exammt1s14solutions(1)

Alsoitisobservedthat50ofall

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Unformatted text preview: 7 P(V) 0.35 c. (3 points) What is the probability that a randomly chosen customer is not a student and did not order a vegetarian meal? S ∩ V = complement(S∪ V) by DeMorgan’s Law P(S∪ V) = P(S) + P(V) − P(S∩ V) = 0.5 + 0.35 − 0.125 = 0.725 So the answer is 1‐0.725=0.275 d. (3 points) Are the events ``customer is a student’’ and ``customer ordered a vegetarian meal’’ mutually exclusive? No since P(S ∩ V) > 0 e. (3 points) Are the events ``customer is a student’’ and ``customer ordered a vegetarian meal’’ independent? No – many ways to show this – here is one: 0.125 = P(S ∩ V) ≠ P(S)P(V) = 0.5(0.35) = 0.175 6. (15 total) A survey is conducted concerning gender and preferences regarding Winter Olympic sports. The sport categories are: a) X games events (ie: halfpipe, slopestyle, short track, skeleton etc), b) Figure Skating, c) Traditional actual sports (alpine skiing, cross country skiing, long track skating, luge, bobsled etc). In the survey 50% of those surveyed are male. Among the females 40% preferred Figure Skating while 40% preferred X games events. Also it is observed that 50% of all surveyed preferred X games sports and 20% of the males preferred Figure Skating. A person is selected at random from the sample represented by these proportions . Find each of the following: Here is the table that can be constructed from the info Income Male Female X games 0.3 0.2 Figure Skating 0.1 0.2 Trad 0.1 0.1 a. (3 points) The probability that the person preferred X games sports given that the person is a male. P(X|M) = P(X∩ M) P(X) − P(X∩ F) 0.5 − (0.4)(0.5) = = = 0.6 P(M) P(M) 0.5 b. (3 points) The probability that the person was a male and preferred traditional sports. P(M∩ Trad) = P(Trad|M)P(M) P(Trad|M) = 1 − P(X|M) − P(FS|M) = 1 − 0.6 − 0.2 = 0.2 c. (3 poi...
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