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Queation:

A box contains 100 balls, of which r are red. Suppose that the balls are drawn from the box one at a time, at random, without replacement. Determine

(a) the probability that the first ball drawn will be red; (b)the probability that the second ball drawn will be red;(c)the probability that the 50th ball drawn will be red; and (d)the probability that the last ball drawn will be red.

Answer:

I know that the answer for each of $(a),(b),(c)$ and $(d)$ is $\frac{r}{100}$.

For (a), it is trivial.

For (b), $Pr$(the second ball drawn will be red)$=Pr$(the first ball drawn is red and second is also red)$+Pr$(the first ball drawn is not red but second is red)$$=\bigg(\frac{r}{100}\frac{r-1}{99}\bigg)+\bigg(\frac{100-r}{100}\frac{r}{99}\bigg)=\bigg(\frac{r}{100}\bigg)$$

But I don't know how to generalize the result.That is how to extend the result for $50$ and $100$ balls as asked in (c) and (d).

Please help.thank you.

Silent
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  • No reason at all to think that the probability of n-th ball to be drawn not the same as of the first ball – drhab Sep 29 '13 at 10:42

3 Answers3

6

Denote $N=100$ (number of balls), $K=r$ (number of red balls);

$P_{N,K}(n,k) = \Pr$ ($k$ balls are red, when $n$ draws are made), where $0\le k \le n$.

Looking at Hypergeometric distribution, one can see that

$$ P_{N,K}(n,k) = \dfrac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}. $$

Then the probability that the $(n+1)$th ball (where $n=1,2,...,N-1$) drawn will be red, is: $$ P_{n+1} = \sum\limits_{k=0}^{n} P_{N,K}(n,k)\cdot \dfrac{K-k}{N-n} $$ $$ = \sum\limits_{k=0}^{n} \dfrac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}\cdot \dfrac{K-k}{N-n} $$ $$ = \sum\limits_{k=0}^{n} \frac{ \frac{K\cdot(K-1)!}{k!(K-1-k)!} \binom{N-K}{n-k}}{ \frac{N\cdot(N-1)!}{n!(N-1-n)!}} $$ $$ =\dfrac{K}{N} \sum\limits_{k=0}^{n} \dfrac{ \binom{K-1}{k} \binom{N-K}{n-k}}{\binom{N-1}{n}} =\dfrac{K}{N} \sum\limits_{k=0}^{n} P_{N-1,K-1}(n,k)= \dfrac{K}{N} \cdot 1 = \dfrac{K}{N}, $$ where $$\sum\limits_{k=0}^{n} \dfrac{ \binom{K-1}{k} \binom{N-K}{n-k}}{\binom{N-1}{n}}=1$$ follows from Vandermonde's identity. So, $P_{2}=P_{50} = P_{100} = \dfrac{K}{N} = \dfrac{r}{100}$.

Silent
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Oleg567
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4

Consider all the possible results of drawing the balls. Each result can be represented by a binary number having 1 when we draw a red ball, so that the number of 1s in the number is $r$. Your question is then equivalent to asking what is the probability of 1 occurring at any particular place of the number and this is easily seen to be $r/100$ independently of the position.

Marek
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1

-Since the value of r is left arbitrary, we can assume its value is updated in the event that it has been chosen. Therefore the probability for all three parts are the same: r/100

-The work is supported by the answer (not directly) above

Marlon
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