Probability of a tail followed by 2 heads on a biased coin

Suppose we have a biased coin for which the probability of heads is 3/4 while the probability of tails is 1/4 . What is the probability of a tail followed by 2 heads on three flips of the coin?

Here is what I have so far: There are 2^3 possible outcomes {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}, and only one way to get THH. I've hit a wall and any pointers would be appreciated.

Re: Probability of a tail followed by 2 heads on a biased coin

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**battery88** Suppose we have a biased coin for which the probability of heads is 3/4 while the probability of tails is 1/4 . What is the probability of a tail followed by 2 heads on three flips of the coin?

If you were to look up the answer in the 'back-of-the-book' it would be:

$\displaystyle \frac{3}{2^6}$.

Now if you can explain to yourself **WHY?** or **HOW?** then you will understand.

HINT: $\displaystyle TTH$ is one out of eight which is a power of two.

Re: Probability of a tail followed by 2 heads on a biased coin

Quote:

Originally Posted by

**battery88** Suppose we have a biased coin for which the probability of heads is 3/4 while the probability of tails is 1/4 . What is the probability of a tail followed by 2 heads on three flips of the coin?

Here is what I have so far: There are 2^3 possible outcomes {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}, and only one way to get THH. I've hit a wall and any pointers would be appreciated.

The probability that the first coin is tails is 1/4. The probability that the second coin is heads 3/4. The probability that the third coin is heads is 3/4. The probability of three independent results, ABC, happening is P(A)P(B)P(C).