Question

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 brown candy. (Round all probabilities below to four decimal places; i.e. your answer should look like 0.1234, not 0.1234444 or 12.34%.)A.) Compute the probability that the first brown candy is the seventh M&M selected.B.) Compute the probability that the first brown candy is the seventh or eighth M&M selected.C.)Compute the probability that the first brown candy is among the first seven M&M’s selected.D.) If every student in a large Statistics class selects peanut M&M’s at random until they get a brown candy, on average how many M&M’s will the students need to select? (Round your answer to two decimal places.)

Answer 1

See explanation.

Explanation:

We are looking at a geometric distribution.

The probability of selecting a brown peanut is .12 = p

The probability of not selecting a brown peanut is .88 = q

The probability mass function is p(y) = (.88)^(y-1) * (.12)

a) p(7) = (.88)^6 * .12 = .0557

b) p(7 <= y <= 8) = p(7) + p(8)

= .0557 + (.88)^7 * .12 = .1048

c) p(y <= 7) = p(0) + p(1) + ... + p(7)

= .12 + (.88)^1 * .12 + (.88)^2 * .12 + ... + (.88)^6 * .12 = .4713

d) The expect value is 1/p. So, 1/(.12) = 8.33 M&M's

Answer 2

a)
`p(7) = 0.06`

b)
`p(7 \leq y \leq 8) = 0.10`

c)
`p(y \leq 7) =0.47`

d)
`(1)/(p) = 8.33 M\&M's`

Explanation:

We are using geometric distribution.

The probability of selecting a brown peanut is p = 0.12

The probability of not selecting a brown peanut is q = 0.88

probability mass function is p(y) = (0.88)^(y-1) * (0.12)

a)
`p(7) = (0.88)^6 *0.12 = 0.06`

b)
`p(7 \leq y \leq 8) = p(7) + p(8)`

`\ = 0.0557 + (0.88)^7 * 0.12 =0 .10`

c)
`p(y \leq 7) = p(0) + p(1) + ... + p(7)`

`= 0.12 + (0.88)^1 * 0.12 + (0.88)^2 * 0.12 + ... + (0.88)^6 * 0.12 = 0.47`

d) The expected value is
`(1)/(p)`.

So,

`(1)/(0.12) = 8.33 M\&M's`