Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I randomly choose 1 of 2 coins. Coin A is fair: It has probability $\frac{1}{2}$ of yielding either heads or tails. Coin B is biased: It has a probability of $\frac{1}{4}$ of yielding heads and a probability of $\frac{3}{4}$ of yielding tails. If I toss my chosen coin 10 times, what's the probability that I will get exactly 3 heads?

I tried 2 different methods to solve this problem. Strangely, they produced different results. Which of these is right, and why do they differ? I thought they did the same thing. One just does it through the mean probability of getting heads in general.

Let $X$ be the number of heads out of 10 tosses.

Method 1:

P(1 head out of 1 toss) = $\frac{1}{2}\frac{1}{2} + \frac{1}{2}\frac{1}{4} = \frac{3}{8}$
Hence, $P(X=3) = \binom{10}{3}(\frac{3}{8})^3(\frac{5}{8})^7 \approx 0.235$

Method 2:

$P(X=3) = P(X=3|\text{ fair coin})P(\text{fair coin}) + P(X=3|\text{ biased coin})P(\text{ biased coin})$
$P(X=3) = \binom{10}{3}(\frac{1}{2})^3(\frac{1}{2})^7(\frac{1}{2}) + \binom{10}{3}(\frac{1}{4})^3(\frac{3}{4})^7(\frac{1}{2}) \approx 0.184$

share|cite|improve this question
up vote 2 down vote accepted

The first method is right if you randomly choose your coin ten times and each time throw it once before choosing anew.

The second method is right if you choose one of the coins and stick with your choice for all ten throws.

share|cite|improve this answer
Ah thank you! I wonder why they differ though... the amount of information you know in the beginning is the same though, right? – John Hoffman Oct 15 '12 at 13:11
The two processes agree in mean, but there's no reason why their distributions must match. For example, if you had one coin that always came up heads and one that always came up tails and you picked a coin randomly, then the expected number of heads'd be five but you'd always get 0 or 10 heads. – anonymous Oct 15 '12 at 13:20
If the numbers were more skewed, it would be more obvious that the results have to differ. Let us say that both coins are biased, that coin A almost always shows heads, and that coin B almost always shows tails. You repeat your experiment, but ask for the probability that all your throws show heads. With the first method you have to be lucky ten times, but with the second method you only have to be lucky once. – Per Manne Oct 15 '12 at 13:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.