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Here is the question I am struggling with:

A box has 16 Balls, of which 8 are Green, 6 are Red, and 2 are Blue. If you draw 2 Balls with replacement, what is the probability of getting 1 Green Ball and 1 Blue Ball in no particular order?

I see three different ways to get an answer to this problem. Please refute my wrong answers with explanations because I am confused.

Method 1:

Probability of getting one green: 8/16

Probability of getting one blue: 2/16

(8/16) * (2/16) = 1/16 final answer

Method 2: Since the question said order does not matter, I still figured order does count into this equation so I approached it by finding the probability that the green ball is selected first, then the blue ball. Then add that probability to selecting the blue ball first, then the green ball.

Probability of getting green first then blue: (8/16) * (2/16) = 1/16

Probability of getting blue first then green: (2/16) * (8/16) = 1/16

therefore, the final answer is 1/16 + 1/16 = 1/8.

Note: This confuses me because we are double counting the answer, the problem said that order does not matter, but why doesn't the Method 1 take this into account?

Method 3 (Combination Method):

There are Comb(16,2) possible ways to select 2 balls out of 16

There are Comb(8,1)*Comb(2,1) ways to select a green and a blue ball

Probability of one green and one blue = Comb(8,1)*Comb(2,1)/Comb(16,2) = 16/120 = 2/15 final answer

Which one of these, if any, is the correct answer? The book says it is 1/8, but can someone please explain more and explain why my other methods are wrong. Thanks!

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  • $\begingroup$ You can use MathJax to format your posts. There are also other formatting tricks, like using > before text to put it in a block quote and surrounding text with ** to make it bold. $\endgroup$ – let's have a breakdown Nov 26 '17 at 18:44
  • $\begingroup$ Please, if you are ok, you can accept the answer and set it as solved. Thanks! $\endgroup$ – gimusi Jan 10 '18 at 10:04
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Another way of thinking of the problem is to actually write down the sample space and compute the probabilities. In this case, there are 9 possible outcomes: $$ \{ GG, GR, GB, RG, RR, RB, BG, BR, BB \}, $$ where, for example $RG$ denotes the event of first drawing a red ball then a green ball. There are two "favorable" events: $GB$ and $BG$. Since the two favorable outcomes are disjoint, and each draw is independent (thanks to the fact that we are replacing the balls), we have $$ P(GB \lor BG) = P(GB) + P(BG) = \frac{8}{16} \cdot \frac{2}{16} + \frac{2}{16}\cdot \frac{8}{16} = 2 \cdot \frac{1}{2}\cdot \frac{1}{8} = \frac{1}{8}. $$


EDIT: Upon reflection, it may also be useful to explain where you went wrong.

Method 1: Here you have computed the probability of first drawing a green ball, then drawing a blue ball. Order matters in this computation, but order does not matter in the answer. Thus you need to also work out the probability of first drawing a blue ball, then drawing a green ball, and add the two together.

Method 2: This approach is correct.

Method 3: Typically, combinations aren't the right tool when trying to work with problems that include replacement. In particular, your denominator $\binom{16}{2}$ represents the number of ways of drawing two balls without replacement. I don't think that there is a really a nice combination-y way of writing out this problem.

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    $\begingroup$ why does the combination function work without replacement? Is it from the formula itself where you use the factorial, say 16*15 instead of 16*16? $\endgroup$ – Sean Nov 26 '17 at 18:14
  • $\begingroup$ Don't rely on formulae, particularly if you don't really understand where they came from. If you have 16 distinct objects, and you want to select two of them while keeping track of order, there are $16\cdot 15$ ways of doing this (i.e. $P(16,2)$, i.e. permutations of 16 taken 2 at a time). If you don't care about the order, then you have to divide out by the number of ways of rearranging the two objects that you picked out. Since there are $2\cdot 1 = 2!$ ways of rearranging two objects, there are $P(16,2)/2! = \binom{16}{2}$ ways of choosing two objects from 16 (without replacement). $\endgroup$ – Xander Henderson Nov 26 '17 at 18:19
  • $\begingroup$ "why does the combination function work without replacement? Is it from the formula itself where you use the factorial, say 16*15 instead of 16*16?" Think it out. If you have replacement, why would you multiply by $15$? You are never choosing from 15 balls and you have nothing that has $15$ options. $\endgroup$ – fleablood Nov 26 '17 at 18:57
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with method one you should add the case you get firstly one blu and then one green, that's your method 2

the third method doesn't work because you can choose the same ball

the correct method by counting is:

  • n° of total cases = 16*16=256
  • n° of events: 8*2+2*8 = 32
  • probability =32/256 = 1/8
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  • $\begingroup$ Can you explain the part where I can use the same ball? $\endgroup$ – Sean Nov 26 '17 at 18:16
  • $\begingroup$ you have replacement, thus after you get a ball you put the one into the box and proceed with the second extraction $\endgroup$ – gimusi Nov 26 '17 at 18:19
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Method 1:

This is specifying order. The $\frac 8{16}$ applies to a specific drawing as does the $\frac 2{16}$. As blue first then green is as equally likely as green first than blue. The probability is $\frac{8}{16}*\frac{2}{16}*2$

Method 2:

This is correcct.

"Note: This confuses me because we are double counting the answer, the problem said that order does not matter, but why doesn't the Method 1 take this into account?"

You aren't double counting. If the blue ball is first then it is impossible that the green ball is first. The two options are mutually exclusive. That is why adding the probabilities is acceptable. If it were possible for both events we'd have to take conditional probability into account.

Method 3:

If you are allowed replacement, then choosing two balls is not choosing two out of sixteen. It is choosing 1 out of sixteen twice.

So there are $16^2$ ways to choose $2$ two balls.

There are $8*2$ ways to choose a green and then a blue ball. And $2*8$ ways to choose a blue and a black ball. WHich is the same as method 2.

Or if you want to be different. There are $10*10$ ways to choose two balls that are either blue or green. There are $8*8$ ways to choose two green balls and $2*2$ ways of choosing two blue balls and $10*10-8*8 - 2*2 = 100 - 64 - 4 = 32$ ways of choosing exactly one blue and one green ball.

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