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, how many ways are there to do this? (b) Not worrying about the order from here on, suppose a visitor wants to try all 6 allotted samples. How many choices of 6 do they have? (c) If a visitor wants to sample 2 of each type of food (poultry, red meat and vegetarian), how many ways are there to do this? (d) If the 6 samples are randomly selected, what is the probability that two of each type of food are chosen? (e) If the 6 samples are randomly chosen, what is the probability that the majority of the choices are meat (poultry or red meat)?

Answer 1

Answer:A)35choices

B)475,020choices

C)132choices

D)0.000278

E)0.000166

Step-by-step explanation:

Given 7poultry items represented by 7p

10 red meat items=10R

12 vegetarian items=12V

@

A)there are 7combination 3ways of selecting 3out of 7 poultry items

From the formula nCr=n!/(n-r)!r!

7C3=7!/(7-3)!3!=7!/4!×3!=35choices

b)total items available=7p+10R+12V(denotation explained above)=29items

Choosing 6items implies

29C6=29!/(29-6)!×6!

29!/23!×6!=475,020 ways

C)7C2+10C2+12C2

7!/5!2!+10!/8!2!+12!/10!2!

21+45+66=132 choices

D) probability (choosing 2items from each class)=

7C2+10C2+12C2/29C6

Where 29C6 represent total probable.

21+45+66/29!/23!6!

=132/475020=0.000278choices

E)for meats to be in the majority on a 6choices selection impliee that vegetarian items will be

0,1 or2i.e combination of vegetarian and meat will be(0,6)or(1,5)or(2,4) .

Prob(0veg)+prob(1veg item) or prob(2veg items)

12C0+12C1+12C2/29C6

=1+12+66/475,020

=0.000166