A farmer ships many cartons of eggs everyday. Two inspectors independently glanc

Question

A farmer ships many cartons of eggs everyday. Two inspectors independently glance at each carton to determine whether there are any cracked eggs, and each of them has a .8 probability of noticing when a carton contains cracked eggs.

a) What is the probability that a carton with cracked eggs will get past both inspectors?
b) what is the probability that a carton with cracked eggs will be noticed by at least one inspector?
c) Suppose a third inspector (inexperienced) is hired who has a probability of only .6 of noticing when a carton contains one or more cracked eggs. What is the probability that a carton with cracked eggs will get past all three inspectors?

d)What is the probability that a carton containing cracked eggs will be noticed by at least one of these three inspectors?

Explanation / Answer

Solution:-

a) The probability that a carton with cracked eggs will get past both inspectors is 0.04.

Probability of noticing when a carton contains cracked eggs = 0.8

Probability of not noticing when a carton contains cracked eggs = 0.2

The probability that a carton with cracked eggs will get past both inspectors = 0.2 × 0.2 = 0.04

b) The probability that a carton with cracked eggs will be noticed by at least one inspector is 0.96.

Different ways by which it can be noticed by atleast one inspector:-

P(Y,Y) = 0.8 × 0.8 = 0.64

P(Y,N) = 0.8 × 0.2 = 0.16

P(N,Y) = 0.2 × 0.8 = 0.16

The probability that a carton with cracked eggs will be noticed by at least one inspector = 0.16 + 0.16 + 0.64 = 0.96

c) The probability that a carton with cracked eggs will get past all three inspectors is 0.016

Probability of not noticing when a carton contains cracked eggs 1 inspector = 0.2

Probability of not noticing when a carton contains cracked eggs 2nd inspector = 0.2

Probability of not noticing when a carton contains cracked eggs 3 inspector = 0.4

The probability that a carton with cracked eggs will get past all three inspectors = 0.2 × 0.2 × 0.4 = 0.016

d) The probability that a carton containing cracked eggs will be noticed by at least one of these three inspectors is 0.984.

Different ways by which it can be noticed by atleast one inspector:-

P(Y, Y, Y) = 0.8 × 0.8 × 0.6 = 0.384

P(Y, Y, N) = 0.8 × 0.8 × 0.4 = 0.256

P(Y, N, Y) = 0.8 × 0.2 × 0.6 = 0.096

P(Y, N, N) = 0.8 × 0.2 × 0.4 = 0.064

P(N, Y, Y) = 0.2 × 0.8 × 0.6 = 0.096

P(N, Y, N) = 0.2 × 0.8 × 0.4 = 0.064

P(N, N, Y) = 0.2 × 0.2 × 0.6 = 0.024

The probability that a carton containing cracked eggs will be noticed by at least one of these three inspectors is 0.984

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