Question

1.Giving a test to a group of students, the grades and gender are summarized below

	A	B	C	Total
Male	6	7	11	24
Female	17	4	8	29
Total	23	11	19	53

If one student is chosen at random,
Find the probability that the student did NOT get an "C"

2.According to Masterfoods, the company that manufactures M&M’s, 12% of peanut M&M’s are brown, 15% are yellow, 12% are red, 23% are blue, 23% are orange and 15% are green. You randomly select peanut M&M’s from an extra-large bag looking for a blue candy. (Round all probabilities below to four decimal places; i.e. your answer should look like 0.1234, not 0.1234444 or 12.34%.)

Compute the probability that the first blue candy is the ninth M&M selected.

Compute the probability that the first blue candy is the ninth or tenth M&M selected.

Compute the probability that the first blue candy is among the first nine M&M’s selected.

If every student in a large Statistics class selects peanut M&M’s at random until they get a blue candy, on average how many M&M’s will the students need to select? (Round your answer to two decimal places.)

? blue M&M’s

3.Giving a test to a group of students, the grades and gender are summarized below

	A	B	C	Total
Male	20	19	4	43
Female	13	7	14	34
Total	33	26	18	77

If one student is chosen at random,

Find the probability that the student was female:   

Find the probability that the student was female AND got a "B":   

Find the probability that the student was female OR got an "C":   

If one student is chosen at random, find the probability that the student got a 'C' GIVEN they are female:

4.Suppose a designer has a palette of 9 colors to work with, and wants to design a flag with 5 vertical stripes, all of different colors.

How many possible flags can be created?

Answer 1

(1) probability that the student did NOT get an "C" = (53 - 19)/53 = 0.6415 (please round this to the required decimal places).

(2) probability that the first blue candy is the ninth M&M selected = (0.77)⁸(0.23) = 0.0284.
probability that the first blue candy is the ninth or tenth M&M selected = (0.77)⁸(0.23) + (0.77)⁹(0.23) = 0.0503.
probability that the first blue candy is among the first nine M&M’s selected = (0.77)⁸(0.23) = 0.2558.
on average how many M&M’s will the students need to select = 1/0.23 = 4.35.

(3) probability that the student was female = 34/77 = 0.4416.
probability that the student was female AND got a "B" = 7/77 = 0.0909.
probability that the student was female OR got an "C" = (34+18-14)/77 = 0.4935.
probability that the student got a 'C' GIVEN they are female = 14/34 = 0.4118.
(please round these to the required decimal places)

(4) No. of possible flags = = 9!/(4! * 5!) = 126.