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Please read carefully how I solved this question:

A bag contains 4 white 2 black 3 red balls. A ball is drawn from the bag 1-by-1.
(i) Find the probability that $5^{th}$ ball is red. Consider both with and without replacement.
(ii) 3^rd and 7^th are red. Find the probability that $9^{th}$ ball is black. Consider both with and without replacement.

(i)

(a) with replacement. $$P=\frac39=\frac13$$

(b) without replacement. Let's make an array of 9 elements where i^th element denote which ball is taken out in i^th turn, then we can have red in 5^th position with probability:$$P=\frac{\binom31\binom82\binom622!2!4!}{9!}=\frac13$$ [select one red ball to put in 5^th position, then select positions on array for black and remaining red then arrange them divided by total arrangements]

(ii)

(a) with replacement. $$P=\frac29$$

(b) without replacement. Let's make an array of 9 elements where i^th element denote which ball is taken out in i^th turn, then we have red in 3rd and 7th position red,we can have black in 9^th position with probability:$$P=\frac{\binom32\binom212!6!2!}{\binom322!7!}=\frac27$$ [select two red balls for 3 and 7 position then select one black ball for last position and then select position for red and remaining black balls on array then arrange]

Why did drawing with or without replacement make no difference in the first and second can be obtained by considering 3^rd and 7^th balls were never there?

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  • $\begingroup$ Not sure why you said we must read your solution so very carefully. But do see if my answer addresses your real question. =) $\endgroup$ – user21820 Nov 29 '14 at 5:54
  • $\begingroup$ @user21820 it would take me some time, thanks $\endgroup$ – RE60K Nov 29 '14 at 5:56
  • $\begingroup$ As turkeyhundt pointed out, the probability for (ii) with replacement is 2/9, not 2/7. It is 2/7 without replacement. $\endgroup$ – user21820 Nov 29 '14 at 6:54
  • $\begingroup$ @user21820 edited. $\endgroup$ – RE60K Nov 29 '14 at 8:14
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We can consider each configuration for without replacement as a sequence of balls. Then permuting the sequence in any way gives another configuration with equal probability because every sequence has equal probability. Each permutation can be used to give a bijection (one-to-one correspondence) between a set of configurations with another that differ simply by that permutation.

For example, in the first question you want #(sequences with 5th ball red)/#(sequences), and #(sequences with 5th ball red) = #(sequences with 1st ball red) by the bijection that switches the 1st and 5th ball in a sequence. Check that every sequence with 5th ball red corresponds to exactly one sequence with 1st ball red, and vice versa. This gives the answer since #(sequences with 1st ball red)/#(sequences) = 3/9 because there are no restrictions on the later balls.

For the second question, to do it rigorously we choose some permutation that sends the 3rd ball to position 1, the 7th ball to position 2 and the 9th ball to position 3. Then we get #(sequences with 3rd and 7th ball red and 9th ball black) = #(sequences with 1st and 2nd ball red and 3rd ball black), and #(sequences with 3rd and 7th ball red) = #(sequences with 1st and 2nd ball red). So the answer is #(sequences with 1st and 2nd ball red and 3rd ball black)/#(sequences with 1st and 2nd ball red) = 2/7 since the first two balls already use up two red and after drawing the first two we are left with 7 balls of which 2 are black.

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  • $\begingroup$ i can't understand "If we switch the 1st and k-th ball in the sequence, it is in both your questions a bijection between valid configurations." $\endgroup$ – RE60K Nov 29 '14 at 6:08
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    $\begingroup$ I should have been more precise. In your first question, you are interested in the 5th ball. Given any sequence drawn, we can switch the 1st and 5th ball. That results in another possible drawn sequence, and this switching process is a bijection between the set of drawn sequences with 5th ball of interest and the set of drawn sequences with 1st ball of interest. Similarly for the second question, but we have to exclude the used balls. I can expand my answer if it's not clear. $\endgroup$ – user21820 Nov 29 '14 at 6:21
  • $\begingroup$ you are very close to explaining me i understand exchnage but what do you mean by bijection, bijective functions which are invertible? $\endgroup$ – RE60K Nov 29 '14 at 6:27
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    $\begingroup$ A bijection is a exact correspondence between two sets, meaning that each element of one set corresponds to exactly one element of the other set. I'll expand my answer to make it more rigorous since it's not so trivial for your second question. $\endgroup$ – user21820 Nov 29 '14 at 6:32
  • $\begingroup$ @eaxdpiotnyeantial: See if the above makes sense now. It should be as rigorous as possible without clutter. Note the important points are that every sequence is equally likely, which you need to prove separately as a lemma, and that a permutation is really a bijection as claimed, which is intuitive but should also be given a proof. $\endgroup$ – user21820 Nov 29 '14 at 6:45
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Without replacement is a word. So consider strings of the form $\{R, B, G\}^{*}$. That is, strings consisting only of letters $R$, $B$, and $G$. If we do replace, there are $3^{n}$ strings of length $n$ over this alphabet. That is, we have an inexhaustable supply of each character.

If we do not replace, we have $3!$ strings of length $3$ (permute the letters), $3!$ strings of length $2$, and $3$ strings of length $1$. We cannot form a string of length $4$.

So consider:

A bag contains 4 white 2 black 3 red balls. A ball is drawn from the bag 1-by-1. (i) Find the probability that 5th ball is red.Consider both w and w/o replacement.

If we do replace, the odds of getting a red ball on the $nth$ character are the same as for the first character. So the red ball on $5$ is $\frac{3}{9}$.

Combinatorially, if we have a string of length $n \geq 5$, we know a character will be drawn for each slot. Let's label each ball to make the visualization a little easier, so we have $9$ characters. We are interested in the 5th being one of $r_{1}, r_{2}, r_{3}$. So there are $9^{n-1} * 3$ such strings with red on character $5$, and $9^{n}$ such strings of length $n$. Do you see how I got that combinatorially based on my explanation before your problem?

Now if we do not replace, the explanation you described in your solution is spot on. This is a permutation problem, not a word problem.

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  • $\begingroup$ what do you mean by "word problem"? $\endgroup$ – RE60K Nov 29 '14 at 6:41
  • $\begingroup$ A "word" problem is where we have an alphabet like $\{R, G, B\}$ and we construct strings or "words" over that alphabet. $\endgroup$ – ml0105 Nov 29 '14 at 6:43
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Consider this simpler problem, conceptually, as an example.

What if you have a blue, green, red, yellow, purple and black marble. What if you draw 3 marbles without replacement. What is the probability the 3rd one is black? Well, no marble has an inherent advantage over the others, so intuitively, it must be a one sixth chance.

Now, do the same problem, but WITH replacement. Again, no marble has an inherent advantage or disadvantage to being chosen, so there must be a $\frac{1}{6}$ chance of the 3rd one being black.

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  • $\begingroup$ this explains the first part very well, could you expand your answer further to include second part as well? $\endgroup$ – RE60K Nov 29 '14 at 6:40
  • $\begingroup$ Hmmm, let me take a look. I'm not convinced WITH REPLACEMENT, that the probability would be $\frac{2}{7}$. It seems like it would be $\frac{2}{9}$ if you are replacing your ball each time. Am I reading the problem correctly? $\endgroup$ – turkeyhundt Nov 29 '14 at 6:46
  • $\begingroup$ Also, one thing that might help on the WITHOUT REPLACEMENT is, instead of imagining them jumbled up in a bag, imagine them hidden, but queued up in a pipe or half pipe and they are taken from the end one at a time. Once they are shuffled, the outcome is set. So regardless of what is chosen ahead of it, each ball's fate is already sealed. $\endgroup$ – turkeyhundt Nov 29 '14 at 6:51
  • $\begingroup$ You're right. I didn't notice that with replacement the known ball colours become irrelevant. Then I guess the real question is why we can sort of see what happens if we have taken out the known balls. Your idea is indeed leads to the method I outline in my answer. $\endgroup$ – user21820 Nov 29 '14 at 6:52

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