2

0

Clearly, the ball color is irrelevant as the problem is posed. So let's look at the general case of draws from a discrete uniform distribution on the set $\{1, 2, \ldots, m\}$. In this case, when draws are with replacement, the minimum number drawn is given by the order statistic $$X_{(1)} = \min(X_1, X_2, \ldots, X_k)$$ where each $X_i \sim \operatorname{DiscreteUniform}(1,m)$. Then $$\Pr[X_{(1)} \ge x] = \prod_{i=1}^k \Pr[X_i \ge x] = \left(\frac{m-x+1}{m}\right)^k,$$ hence $$\Pr[X_{(1)} = x] = \Pr[X_{(1)} \ge x] - \Pr[X_{(1)} \ge x+1] = \frac{(m-x+1)^k-(m-x)^k}{m^k}, \quad x=1, 2, \ldots, m.$$

In the case without replacement, let $Y_k$ represent the minimum value observed from a sample of size $k$, with $1 \le k \le m$. Clearly, there are $\binom{m}{k}$ possible distinct (irrespective of order) $k$-subsets that could be observed. Suppose a randomly chosen $k$-subset has the form $$S_k = \{s_1, s_2, \ldots, s_k\}, \quad 1 \le s_1 < s_2 \ldots < s_k \le m.$$ Then $Y_k = s_1$ by construction. How many subsets satisfy $s_1 = y$? There is only one way to choose the first element. Of the rest, there are $\binom{m-y-1}{k-1}$ ways to choose the others; therefore, $$\Pr[Y_k = y] = \frac{\binom{m-y}{k-1}}{\binom{m}{k}}, \quad y = 1, 2, \ldots, m-k+1.$$

0

Most of what you have in part (a) looks correct; however, I have a couple of points to correct. If $x=\min{X_i}$, then $\mathbb{P}(x\geq k)=\prod\mathbb{P}(X_i\geq k)$. Additionally, $\mathbb{P}(X_i\geq k)=1-\frac{k-1}{13}=\frac{14-k}{13}$, so $\mathbb{P}(x\geq k)=\left(\frac{14-k}{13}\right)^5$.

Do you have any thoughts on (b)? I have a few helpful computations (maybe more than helpful). I start by computing the highest possible value of $x=9$. The value $\binom{13}{5}=1287$ is useful here. \begin{align*} \mathbb{P}(x=9)&=\frac{1}{1287}\\ \mathbb{P}(x=8)&=\frac{1}{1287}\cdot\binom{5}{4}\\ \mathbb{P}(x=7)&=\frac{1}{1287}\cdot\binom{6}{4}\\ \mathbb{P}(x=6)&=\frac{1}{1287}\cdot\binom{7}{4}... \end{align*}

If you continue in this pattern, you will have all of the outcomes! The binomial terms come from fixing the minimum value and allowing four free choices larger than the minimum.