The first one is that B arrives within the first 22 hours, which is relatively easy to calculate. It is (22 * 21)/(24 2).

The second is that B arrives in the last two hours, and the last two hours are subdivided into N segments. Let's assume that n is large enough so that the probability that B will arrive at each segment is a fixed value (2/n)/24, and the probability that A will not meet B at this time (assuming K segment) is

(23-(2k/n))/24, add and simplify these n probabilities, and get the approximation of the second case.

(44-(2/n))/(24 2), the probability that n tends to positive infinity is 44/(24 2).

The sum of the two cases is 253/288=0.879.

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