I have $k_1$ red balls and $k_2$ blue balls in a box. Let $p_r$ be probability that randomly select $r$ balls in the box without replacement, $\sum_{r=1}^{m} p_r=1$. A trial is succeeds if the trial can draw at least one red ball in the trial. Find expected number of successful trail in $N$ times.

This is my solution

Let $X$ be event that draw $r$ ball in the box which has at least one red ball.

$$P(X)=p(r) \times\sum_{i=0}^{k_1}\frac{\binom{k_1}{i}\binom{k_2}{r-i}}{\binom{k_1+k_2}{n}}$$

Finally, we got the expected value as $$E(X)= N \times \sum_{r=1}^{m} p(r) \sum_{i=0}^{k_1}\frac{\binom{k_1}{i}\binom{k_2}{r-i}}{\binom{k_1+k_2}{n}}$$

Is it right?

  • $\begingroup$ What is $m$ in the sum? What exactly do you mean by $p_r$? Picking $r$ balls out of what? $\endgroup$ – BCLC Aug 23 '16 at 10:49
  • $\begingroup$ Thank for regarding my question. $m$ just a constant, and it is upper range of number selection, and given. The $p_r$ is used to determine the number of selection $r$ in each trial. Pick $r$ ball from $k_1+k_2$ balls in the box. For example, $p_r=[0.1, 0.5, 0.2, ...]$, $p_r(2)=0.5$ means prob. that we draw 2 ball is 0.5 $\endgroup$ – Jame Aug 23 '16 at 10:54
  • $\begingroup$ There are a few errors in your solution. For one, $X$ is an event, not a random variable, so what does $E(X)$ mean? For another, your sum starting with $i=0$ implies a success if you draw $0$ red balls, which isn't a success. $\endgroup$ – Tom Aug 23 '16 at 11:19
  • $\begingroup$ You are right. $r$ must from 1. $\endgroup$ – Jame Aug 23 '16 at 11:21
  • $\begingroup$ @user2938494 $r$ starting at $0$ is OK, since $r$ represents how many balls you draw and if $p_0 \neq 0$, then there is a probability of drawing no balls. However, your sum from $i=0$ to $k_1$ (the inner sum) is incorrect since here $i$ represents how many red balls you grab, right? $\endgroup$ – Tom Aug 23 '16 at 11:26

Let $Y$ be the random variable counting how many successes you have in $N$ trials. If $X_i$ is the Bernoulli random variable giving $1$ if the $i$th draw is a successs and $0$ otherwise, then $Y = \sum\limits_{i=1}^N X_i$.

Thus, $Y$ is a sum of presumably-independent Bernoulli random variables, hence a binomial random variable with probability of success $p = \mathbb{P}(X_i = 1)$. For example, the mass function is $$\mathbb{P}(Y = s) = \mathbb{P}\left(\sum\limits_{i=1}^N X_i = s\right) = {{N}\choose{s}} \mathbb{P}(X_i = 1)^s\, \mathbb{P}(X_i = 0)^{N-s} = {{N} \choose{s}} p^s \,(1-p)^{N-s}$$ for $0 \leq s \leq N.$

At this point, you probably know that the expected value of a Binomial random variable with parameters $N$ and $p$ is $\mathbb{E}[Y] = Np$, or you can go through the calculation yourself easily since the expected value is linear to find $$ \mathbb{E}[Y] = \sum\limits_{i=1}^N \mathbb{E}[X_i] = \sum\limits_{i=1}^N \Big[ 0 \cdot \mathbb{P}(X_i = 0) + 1 \cdot \mathbb{P}(X_i =1) \Big] = N \, \mathbb{P}(X_i = 1) $$ So, what you have left is to find $p = \mathbb{P}(X_i = 1)$.

For this, it is helpful to consider the following: $$ \mathbb{P}(X_i = 1) = \sum\limits_{r = 0}^m \mathbb{P}(X_i = 1 \mid r \text{ draws})\,p_r $$ with \begin{align*} \mathbb{P}(X_i =1 \mid r \text{ draws}) = \begin{cases} 1 & r > k_2 \\ 1 - \frac{{k_2} \choose {r}}{{k_1 + k_2} \choose {r}} & r \leq k_2 \end{cases} \end{align*} where the last case with $r \leq k_2$ used: $\mathbb{P}(X_i = 1 \mid r) = 1 - \mathbb{P}(X_i = 0 \mid r)$. From here, you can put these pieces together.

  • $\begingroup$ Is range of $r$ from $0$ or $1$ in the term $P(X_i=1)$? $\endgroup$ – Jame Aug 23 '16 at 11:24
  • $\begingroup$ $r$ starting from $0$ is ok. Looking at the expression, you will get $$\mathbb{P}(X_i = 1 \mid 0 \text{ draws}) = 1 - \frac{{k_2}\choose{0}}{{k_1+k_2}\choose{0}} = 1 - 1 = 0$$ which is correct (you can't succeed if you don't try!). $\endgroup$ – Tom Aug 23 '16 at 11:28
  • $\begingroup$ Sorry Tom. I edited my question. $r$ must be from 1, it means each trial I, the number of drawn ball must be bigger than 1 $\endgroup$ – Jame Aug 23 '16 at 11:30
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    $\begingroup$ @user2938494 Since $\mathbb{P}(X_i = 1 \mid r =0) = 0$ (as I just remarked above), the answer is still valid since the $r=0$ case doesn't not contribute to the overall probability. $\endgroup$ – Tom Aug 23 '16 at 11:31
  • $\begingroup$ Yes. Thank you. I am reading and understanding your answer. It is so nice $\endgroup$ – Jame Aug 23 '16 at 11:31

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