I need help finishing this problem (1st half solved)

Hello all. I'm taking a college level Stats and Probabilities class. This was an example of a homework problem for this week. I have the majority of it figured out but I need someone to help me finish it.

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?

There is an easy way to solve this- a coin has no "memory". This is NOT a "conditional probability" problem. One flip is independent of any other. If the coin comes up tails on the first two flips, all that means is that to get six tails, you need to get tails on the last 4. The probability that you get "tails on all 6 flips given that you got tails on the first two" is exactly the probability the coin comes up tails on four consecutive flips.

so would my final answer be 1/64?

No, it wouldn't. \(\displaystyle \frac{1}{64}= (1/2)^6\) but you are " given that tails appears on the first two tosses" so the only "probability" occurs on the last four tosses.

HallsofIvy said:

No, it wouldn't. \(\displaystyle \frac{1}{64}= (1/2)^6\) but you are " given that tails appears on the first two tosses" so the only "probability" occurs on the last four tosses.

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Would 1/64 - 1/4 = the answer since 1/64 is the odds all six are tails and 1/4 the odds of the first two tosses being tails?

HallsofIvy said:

There is an easy way to solve this- a coin has no "memory". This is NOT a "conditional probability" problem. One flip is independent of any other. If the coin comes up tails on the first two flips, all that means is that to get six tails, you need to get tails on the last 4. The probability that you get "tails on all 6 flips given that you got tails on the first two" is exactly the probability the coin comes up tails on four consecutive flips.

Click to expand...

the probability of four straight tails is 1/16. Is this the answer to the original intended question?