N = 6
For N = 1 & 2, we already know that the result is tails. So the result now depends only on the last four tosses.
So, probability of getting 6 tails is = (1).(1).(1/2).(1/2).(1/2).(1/2) = 1/16
Can you help me finish off this probability question?
A coin is tossed 6 times. What is the probability that tails appears on every toss, given that tails appears on the first two tosses?
This is what I have so far:
n = intersection symbol
The outcomes of each individual toss are dependent upon the toss prior so the odds of a heads or tail on any given toss is 1/2. We are going to use the formula Pr(EIF) = Pr(E n F)/Pr(F.
F = "tails appears on the first two tosses"
E = "tails appears on every toss"
Find the probabilities:
Pr(F) = 1/4 - odds of tails on first two tosses
Pr(E) = 1/64 - odds for tails on all six tosses
Now I need to use the rule of Conditional Probability to solve. This is where I'm stuck. How to finish off this problem?