# Probability of a tail followed by 2 heads on a biased coin

1. Feb 12, 2013

### battery88

1. The problem statement, all variables and given/known data

Suppose we have a biased coin for which the probability of heads is 3/4 while the probability of tails is 1/4 . What is the probability of a tail followed by 2 heads on three flips of the coin?

2. Relevant equations

3. The attempt at a solution

Here is what I have so far: There are 2^3 possible outcomes {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}, and only one way to get THH. I've hit a wall and any pointers would be appreciated.

2. Feb 12, 2013

### bossman27

Alright, just think about the chances of getting a permutation with one tail and two heads: {THH, HTH, HHT}. Hint: each flip outcome is independent of the others. Then, how likely is it you get a particular one of these outcomes?

Last edited: Feb 12, 2013
3. Feb 12, 2013

### battery88

Okay, thanks for the help. I've been stuck on this for hours.

4. Feb 12, 2013

### battery88

Isn't there only one way to get THH, since it says "a tail followed by two heads on three flips"

5. Feb 12, 2013

### haruspex

Yes. What is the probability of each flip outcome individually? How would you combine them?

6. Feb 12, 2013

### bossman27

Yes, I was just asking you to take the problem in steps. First to figure out how many possible ways you could get one tail and two heads in any order, with three flips. Then, what is that chance that out of those possibilities you get one particular one.

7. Feb 12, 2013

### battery88

Multiply them.

8. Feb 12, 2013

### haruspex

So go ahead and multiply, to coin a phrase.

9. Feb 12, 2013

### battery88

1/4 * 3/4 * 3/4 = 9/64

10. Feb 12, 2013

### bossman27

What I was asking you to do is multiply the odds to find the probability of one tail and two heads, out of three flips. This, however, does not account for order; you are finding the probability of THH, HTH, or HHT.

Maybe it would help to write formally:

A = getting THH in particular
B = occurrence of getting 1 head and 2 tails in any order

P(A and B) = P(A|B) * P(B)

That is to say, the probability of getting THH, is equal to the probability of getting THH out of {THH, HTH, HHT} times the probability of getting two heads and one tail in any order

I can be a little more explicit if you need, but ideally I'd like to get you to figure it out yourself from that hint if you're able.

Edit: I mistakenly typed TTH. Obviously, I meant THH.

11. Feb 12, 2013

### bossman27

If you can come up with an answer, I'll tell you if it's correct or not.

12. Feb 12, 2013

### battery88

Okay, the probability of getting a tail and two heads in any order = 3/8.

And, the probability of getting THH out of {THH, HTH, HHT} = 1/3.

So, 3/8 * 1/3 = 1/8

13. Feb 12, 2013

### battery88

But, I'm not accounting for the bias.

14. Feb 12, 2013

### haruspex

Yes.

15. Feb 12, 2013

### haruspex

The OP says "What is the probability of a tail followed by 2 heads on three flips of the coin?" I.e. just THH.

16. Feb 12, 2013

### battery88

Let me try again.

the probability of getting a tail and two heads in any order = 3 * 3/4 * 3/4 * 1/4 = 27/64.

And, the probability of getting THH out of {THH, HTH, HHT} = 1 * 3/4 * 3/4 * 1/4 = 9/64.

So, 27/64 * 9/64 = .059

17. Feb 12, 2013

### bossman27

Right, I was just referencing what I recommended to him for a first step. I clarified that he must then account for the fact that he's looking for a particular order.

18. Feb 12, 2013

### bossman27

Not quite, 3/4 * 3/4 * 1/4 is the probability of getting a tail and two heads in any order.

What's the probability of picking one option out of three?

19. Feb 12, 2013

### bossman27

To maybe explain a little more:

Notice that 3/4 * 3/4 * 1/4 = 1/4 * 3/4 * 3/4 = 3/4 * 1/4 * 3/4

This is giving the probability that in three independent trails, where in each trail one outcome has a probability of .25, and the other has a probability of .75, the outcome with .25 odds will happen once, and the one with .75 odds will happen twice.

You have found the odds of {THH, HTH, HHT}. Now that you have that probability, look back at my other formula/post and notice that you have only one simple step left...

Namely, given {THH, HTH, HHT}, what is the probability of picking THH? It's probably much simpler than you think.

20. Feb 12, 2013

1/3.