CSProbabilitypartav.pptVIP

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CSProbabilitypartav.ppt

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CSProbabilitypartav

* * * * * * * * * * * * * * * * * * * * * * * * 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111 Sample space 0000 0001 0010 0011 0100 0101 0110 0111 F P(F) = 1/2 0000 0001 0010 0011 0100 1000 1001 1100 E P(E) = 1/2 0000 0001 0010 0011 0100 0101 0110 0111 P(E|F) = 5/8 0000 0001 0010 0011 0100 Pr(E?F) = 5/16 E?F) 0000 0001 0010 0011 0100 1000 1001 1100 P(F|E) = 5/8 Independence The events E and F are independent if and only if Pr(E?F) = Pr(E) x Pr(F). Note that in general: Pr(E?F) = Pr(E) x Pr(F|E) (defn. cond. prob.) So, independent iff Pr(F|E) = Pr(F). (Note: Pr(F|E) = Pr(E ? F) / P(E) = (Pr(E)xPr(F)) / P(E) = Pr(F) ) Example: P(“Tails” | “It’s raining outside”) = P(“Tails”). Independence The events E and F are independent if and only if Pr(E?F) = Pr(E) x Pr(F). Let E be the event that a family of n children has children of both sexes. Let F be the event that a family of n children has at most one boy. Are E and F independent if n = 2? No Hmm. Why? BB BG GB GG P(E) = 1/2 P(F) = 3/4 P(E ? F) = 1/2 =/= 1/2 x 3/4 Independence The events E and F are independent if and only if Pr(E?F) = Pr(E) x Pr(F). Let E be the event that a family of n children has children of both sexes. Let F be the event that a family of n children has at most one boy. Are E and F independent if n = 3? Yes !! BBB BBG BGB BGG GBB GBG GGB GGG P(E) = 6/8 P(F) = 4/8 P(E ? F) = 3/8 = 6/8 x 1/2 Independence The events E and F are independent if and only if Pr(E?F) = Pr(E) x Pr(F). Let E be the event that a family of n children has children of both sexes. Lef F be the event that a family of n children has at most one boy. Are E and F independent if n = 4? No So, dependence / independence really depends on detailed structure of the underlying probability space and events in question!! (often the only way is to “calculate” the probabilities to determine dependence / independence.) n = 5?