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I've been trying to apply the same knowledge from a previous post, but perhaps my reasoning is wrong.

"Errors in a computer surfaces according to a Poisson process with rate 0.4 per day. If there has surfaced three errors during 15 days, what is then the probability of one error surfacing the first 5 days?"

my solution thus far:

\begin{align*} P &= P (X(5)=1 \;\; \cap \;\; X(15) - X(10) = 2) \\ &= P (X(5)=1) \cdot P(X(15) - X(5) = 2) \;\; \;\; \text{by independent increments} \\ &= P (X(5)=1) \cdot P(X(10) = 2) \\ &= 3.96\% \end{align*}

but the answer is supposed to be 44%, any thoughts?

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DOGOFWALLSTREET
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1 Answers1

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You are looking for the wrong probability. Using your notation, what the problem asks of you is: $$ \mathbb{P}(X(5) = 1 \mid X(15) = 3) = \frac{\mathbb{P}(X(5) = 1, X(15) = 3)}{\mathbb{P}(X(15) = 3)} = \frac{\mathbb{P}(X(5) - X(0) = 1, X(15) - X(5) = 2)}{\mathbb{P}(X(15) = 3)} = \ldots $$ I believe you can finish the rest (you already computed the numerator correctly).

d125q
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  • thank you! i did try your version but my denominator was P(X(10)=2) which would only cancel out. Your version makes more sense :) – DOGOFWALLSTREET May 15 '15 at 15:59