Help with Conditional Probability Question

Question

I'm trying to solve the following problem:

Susan tosses a fair coin three times. What is the probability that she tosses "all tails" if she tosses at least two tails?

The probability space is:

HHH
HHT
HTT
TTT
THH
TTH

The conditional probability formula states that $P(A|B)=\dfrac{P(A \cap B)}{P(B)}$.

In this case, "$A$" is tossing three tails. I believe the probability of tossing $3$ tails is $\dfrac{1}{6}$. In this case, "$B$" is tossing at least two tails. I believe the probability of this event is $\dfrac{1}{2}$.

Then, according to the formula above, the conditional probability of tossing $3$ tails given that she tosses at least $2$ tails is:

$$\frac{\left(\frac{1}{6} \cdot \frac{1}{2}\right)}{\frac{1}{2}}=\frac{1}{6}$$

Is my understanding correct? Thanks in advance!

Answer 1

You have two mistakes, but based on your current work I am confident you can fix them.

1. You used $P(A\cap B) = \frac16 \cdot \frac12$ for some reason. Actually, it can be calculated that $A\cap B=A$. Thus, you should have used $P(A\cap B)=P(A)=\frac16$, which is different than $\frac16\cdot\frac12$.

2. Your probability space is wrong. You forgot

   HTH THT

So, the probability spaces has 8 elements instead of 6. Unfortunately, this means you have to redo all calculations (and even my numbers from the first mistake are wrong).