I'm taking an MIT OCW course on Probability.


Al performs an experiment comprising a series of independent trials. On each trial, he simultaneously flips a set of three fair coins. Whenever all three coins land on the same side in any given trial, Al calls the trial a success. Find the PMF for K, the number of trials up to, but not including, the second success.

My solution: Success occurs only when we get 3 heads or 3 tails

$P(success) = 1/4$

In $k$ trails, we will 1 success, so the PMF is -

$PMF = {{k}\choose{1}} * 1/4 * (3/4)^{k-1}$

Solution Given:

$PMF = {{k}\choose{1}} * (1/4)^2 * (3/4)^{k-1}$

Can anyone explain what is wrong in my answer?

Link to the solution: 2 (b)


Let $X_n=1$ if there is a succes on trial $n$ and $X_n=0$ otherwise.

Let $S_n=X_1+\cdots+X_n$, i.e. the number of successes in the first $n$ trials.

Then: $$K=k\iff S_k=1\wedge X_{k+1}=1$$

The events $\{S_k=1\}$ and $\{X_{k+1}=1\}$ are independent so that: $$\mathbb P\{K=k\}=\mathbb P\{S_k=1\}\times \mathbb P\{X_{k+1}=1\}$$

You calculated $\mathbb P\{S_k=1\}$ but forgot the multiplication with $\mathbb P\{X_{k+1}=1\}=\frac14$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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