anonymous
  • anonymous
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?
Mathematics
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SOLVED
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katieb
  • katieb
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amistre64
  • amistre64
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
  • amistre64
im curous how to approach this since 5/7 + 3/7 does not add to 7/7
anonymous
  • anonymous
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?

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amistre64
  • amistre64
that was a thought .. jsut wasnt sure on it :) if A = 3/7 then notA = 1-3/7 right?
anonymous
  • anonymous
yes
amistre64
  • amistre64
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
  • anonymous
yeah thanks!
amistre64
  • amistre64
youre welcome

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