According to Master foods, the company that nan tures M&M;\'s 12% of peanut M&M;

Question

According to Master foods, the company that nan tures M&M;'s 12% of peanut M&M;'s are bro 15% are yellow, 12%are red, 23% are green. You randomly select five peanut M&M;'s from an extra-large bag of the candies. (Round all probabilities below to four decimal places, i e. your answer should look like 0.1234, not 0.1234444 or 12.3496) are blue, 23% are orange and 15% Compute the probability that exactly two of the five M&M;'s are red Compute the probability that two or three of the five M&M;'s are red Compute the probability that at most two of the five M&M;'s are red. Compute the probability that at least two of the five M&M;'s are red If you repeatedly select random samples of five peanut M&M;'s, on average how many do you expect to be red? (Round your answer to two decimal places ) red M&M;'s with what standard deviation? (Round your ans red M&M;'s wer to two decimal places .) Points possible: 20

Explanation / Answer

Ans:

Probability of being red,p=0.12

Binomial distribution with n=5,p=0.12

Let x be the number of red M&M 's in the sample.

P(x=k)=5Ck*0.12k*0.885-k

a)

P(x=2)=5C2*0.122*0.883=0.0981

b)

P(x=2 or 3)=P(x=2)+P(x=3)=5C2*0.122*0.883+5C3*0.123*0.882

=0.0981+0.0134=0.1115

c)P(x<=2)=P(x=0)+P(x=1)+P(x=2)

=5C0*0.120*0.885+5C1*0.121*0.884+5C2*0.122*0.883

=0.5277+0.3598+0.0981=0.9857

d)P(x>=2)=1-P(x=0)-P(x=1)=1-0.5277-0.3598=0.1125

e)Average=np=5*0.12=0.6

f)standard deviation=sqrt(5*0.12*0.88)=0.73

x p(x) 0 0.5277 1 0.3598 2 0.0981 3 0.0134 4 0.0009 5 0.0000

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