Question

1. A quality control specialist at Albertsons grocery knows that there is an 11% chance that a given head of lettuce that is delivered to a grocery store will have significant defects that preclude the head of lettuce from being sold. Assume, 12 heads of lettuce are delivered to a store.
   • (a) What is the probability that exactly 2 heads of lettuce are not able to be sold due to defects?
   • (b) What is the probability that 2 or less heads of lettuce are not able to be sold due to defects?
   • (c) What is the probability that 3 or greater heads of lettuce are not able to be sold due to defects?
   • (d) What is the probability that no heads of lettuce are not able to be sold due to defects?
   • (e) What is the probability that at least one head of lettuce is not able to be sold due to defects?

Need help walking through these.

Answer 1

Given :

n = number of heads of lettuce are delivered =12

p = P( Given head of lettuce will have significant defect) = 0.11

X: Number of heads of lettuce are not able to be sold due to defects.

Since p is constant for each heads of lettuce and heads of lettuce are independent to each other.

The probability distribution of X is binomial with parameter n =12 and p = 0.11

X ~ Bin ( n=12, p = 0.11)

The p.m.f. of X is

a) Required probability = P(X=2)

b) Required probability = P ( X<=2) = P(X=0) + P(X=1) + P(X=2)

c) Required probability = P( X>=3) = 1- P(X<=2)

From part (b) P(X<=2) = 0.8623

P(X>=3) = 1-0.8623 =0.1377

d) Required Probability = P(X=0)

P(X=0) = 0.2469

e) Required probability = P(X>=1) =1 - P(X=0)

= 1- 0.2469

P(X>=1) =0.7531.