The Student Room Group

Stats 1 probability question

Hey, i'm yet again stuck with a stats question!

The question is:

Six fair coins are tossed and those landing heads uppermost are eliminated. The remainder are tossed again and the process of elimination is repeated. Tossing and elimination continue in this way until no coins are left.

Find the probability that:

exactly 2 coins are eliminated in the first round and exactly 2 coins are eliminated in the third round

I just dont really know how to go about working this one out, i'm not entirely sure how to work out the probability for the second round.

Thanks in advance for any help :smile:
Reply 1
Nattynoo101

exactly 2 coins are eliminated in the first round and exactly 2 coins are eliminated in the third round

For two coins in the first round to be eliminated: 6C2(12)2(12)4^6\mathrm{C}_2 \left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^4
For the second round, because there are four coins and the probability is 0.5, you will probably have two coins remaining (on re-examining that statement I'm not sure about it ...damn!)
So on the third round:2C2(12)2(12)0^2\mathrm{C}_2 \left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^0
So overall, you'll have 6C2(12)2(12)4×2C2(12)2(12)0^6\mathrm{C}_2 \left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^4 \times ^2\mathrm{C}_2 \left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^0

EDIT: Rereading that solution, I'm not so sure now, sorry :frown:
Reply 2
Thanks for trying anyway :smile: its a bugger of a question!
Reply 3
corney91
For two coins in the first round to be eliminated: 6C2(12)2(12)4^6\mathrm{C}_2 \left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^4
For the second round, because there are four coins and the probability is 0.5, you will probably have two coins remaining (on re-examining that statement I'm not sure about it ...damn!)
So on the third round:2C2(12)2(12)0^2\mathrm{C}_2 \left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^0
So overall, you'll have 6C2(12)2(12)4×2C2(12)2(12)0^6\mathrm{C}_2 \left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^4 \times ^2\mathrm{C}_2 \left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^0

EDIT: Rereading that solution, I'm not so sure now, sorry :frown:


The first bit sounds ok....
For the second round you will have 4 coins remaining so it will be 4C2^4\mathrm{C}_2 times [(0.5)(0.5) + (0.5)(0.5)] because the probability of getting a head is the same as not getting it....
I think that's how I would do it. :smile:
Reply 4
I've just drawn a tree diagram with probabilities for each round using Binomials:
1st round 6C2(12)2(12)4^6\mathrm{C}_2 \left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^4
2nd round:4C0(12)0(12)4^4\mathrm{C}_0 \left(\frac{1}{2}\right)^0\left(\frac{1}{2}\right)^4
4C1(12)1(12)3^4\mathrm{C}_1 \left(\frac{1}{2}\right)^1\left(\frac{1}{2}\right)^3
4C2(12)2(12)2^4\mathrm{C}_2 \left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^2
Just those three in this round because you can't eliminate more than 2 otherwise you can't eliminate 2 in the third round :wink:

Use the same method for the third round and can follow each of the paths along the tree diagram to work out all the probabilities... I'm pretty sure this should work:smile:

BTW, nice Ed Byrne quote in your sig :biggrin:
We have three posibilities
Either
1st round: 6C2 (1/2)^2 (1/2)^4 = 15/64 exactly 2 in the first round,
2nd round:4C0 (1/2)^0 (1/2)^4 = 1/16 exactly 0 in the second round,
3rd round: 4C2 (1/2)^2 (1/2)^2 = 3/8 exactly 2 in the third round.
multiply all 3 rounds we get the first possibility: 15/64X1/16X3/8=45/8192.
OR
1st round: 6C2 (1/2)^2 (1/2)^4 = 15/64 exactly 2 in the first round,
2nd round:4C1 (1/2)^1 (1/2)^3 = 1/4 exactly 1 in the second round,
3rd round: 3C2 (1/2)^2 (1/2)^1 = 3/8 exactly 2 in the third round.
multiply all 3 rounds we get the second possibility: 15/64X1/4X3/8=45/2048.
OR
1st round: 6C2 (1/2)^2 (1/2)^4 = 15/64 exactly 2 in the first round,
2nd round:4C2 (1/2)^2 (1/2)^2 = 3/8 exactly 2 in the second round,
3rd round: 2C2 (1/2)^2 (1/2)^0 = 1/4 exactly 2 in the third round.
multiply all 3 rounds we get the third possibility: 15/64X3/8X1/4=45/2048.

so what you need is the probability that you get exactly 2 in the first round and exactly 0 or 1 or 2 in the second round(no more then this because you need 2 in the third round), and exactly 2 in the third round.The rounds are each (and) functions and we multiply them by each other, and the possibility are an OR function so we add each of the three possibilities
So: 45/8192+45/2048+45/2048=405/8192. tadaaaa