conditional probability

[Mogie]

Sorry that this is a dumb question.

Sequence of three coin tosses.

Let P(A) = the probability of three heads = 1/8.

Let P(B) = the probability that the first two are heads = 1/4.

What is the probability of three heads given that the first two are heads?

The answer should be 1/2, right?

But using conditional probability we get:

P(A|B) = (P(A)*P(B))/P(B) = 1/8.

Where did I go wrong?

 

[Zermelo]

There are no dumb questions about math! And that's not the formula for conditional probability buddy. If it were, you could just cancel out the P(B) in the numerator and denominator, and there would be no need for introducing the formula, right? The real formula is [MATH]P(A\lvert B) = \frac{P(A \cap B)}{P(B)}[/MATH], aka the probability of A AND B happening divided by the probability of B happening. So, what is the probability of A and B, aka that there are 3 heads in total and that the first two tosses were heads? Can one event happen without the other?
Also, did you learn about independent/dependent events? When are events independent or dependent, and what does that mean in the sense of computing the probability?

 

[Steven G]

P(A and B) is NOT P(A)*P(B)

 

[pka]

Sequence of three coin tosses.
Let P(A) = the probability of three heads = 1/8.
Let P(B) = the probability that the first two are heads = 1/4.
What is the probability of three heads given that the first two are heads?
The answer should be 1/2, right?

\(\begin{array}{*{20}{c}} H&H&H \\ H&H&T \\ H&T&H \\ H&T&T \\ T&H&H \\ T&H&T \\ T&T&H \\ T&T&T \end{array}\)
The above is the outcome space for tossing a coin three times. How many have two heads first?
Of those how many have three heads?

 

[Mogie]

I see now that I was mistaken in thinking that intersection implies multiplication.

Thanks all for your help.