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

  • Thread starter Thread starter battery88
  • Start date Start date
AI Thread Summary
The discussion centers on calculating the probability of specific outcomes when flipping a biased coin with a heads probability of 2/3. The initial calculation for getting exactly four heads in seven flips is confirmed as 560/2187. A question arises about finding the probability of getting exactly one tail followed by three heads, which involves understanding the independence of flips. It is clarified that the sequence of one tail followed by three heads would be considered out of four flips. The conversation emphasizes the importance of defining the total number of flips when calculating probabilities for specific sequences.
battery88
Messages
18
Reaction score
0

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.
 
Physics news on Phys.org
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.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top