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I'm taking an MIT OCW course on Probability.

Question:

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)

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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 $\mathbb P\{X_{k+1}=1\}=\frac14$.

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