Figuring out probability in a sequence of coin tosses

I have a question I have worked out and I would like to check my solution. I am told that 4 fair coins are tossed in succession. I am to find the probability of getting 4 tails given that the first 2 tosses were all tails.

I got $$\frac{1/16}{1/4} = \frac{1}{4}.$$ Is this right?

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That’s right, though you could get the answer more easily by noticing that what you want is simply the probability that the last two tosses are both tails, which is $\left(\frac12\right)^2=\frac14$. – Brian M. Scott Jun 7 '12 at 9:42
yeah...that's right. I didn't notice that. Thanks. – count Jun 7 '12 at 9:53
@count: You can answer your own question now that you know the right answer. – Gigili Jun 7 '12 at 10:19

As Brian said in the comment, what we want is the probability that last two tosses are both tails. Since the probability of getting a tail is $\frac{1}{2}$, we get $$\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}.$$