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?

Question

amistre64 · Answer

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

amistre64 · Answer

im curous how to approach this since 5/7 + 3/7  does not add to 7/7

anonymous · Answer

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?

amistre64 · Answer

that was a thought .. jsut wasnt sure on it :)

if A = 3/7  then notA = 1-3/7  right?

anonymous · Answer

yes

amistre64 · Answer

probability of A and notB = 3/7(1-5/7)

probability of B and notA = 5/7(1-3/7)

the probabilty of one OR the other event occuring is the sum of the two

anonymous · Answer

yeah thanks!

amistre64 · Answer

youre welcome