Probability and Statistics - uniform and binomial distributions

Question

Say 10% of peanut M&Ms are brown, 10% are yellow, 20% are red, 40% are blue, 10% are orange and 10% are green. You grab 10 peanut M&Ms from an extra-large bag of the candies.

Round all probabilities below to at least 3 decimal places.

What is the probability that

a) exactly three M&Ms are blue?    

b) three or four are blue?    

c) at most three are blue?    

d) at least three are blue?    

e) three are blue given at least three are blue?

f) On average, how many M&Ms do you expect to be blue? (Round your answer to two decimal places)

g) What is the standard deviation of the number of blue M&Ms? (Round your answer to two decimal places.)

Answer 1

Pr(B) = .4

Pr(not B) =1 - .4 = .6

Pr(B=3) = 10C3(.6)^7(.4)^3 = 10!/3!7!(.6)^7(.4)^3 = 10x9x8(.0279936)(.064)/3x2 = .21499

= about 21.5% chance of exactly 3 blue

Pr(B=3 or 4) = 10C3(.6)^7(.4)^3 + 10C4(.6)^6)(.4)^4 =.21499 + 10C4(.6)^6(.4)^4

Pr(B<or=3) = Pr(0,1,2, or 3) = 10C0(.6)^10 + 10C1(.6)^9(.4)+ 10C2(.6)^8(.4)^2 + 10C3(.6)^7(.4)^3

Pr(B=or>3) = 1-Pr(0,1,2) = 1-Pr(0,1,2,or3) +Pr(3)

Pr(3B/=3or>3) =1-Pr(B=3 or B>3)

on average expected number of blue = mean = np = 10(.4) = 4 blue M&Ms

n= number randomly selected

p = Pr(B)

q= Pr(not B)

variance = npq = 4(.6) = 2.4

standard deviation = sqr2.4 = about 1.55

which leads to an alternative method to solve the questions.

use the binomial distribution to approximate the normal distribution

find the difference in z scores between x = 2.5 and 3.5 (with 3 as the midpoint)

3.5 - 4 = -.5

.5/1.55 = -.3226

2.5-4 = -1.5

-1.5/1.55 = -.9677

use a z calculator to find the area between -.3226 < z < -.9677

= .2069 = about 21% probability that x is "exactly" = 3