Answer to Question #152697 in Macroeconomics for Devi Uli

Question

Q2 i) Suppose that four persons working in a factory are given the job of stamping expiry  date on the packet of buns. Person A, who stamps 20% of packets, fails to stamp the expiry date once in every 200 packets. Person B who stamps 60% of packets, fails to  stamp the expiry date once in every 100 packets. Person C who stamps 15% of packets,  fails to stamp the expiry date once in every 90 packets and Person D who stamps 5% of 

packets, fails to stamp the expiry date once in every 200 packets. If a buyer complains  that her packet of buns does not show the expiry date, what is the probability that,

a) It was Person A who failed to stamp?

b) It was stamped by Person B or C?

c) What is the probability that a packet bought shows an expiry date and it was  not stamped by B? 

Answer 1

P(person A) = P(A) = 20% = 0.2

P(person B) = P(B) = 60% = 0.6

P(person C) = P(C) = 15% = 0.15

P(person D) = P(D) = 5% = 0.05

P(failed to stamp | person A) = P(failed | A) = 1 in every 200 $= \frac{1}{200} = 0.005$

P(failed to stamp | person B) = P(failed | B) = 1 in every 100 $= \frac{1}{100} = 0.01$

P(failed to stamp | person C) = P(failed | C) = 1 in every 90 $= \frac{1}{90} = 0.011$

P(failed to stamp | person D) = P(failed | D) = 1 in every 200 $= \frac{1}{200} = 0.005$

a) Required probability = P(A | failed to stamp)

Bayes theorem

$P(A | failed) = \frac{P(A) \times P(failed | A)}{P(A) \times P(failed | A) + P(B) \times P(failed | B) + P(C) \times P(failed | C) + P(C) \times P(failed | C)} \\ = \frac{0.2 \times 0.005}{0.2 \times 0.002 + 0.6 \times 0.01 + 0.15 \times 0.011 + 0.05 \times 0.005} \\ = 0.2857$

b) Required probability = P(B | failed to stamp) + P(C | failed to stamp)

P(B | failed to stamp) $= \frac{P(B) \times P(failed | B)}{P(A) \times P(failed | A) + P(B) \times P(failed | B) + P(C) \times P(failed | C) + P(C) \times P(failed | C)} \\$

$= \frac{0.6 \times 0.01}{0.2 \times 0.002 + 0.6 \times 0.01 + 0.15 \times 0.011 + 0.05 \times 0.005} \\ = 0.1714$

P(C | failed to stamp) $= \frac{P(C) \times P(failed | C)}{P(A) \times P(failed | A) + P(B) \times P(failed | B) + P(C) \times P(failed | C) + P(C) \times P(failed | C)} \\$

$= \frac{0.15 \times 0.011}{0.2 \times 0.002 + 0.6 \times 0.01 + 0.15 \times 0.011 + 0.05 \times 0.005} \\ = 0.4714$

Required probability = P(B | failed to stamp) + P(C | failed to stamp)

= 0.1714 + 0.4714

= 0.6428

c) Required probability = P[B^c | failed to stamp]

= 1 - P(B | failed to stamp)

= 1 - 0.1714

= 0.8286