Unformatted text preview: ected’’. Increase – no partial credit d. (3 points) How would your answer to a. and b. change if the ipod shuffle was only allowed to play exactly 2 different songs from each of the 3 bands. For a. 20 C2 •15 C2 •10 C2 For b. 20 C2 •15 C2 •10 C2 •6 P6 4. (9 total) Kevin Durant is a streaky basketball 3‐point shooter. When he starts a game the probability that he scores on his first 3‐point attempt is 0.3. If he scores on his first attempt then the probability that he scores increases to 0.4 on all his subsequent attempts. If he misses on his first attempt the probability that he scores decreases to 0.2 on all his subsequent attempts. a. (3 points) Find the probability that Durant scores on all of his first three attempts. (0.3)(0.4)(0.4)= 0.048 – no need to compute b. (3 points) Find the probability that Durant scores on his first two attempts and then misses the third attempt? (0.3)(0.4)(0.6) = 0.072 ‐‐ no need to compute Solution used in grading had (0.3)(0.3)(0.6) which is wrong ‐ I will give everyone 3 points to correct for this error. c. (3 points) Is the event ``Durant scores on his third attempt’’ independent of the event ``Durant scores on his first attempt’’? Why/Why not? No since …. P(S3 S1 ) = 0.4
P(S3 ) = P(S3 S1 )P(S1 ) + P(S3 S1 )P(S1 ) = (0.4)(0.3) + (0.2)(0.7) ≠ 0.4 5. (15 total) At a restaurant on the ``drag’’ 50% of customers are students. Also 35% of all customers order vegetarian meals while 25% of the students order vegetarian meals. a. (3 points) A customer is chosen at random. What is the probability that this customer is a student who ordered a vegetarian meal? Let S be student and V be vegetarian. We want : P(S ∩ V) = P(VS)P(S) = 0.25(0.5) = 0.125 b. (3 points) If a randomly chosen customer ordered a vegetarian meal what is the probability that they are a student? P(SV) = P(S ∩ V) 0.125
=
= 0.35...
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 Spring '08
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