Answer to Question #203691 in Statistics and Probability for Madimet

Question

A shopper goes to the grocery store and purchases a bag of 10 apples. Based on previous

experience, he knows that 5% of the apples of this brand will be bruised and inedible. Assume

that the apples are randomly assigned to the bag.

(i) State the mean and variance of the distribution of the number of bad apples in the bag.

(2)

(ii) What is the probability that in his bag he has either no bad apple or only one bad apple?

(4)

(iii) What is the probability that in his bag he has at least seven good apples?

Answer 1

Solution:

Let $p$ denotes success, i.e., getting bad apples.

$p=5\%=0.05$

$q=0.95$

$n=10$

$X\sim Bin(n,p)$

(i) Mean$=\mu=np=10(0.05)=0.5$

Variance$=\sigma=\sqrt{npq}=\sqrt{10(0.05)(0.95)}=0.6892$

(ii) Required probability = $P(X=0)+P(X=1)$

$=^{10}C_0(0.05)^0(0.95)^{10}+^{10}C_1(0.05)^1(0.95)^{9} \\=(0.95)^9(0.95+10\times0.05) \\=0.9138$

(iii) Now, Y denotes number of good apples.

$Y\sim Bin(10,0.95)$

$P(Y\ge7)=P(Y=7)+P(Y=8)+P(Y=9)+P(Y=10)$

$=^{10}C_7(0.95)^7(0.05)^{3}+^{10}C_8(0.95)^8(0.05)^{2}+^{10}C_9(0.95)^9(0.05)^{1} \\+^{10}C_{10}(0.95)^{10}(0.05)^{1}$

$=0.99897$