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

[battery88]

Homework Statement

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?

Homework Equations

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.

 

[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?

 

[battery88]

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

 

[battery88]

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

 

[haruspex]

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

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

 

[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.

 

[battery88]

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

Multiply them.

 

[haruspex]

Multiply them.

So go ahead and multiply, to coin a phrase.

 

[battery88]

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

 

[bossman27]

Multiply them.

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.

 

[bossman27]

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

 

[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

 

[battery88]

But, I'm not accounting for the bias.

 

[haruspex]

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

Yes.

 

[haruspex]

you are finding the probability of THH, HTH, or HHT.

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

 

[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

 

[bossman27]

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

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.

 

[bossman27]

Let me try again.

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

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

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?

 

[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.

 

[battery88]

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?

1/3.

 

[battery88]

So, would the answer be 27/64 * 1/3?

 

[bossman27]

No, you multiplied by an extra 3 in there somewhere. It should be (9/64) * (1/3)

 

[battery88]

Thanks for all your help. We're working on these more in class this week so it'll be nice to know what's going on.

 

[bossman27]

No problem at all, just make sure make sure you try to understand what you're actually calculating at each step. It's a much better way to learn than just trying to plug in numbers.

 

[vela]

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

No, it's not. Consider the case of an unbiased coin. You'd be saying that 1/2 * 1/2 * 1/2 = 1/8 is the probability of getting a tail and two heads in any order. That's obviously wrong.

Battery88 got the correct answer way back in post 9.

 

[bossman27]

No, it's not. Consider the case of an unbiased coin. You'd be saying that 1/2 * 1/2 * 1/2 = 1/8 is the probability of getting a tail and two heads in any order. That's obviously wrong.

Battery88 got the correct answer way back in post 9.

Woops , looks like I'm the one who needs to understand what they're doing. Apologies for the wrongheaded answer, I guess I'll need to do some brushing up of my own on stats.

 

[battery88]

Yeah, that's definitely true. When you don't know the how and why it's hard to figure out which formulas to use in the first place.

 

[Ray Vickson]

Yeah, that's definitely true. When you don't know the how and why it's hard to figure out which formulas to use in the first place.

That is why it is a good idea to avoid canned formulas until you *do* know what is happening.

You can work it out from first principles, but you need to forget about binomnial coefficients and all that, and just start from basics. So, if T1 = tails on toss 1, H2 = heads on toss 2 and H3 = heads on toss 3, you want to calculate P(T1 H2 H3) (and by AB I mean ##A \cap B## for events A and B). Let p be the probability of heads on a single toss. We have P(T1 H2) = P(T1)*P(H2|T1) = (1-p)*P(H2|T1). Since the two tosses are independent we have P(H2|T1) = P(H2) = p, so we finally get P(T1 H2) = (1-p)*p. Similarly, P(T1 H2 H3) = P(T1)*P(H2|T1)*P(H3|T1,H2) = P(T)*P(H)*P(H) = (1-p)*p^2.

So, why do we write P{1T, 2H} = C(3,1)*p^2*(1-p) = 3*p^2*(1-p)? The reason is that {1T,2H} = {H1 H2 T3, H1 T2 H3, T1 H2 H3}, and all three of these points have the same probability p^2*(1-p), so when we add them up we get 3*p^2*(1-p).