Tricky probability question

xadoxic · **Joined:** Wed, 6 Jun 2012 09:36:32 UTC **Posts:** 1

Hey guys, im new to this community ! I have a problem with a probability question that does not provide much data -

A box may contain any combination of bottles of drinks (orange or apple) . All boxes were assumed to contain the same number of bottles of drinks and all combinations were equally probably. Sam bought a box of drinks and pulled two bottles out at random, one after another without replacement. What was the probability that the second bottle was orange, given that the first bottle was orange ?

Is there a possible way to get a numeric answer to this problem?

Thanks a ton

Denis · **Posted:** Wed, 6 Jun 2012 10:42:23 UTC

Don't think so; if 2n = number of bottles:
probability of 1st two being same = (n-1) / (2n-1)
So if 20 orange and 20 apple: 19/39.

torbye · **Joined:** Thu, 7 Jun 2012 08:01:59 UTC **Posts:** 2

There are 2^n ways of arranging n bottles of orange and applejuice. Half of them starts with a bottle of orange juice. The probability that Sams first bottle has orangejuice is 1/2. There are 2^(n-1) ways of arranging the remaining n-1 bottles and half of them starts with a bottle of orangejuice. The probability of the second bottle having orangejuice is then 1/2, and the total probability of pulling two bottles of orangejuice is 1/4

alstat · **Posted:** Thu, 7 Jun 2012 09:32:30 UTC

I think $P = \frac{1}{2}$ .

For every combination of bottles where the first two are orange, there is a combination that is identical, except that the second bottle is apple.

torbye · **Joined:** Thu, 7 Jun 2012 08:01:59 UTC **Posts:** 2

I agree. The probability of pulling two bottles of orange juice is 1/4. But the probability of pulling a second bottle of orange juice when you have already pulled anothe one is 1/2

Soroban · **Posted:** Thu, 7 Jun 2012 15:03:11 UTC

Hello, xadoxic!

Quote:

A box may contain any combination of bottles of drinks (orange or apple).
All boxes were assumed to contain the same number of bottles of drinks and all combinations were equally probable.
Sam bought a box of drinks and pulled two bottles out at random, one after another without replacement.
What was the probability that the second bottle was orange, given that the first bottle was orange?

Is there a possible way to get a numeric answer to this problem?

There is no way to get an exact answer, but we can construct a formula.

Suppose the boxes contain

bottles.

Let (x,y)

represent $\begin{Bmatrix} x &=& \text{no. of orange drinks} \\ y &=& \text{no. of apple drinks} \end{Bmatrix}$

Then there are n+1

possible contents:
. . $\begin{array}{ccccccccc}(n,0) & (n-1,1) & (n-2,2) & (n-3,3) & \hdots & (2,n-2) & (1,n-1) & (0,n) \end{array}$

One orange is drawn from the box.
The probability that the second bottle is also orange is:

. . $\begin{array}{cccccccccccccccc}\frac{n-1}{n-1}&& \frac{n-2}{n-1} && \frac{n-3}{n-1} && \hdots && \frac{2}{n-1} && 0 && 0 \end{array}$

$$\text{Therefore: }\:P(\text{2nd Or}\,|\,\text{1st Or}) \;=\;\frac{1}{n+1}\left[\frac{2}{n-1} + \frac{3}{n-1} + \frac{4}{n-1} + \cdots + \frac{n-2}{n-1} + \frac{n-1}{n-1}\right]$

. . . . . . . . . . . . . . . . . . . . . . . . $$=\;\frac{1}{(n+1)(n-1)}\big[2 + 3 + 4 + \hdots + (n-1)\big]$

. . . . . . . . . . . . . . . . . . . . . . . . $$=\;\frac{1}{(n+1)(n-1)}\cdot \frac{(n+1)(n-2)}{2}$

. . . . . . . . . . . . . . . . . . . . . . . . $$=\;\frac{n-2}{2(n-1)}$
And that's all we can do.

However, we note that: . $$\lim_{n\to\infty}\frac{n-2}{2(n-1)} \:=\:\frac{1}{2}$

alstat · **Posted:** Thu, 7 Jun 2012 19:40:47 UTC

But if an orange is drawn first, then (0,n)

was not a possibility.

Then you have n-1 bottles remaining, with the following

possible combinations:
(n-1,0), (n-2,1), (n-3,2),... (1,n-2), (0,n-1)

.

Then P(2nd = orange) = $\frac{1}{n}[\frac{n-1}{n-1} + \frac{n-2}{n-1} + ... + \frac{0}{n-1}]$
$= \frac{1}{n(n-1)}[(n-1) + (n-2) + ... + 1] = \frac{1}{n(n-1)}\cdot\frac{n(n-1)}{2}) = \frac{1}{2}$

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Tricky probability question

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