Question

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

	A	B	C	Total
Male	14	11	3	28
Female	15	8	19	42
Total	29	19	22	70

If one student is chosen at random,

find the probability that the student was female OR received a grade of "C". (Leave answer to four decimal places.)

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. [Round your answers to three decimal places, for example: 0.123]

Compute the probability that three randomly selected peanut M&M’s are all brown.

If you randomly select six peanut M&M’s, compute that probability that none of them are blue.

If you randomly select six peanut M&M’s, compute that probability that at least one of them is blue.

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

	A	B	C	Total
Male	12	9	20	41
Female	7	13	10	30
Total	19	22	30	71

If one student is chosen at random,

Find the probability that the student was male OR got an "C".

Answer 1

1)The probability that the student was female OR received a grade of C

female and C = 42+22 = 64

female and C counted 19 two times so remove from 64 = 64 - 19 = 45

The probability that the student was female OR received a grade of C is 45/70 = 0.6429 = 64.29%

2)

a)The probability that three randomly selected peanut M&M’s are all brown

Given 12% of peanut M&M’s are brown = 0.12

So, Probability is 0.12*0.12*0.12 = 0.12^3 = 0.002

b)If you randomly select six peanut M&M’s, the probability that none of them are blue

Given 23% of peanut M&M’s are blue

Probability = (1-0.23)^6 = 0.77^6 = 0.208

c)If you randomly select six peanut M&M’s, the probability that at least one of them is blue

23% are blue = 0.23

P(X ≥ 1) = P(1) + P(2) + P(3) + P(4) + P(5) + P(6)

P(1) = (6!/1!*(6−1)!) * 0.23^1 * (1−0.23)^(6−1) = 0.374

Similarly

P(1) = 0.374
P(2) = 0.279
P(3) = 0.111
P(4) = 0.025
P(5) = 0.003
P(6) = 0.000

P(X ≥ 1) = P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 0.792

3)

The probability that the student was male OR got an "C".

male and C = 30+41 = 71

male and C counted 20 two times. so, remove from 71 = 71-20 = 51

The probability that the student was male OR got an "C" = 51/71 = 0.7183 = 71.83%