Probabilty of one tail followed by three heads with a biased coin

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Homework Help Overview

The discussion revolves around calculating probabilities involving a biased coin, specifically focusing on scenarios with a probability of heads at 2/3. The original poster presents a problem regarding the probability of obtaining exactly four heads in seven flips and then shifts to a related question about finding the probability of one tail followed by three heads.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of independence in coin flips and question how the setup of one tail followed by three heads relates to the original problem of four heads in seven flips.

Discussion Status

The discussion is ongoing, with participants clarifying the conditions of the problem and examining the relationship between the two scenarios. There is no explicit consensus yet, but questions are being raised about the interpretation of the tosses involved.

Contextual Notes

There is some ambiguity regarding whether the one tail followed by three heads is considered within the context of four total tosses or if it relates to a larger number of flips. This distinction is under discussion.

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Homework Statement


A coin is biased so that the probability of heads is 2/3. What is the probability that exactly four heads come up when the coin is flipped seven times, assuming that the flips are independent?


Homework Equations





The Attempt at a Solution


C(7,4)(2/3)^4(1/3)^3 = (35*16)/3^7 = 560/2187

Now, I know this answer is correct, but what if we were asked to find probability of exactly one tail followed by three heads rather than exactly four heads: How would I calculate the answer for that?

Thank you.
 
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battery88 said:

Homework Statement


A coin is biased so that the probability of heads is 2/3. What is the probability that exactly four heads come up when the coin is flipped seven times, assuming that the flips are independent?


Homework Equations





The Attempt at a Solution


C(7,4)(2/3)^4(1/3)^3 = (35*16)/3^7 = 560/2187

Now, I know this answer is correct, but what if we were asked to find probability of exactly one tail followed by three heads rather than exactly four heads: How would I calculate the answer for that?

Thank you.

How would you calculate the probability of getting T on toss 1? If your first toss is T, does that affect the probability that the next three tosses all give H?
 
No, it doesn't since each toss is independent. So, it would be calculated the same way?
 
battery88 said:
Now, I know this answer is correct, but what if we were asked to find probability of exactly one tail followed by three heads rather than exactly four heads: How would I calculate the answer for that?
The four heads were out of 7 tosses. Are your 1T3H out of just four tosses? If not, what do you mean?
 
haruspex said:
The four heads were out of 7 tosses. Are your 1T3H out of just four tosses? If not, what do you mean?

I believe he was referencing another problem he posted on PF. But yeah, I assume that THHH would be out of four tosses; obviously any later ones would be superfluous.
 

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