The probabilities of two students A and B coming to the school in time are 3/7 and 5/7 respectively. Assuming that the events A & B are independent,find the probability of ONLY ONE of them coming to school in time?
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there are eight ways that I see this happening
ab are on time
ba are on time
ab are both late
ba are both late
ab a is late
ba a is late
ab b is late
ba b is late
im curous how to approach this since 5/7 + 3/7 does not add to 7/7
The answer is 26/49 and the solution says:
P(A)*(1 - P(B)) + P(B)*(1 - P(A)) = 26/49
I just odnt know know how it happened?