I have a question about probability that need your help. I assume that I have balls that are numbered from 1 to 100. The probability selection each ball is followed uniform distribution. I divide balls into two classes as the figure enter image description here

The selection of each class follows the distribution $P_{class1}=\frac {i-1}{100}$ and $P_{class 2}=1-\frac {i-1}{100}$. I create 2 windows: first window contains class 1. And second window contains class 1 and class 2. I make a rule to select balls in each window. These balls are choose randomly from:

  1. Window 1
  2. Window 2

For example, if I select window 1, then only balls from 1 to $i-1$ are selected. If I select second option (class 1+ class2), then balls from 1 to 100 are selected.

Number of selected balls is followed the special distribution as


For example $x^2=0.5$ that mean prob. select 2 balls is $50\%$.

Assume that I choose window 2, then What is probability selection these balls (follow $\Omega$) that index from $i^{th}$ ball to last ball (that means only choose balls in class 2)?

enter image description here

  • 1
    $\begingroup$ To clarify: You draw one ball with probability $0.1$, two balls with probability $0.5$, three balls with probability $0.3$, and ten balls with probability $0.1$. These balls are drawn from the entire set of $100$ balls. First question: You want the probability that all of the balls you draw happen to be from the second class (from $i$ through $100$), is that right? And second question: Are the balls selected with replacement (that is, you draw one ball, record its number, put it back, draw the second ball, record its number, etc.), or without replacement (draw all balls at once)? $\endgroup$ – Brian Tung May 14 '15 at 18:43
  • $\begingroup$ First question: Right, Second question: Without replacement. $\endgroup$ – user3051460 May 14 '15 at 18:44

With replacement, the problem is rather nice: The probability of any given ball being in class 2 is $1-(i-1)/100 = (101-i)/100$, so the probability of all of $k$ balls being in class 2 is $[(100-i)/100]^k$, and the overall probability would just be given by

$$ \Omega((101-i)/100) $$

The fact that they are drawn without replacement makes the analysis somewhat messier. We can apply the same reasoning, but now the distribution is hypergeometric, and the probability that all of $k$ balls is in class 2 is given by

$$ p_k = \frac{\binom{101-i}{k}}{\binom{100}{k}} $$

and then the probability we want is

$$ 0.1p_1 + 0.5p_2 + 0.3p_3 + 0.1p_{10} $$


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