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With a Bernoulli process success probability $p$ and failure probability $q = 1 - p$, the probability for $T_k$ time of the $k^{th}$ success is $$P(T_k \leq n) = \sum_{j=k}^n \binom{n}{j}p^jq^{n-j}$$

My question is how does this work for $k=1$? According to the binomial theorem, the probability would be 1 for any $n$, and this doesn't make sense to me. For example, consider $n=7$. What if the first seven steps are failures, and $T_1 = 8$? It wouldn't make sense that $P(T_1 \leq 7) = 1$.

Vahan
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    Your statement puzzles me. The R.H.S. is the probability of at least k successes out of n trials. What does $P(T_k\le n)$ mean? – herb steinberg May 11 '19 at 21:40
  • @herbsteinberg There are $\ge k$ successes in the first $n$ trials if and only if the $k$th success occurs before/on the $n$th trial. – angryavian May 11 '19 at 21:48
  • Binomial theorem tells that $\sum_{j=0}^{n}\binom{n}{j}p^j(1-p)^{n-j}=1$. Using this, we get $$P(T_k \leq n) = 1 - \sum_{j=0}^{k-1} \binom{n}{j}p^j(1-p)^{n-j}.$$ In particular, when $k = 1$, we get $P(T_1 \leq n) = 1 - (1-p)^n$, which matches the CDF of the geometric distribution with parameter $p$. – Sangchul Lee May 11 '19 at 21:51

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No, $\sum_{j=1}^n \binom{n}{j} p^j q^{n-j} \ne 1$. You are conflating it with $\sum_{j=0}^n \binom{n}{j} p^j q^{n-j} = 1$.

angryavian
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