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

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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.