(Sec. 2.3) A restaurant having a grand opening is allowing visitors to try 6 dif
ID: 3353254 • Letter: #
Question
(Sec. 2.3) A restaurant having a grand opening is allowing visitors to try 6 different free samples from a selection. Options for sampling include 7 poultry items, 10 red meat items and 12 vegetarian items. (a) If a visitor only wants to taste 3 of the poultry items, and if the order of tasting is important, (b) Not worrying about the order from here on, suppose a visitor wants to try all 6 allotted samples. (c) If a visitor wants to sample 2 of each type of food (poultry, red meat and vegetarian), how many (d) If the 6 samples are randomly selected, what is the probability that two of each type of food are (e) If the 6 samples are randomly chosen, what is the probability that the majority of the choices are how many ways are there to do this? How many choices of 6 do they have? ways are there to do this? chosen? meat (poultry or red meat)?Explanation / Answer
a) number of ways to select 3 out of 7 poutry iterms =7P3 =7!/(7-3)! =210
b) number of samples available =7+10+12 =29
therfore choices of 6 =selecting 6 out of 29 =29C6 =29!/(6!*(29-6)!) =475020
c)
number of ways =7C2 *10C2 *12C2 =21*45*66 =62370
d)
probability =62370/475020 =0.1313
e)
number of ways that majority is meat =number of ways vegetarian die are 2 at most
=17C4*12C2 +17C5*12C1 +17C6*12C0 =243712
hence probability =243712/475020 =0.5131
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