anthropos Posted June 1, 2008 Posted June 1, 2008 Two friends Jane and Mary often do their shopping on Saturday afternoons. If Mary goes shopping, the probability that Jane also goes is 0.96. If Jane goes, the probability that Mary also goes is 0.8. The probability that neither goes is 0.07. Find the probability that both go shopping. How do I approach the question? I tried to do conditional probability by doing all the fractions with eg. R(Jane goes intersect Mary goes)/P(Mary goes) = 0.96 and a couple of these but didn't really work.
Bignose Posted June 1, 2008 Posted June 1, 2008 When I write everything out, I get 4 equations and 4 unknowns. Here are the statements in words (I'll leave the math to you) 1) The probability neither goes is 0.07. 2) The probability that Mary either goes shopping or doesn't is 1. 3) The probability that Jane either goes shopping or doesn't is also 1. 4) Bayes' theorem that relate conditional probabilities http://en.wikipedia.org/wiki/Bayes'_theorem 4 statements, 4 unknowns (I'll leave you to figure out the 4 unknowns, unless you get really stuck and ask for help later). Should be solvable. Get to know Bayes' Theorem. It is an exceptionally powerful tool. 1
anthropos Posted June 2, 2008 Author Posted June 2, 2008 I managed to solve the problem by using a Venn diagram. How do you do it by the Baye's Theorem? I tried but I keep getting the same equations. Thanks for your help! Haha I need somebody to help my Probability and P and C. I tend to um be rather bad at the logic stuff.
Bignose Posted June 2, 2008 Posted June 2, 2008 When I write everything out, I get 4 equations and 4 unknowns. Here are the statements in words (I'll leave the math to you) 1) The probability neither goes is 0.07. 2) The probability that Mary either goes shopping or doesn't is 1. 3) The probability that Jane either goes shopping or doesn't is also 1. 4) Bayes' theorem that relate conditional probabilities http://en.wikipedia.org/wiki/Bayes'_theorem 4 statements, 4 unknowns (I'll leave you to figure out the 4 unknowns, unless you get really stuck and ask for help later). Should be solvable. Get to know Bayes' Theorem. It is an exceptionally powerful tool. Since you have solved it, here's the equations I had: Let A = Mary goes shopping. Let A' ("A prime") denote she doesn't. Let B and B' denote Jane does or does not go shopping, respectively. Statement 1 mathematically is: P(A') + P(B') = 0.07 Statement 2 is: P(A) + P(A') = 1 Statement 3 is: P(B) + P(B') = 1 and Statement 4, Bayes' (please note the correct spelling, the guy who came up with it, his last name was Bayes, NOT Baye, so the possessive is Bayes') theorem is P(B|A)P(A) = P(A|B)P(B) Since you have been given the conditional probabilities, P(A|B) and P(B|A) in the problem statement, there are 4 equations with only 4 unknowns, which should be solvable.
Bignose Posted June 7, 2008 Posted June 7, 2008 Just for the record, my mathematical translation of statement 1 is wrong... It should be P(A')P(B') = 0.07 multiplication, not addition (must have been thinking of something else)
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