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A is a typist who makes an average of 2.5 mistakes per letter. B is a typist who makes an average of 4.1 mistakes per letter. Assume that the number of mistakes made by any typist follows a Poisson distribution.

Q) A and B type one letter each. Given that the letters contain a total of three mistakes, find the probability A made more mistakes than B.

I tried finding Probability A makes 3 mistakes and B makes 0 plus probability A makes 2 mistakes and B makes 1 but this doesn't give 0.322, the correct answer/

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You're missing a buzzword here: given. In probability, this is a signal that you are looking at the conditional probability.

So, here, you want to compute $$ P(M_A>M_B\mid M_A+M_B=3), $$ where $M_A$ and $M_B$ are the number of mistakes made by $A$ and $B$, respectively.

You can rewrite this as $$ \frac{P(M_A>M_B\text{ and }M_A+M_B=3)}{P(M_A+M_B=3).} $$ The top probability is what you've already computed: $$ P(M_A>M_B,M_A+M_B=3)=P(M_A=3,M_B=0)+P(M_A=2,M_B=1). $$ So, you just have to compute the denominator. To that end: note that the sum of two independent Poisson variables is itself Poisson.

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    Thanks. If the 'given' word wasn't explicitly used in the question, how do you decide whether it is conditional or just intersection? – 2017-02-15
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    Fundamentally, the question is this: do you want to know whether $A$ and $B$ BOTH happen, or do you want to know whether $B$ happens when you already know that $A$ does. The clues are all either linguistic ('given ...', 'if you know that...', etc), or based on what you're actually trying to accomplish. – 2017-02-15