Please help with a probability question (1 Viewer)

Elise8842 · Feb 4, 2015

Hi, could someone please explain the second step of the solution to me? [emoji24][emoji24]
I can't get why is it 2n-mCn , because C is for combination isn't it? For box A to be drawn and found empty, box A must have be drawn n times and box B be drawn (n-m) times, so it's only this ONE combination that satisfied the statement, I don't get what's the significance/meaning of 2n-mCn

Your help will be much appreciated [emoji38]

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InteGrand · Feb 4, 2015

Isn't this saying that, in the first $2n-m$ draws, Box A is drawn $n$ times, and Box B is drawn $n-m$ times, and we are given that on the next one, Box A is drawn? (Or we could interchange A and B, which is why we multiply by 2 at the end.)

Then the probability of having $n$ A's and $n-m$ B's in the first $2n-m$ draws is just $\binom{2n-m}{n}\cdot \left (\frac{1}{2}\right )^n\cdot\left ( \frac{1}{2} \right )^{n-m}=\frac{\binom{2n-m}{n}}{2^{2n-m}}$ (this is just a Bernoulli trial).

Multiplying by 2 should give the overall probability as being $\frac{2\binom{2n-m}{n}}{2^{2n-m}}=\frac{\binom{2n-m}{n}}{2^{2n-m-1}}$ , which is double the answer in your picture.

InteGrand · Feb 4, 2015

The step 1 in your picture seems to assume that the $(2n-m+1)^{\text{th}}$ can be any of A or B, but it can only be one of them, depending on which one we're saying is the one that will be empty.

InteGrand · Feb 4, 2015

(I'm assuming the question is saying "Given the man notices the box he has just picked is empty, what is the probability that m matches are left in the other box now?")

funnytomato · Feb 4, 2015

As you said,"box A must have be drawn n times and box B be drawn (n-m) times" before we draw from box A again and realise it's empty.

Then $\binom{2n-m}{n} =\frac{(2n-m)!}{n! \times(n-m)!}$ is the number of ways you can arrange these (2n-m) items [ with n A's and (n-m) B's. ]
It's like those kind of questions where they ask things such as : In how many ways can you arrange 10 textbooks given 6 of them are (identical) maths books and 4 are chemistry books.

And each arrangement represents a different branch on the tree diagram.

funnytomato · Feb 4, 2015

Consider the case where A and B both have 2 matches in them. So n=2.

And we want to find the number of ways such that the box B has 1 left(i.e. m=1 ) when A is found to be empty.
Then before we open A and find it's empty(taking a match for the (2n+1-m)= 4th time), we've drawn 2 matches from A and 1 from B.

So there are $\binom{2n-m}{n} =\frac{(2n-m)!}{n! \times(n-m)!}=\frac{3!}{2! \times 1!}=3$ ways to do this.
Namely AAB, ABA and BAA.

funnytomato · Feb 4, 2015

InteGrand said:
Isn't this saying that, in the first $2n-m$ draws, Box A is drawn $n$ times, and Box B is drawn $n-m$ times, and we are given that on the next one, Box A is drawn? (Or we could interchange A and B, which is why we multiply by 2 at the end.)

Then the probability of having $n$ A's and $n-m$ B's in the first $2n-m$ draws is just $\binom{2n-m}{n}\cdot \left (\frac{1}{2}\right )^n\cdot\left ( \frac{1}{2} \right )^{n-m}=\frac{\binom{2n-m}{n}}{2^{2n-m}}$ (this is just a Bernoulli trial).

Multiplying by 2 should give the overall probability as being $\frac{2\binom{2n-m}{n}}{2^{2n-m}}=\frac{\binom{2n-m}{n}}{2^{2n-m-1}}$ , which is double the answer in your picture.

When you compute the probability for getting n A's and (n-m) B's, the next draw could be either from A or B.
Say $P =\frac{\binom{2n-m}{n}}{2^{2n-m}}$ , then for this case where we need an A on the next draw, we get $\frac{P}{2}$

And similarly for the case where we find B empty , we get $\frac{P}{2}$

So P(one of the boxes is empty)=P(A empty)+P(B empty)= P, which is in agreement with OP's handwritten notes.

InteGrand · Feb 4, 2015

funnytomato said:
When you compute the probability for getting n A's and (n-m) B's, the next draw could be either from A or B.
Say $P =\frac{\binom{2n-m}{n}}{2^{2n-m}}$ , then for this case where we need an A on the next draw, we get $\frac{P}{2}$

And similarly for the case where we find B empty , we get $\frac{P}{2}$

So P(one of the boxes is empty)=P(A empty)+P(B empty)= P, which is in agreement with OP's handwritten notes.

I interpreted the question as we are given that he opened one to find it was empty. So in this case, we don't "need" A on the next draw, because it is given. This is why my answer differed by a factor of 2 from the one in the picture. (The question's wording was ambiguous.)

funnytomato · Feb 4, 2015

InteGrand said:
I interpreted the question as we are given that he opened one to find it was empty. So in this case, we don't "need" A on the next draw, because it is given. This is why my answer differed by a factor of 2 from the one in the picture. (The question's wording was ambiguous.)

Or we could interchange A and B, which is why we multiply by 2 at the end.

self contradictory ?

Just a quick check:

if you sum up the probabilities for m=0,1,2,....,n , shouldn't they add up to 1?

InteGrand · Feb 4, 2015

funnytomato said:
self contradictory ?

Just a quick check:

if you sum up the probabilities for m=0,1,2,....,n , shouldn't they add up to 1?

Ah right, my bad, if I assume it is given that the $(2n-m+1)^{\text{th}}$ box is empty, the probability that there are m remaining from the other box is just $\binom{2n-m}{n}\cdot\frac{1}{2^{2n-m}}$ , and multiplying by 2 just doubles up on cases in this assumption.

FrankXie · Feb 4, 2015

In the case of n=m=1, which is very easy, do u guys think what the probability asked in the question should be? 1 or 1/2?

InteGrand · Feb 4, 2015

FrankXie said:
In the case of n=m=1, which is very easy, do u guys think what the probability asked in the question should be? 1 or 1/2?

But the initial question seems dodgy. It seems that when he picks out a match and empties the matchbox, he doesn't know it's empty until he picks that matchbox again (unrealistic).

So in the case where n = 1, he could pick one matchbox (and so he doesn't know it's empty yet), then pick the other one (so both have 0 matches, but he doesn't know either is empty, unrealistically), which means on the next one, he'll know the matchbox is empty, and we will have m = 0 in this case. So the probability in your above post may not be 1.

This is all based on the man not knowing a box is empty until picking it after it has become empty.

Elise8842 · Feb 8, 2015

Thank you soooooo much!!!!!!!!!!! I finally got this XD

Elise8842 · Feb 8, 2015

InteGrand said:
Isn't this saying that, in the first $2n-m$ draws, Box A is drawn $n$ times, and Box B is drawn $n-m$ times, and we are given that on the next one, Box A is drawn? (Or we could interchange A and B, which is why we multiply by 2 at the end.)

Then the probability of having $n$ A's and $n-m$ B's in the first $2n-m$ draws is just $\binom{2n-m}{n}\cdot \left (\frac{1}{2}\right )^n\cdot\left ( \frac{1}{2} \right )^{n-m}=\frac{\binom{2n-m}{n}}{2^{2n-m}}$ (this is just a Bernoulli trial).

Multiplying by 2 should give the overall probability as being $\frac{2\binom{2n-m}{n}}{2^{2n-m}}=\frac{\binom{2n-m}{n}}{2^{2n-m-1}}$ , which is double the answer in your picture.

funnytomato said:
Consider the case where A and B both have 2 matches in them. So n=2.

And we want to find the number of ways such that the box B has 1 left(i.e. m=1 ) when A is found to be empty.
Then before we open A and find it's empty(taking a match for the (2n+1-m)= 4th time), we've drawn 2 matches from A and 1 from B.

So there are $\binom{2n-m}{n} =\frac{(2n-m)!}{n! \times(n-m)!}=\frac{3!}{2! \times 1!}=3$ ways to do this.
Namely AAB, ABA and BAA.

Thank you guys soooooo much!!!!!!!!!!! I finally got this XD

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Please help with a probability question (1 Viewer)

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