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Unit 3: Conditional Probability and Independence Baye’s Theorem

Let A denote the event that a gold coin is located. The composition of the boxes implies that Notes

P(A | B ) = 1, P(A | B ) = 1/2, P(A | B ) = 0.
1 2 3
Since one gold coin is observed, we’ will have a gold coin in the other unobserved compartment
of the box only. if we have selected Box 1. Thus, we need to obtain P(B 1 A).

Now by Bayes Theorem

P(B )P(A|B )
P(B1 | A) = 1 1
P(B )P(A|B ) P(B )P(A|B ) P(B )P(A|B )


1 1 2 2 3 3
Do you feel confident enough to try and solve these exercises now? In each of them, the crucial
step is to define the relevant events properly. Once you do that, the actual calculation of
probabilities is child’s play.

Notes This is an example of a Markov chain, named after the Russian mathematician.
A, Markov (1856-1922) who initiated their study. This procedure is called Poly’s urn scheme.

Example 1: A manufacturing firm purchases a certain component, for its manufacturing
process, from three sub-contractors A, B and C. These supply 60%, 30% and 10% of the firm's
requirements, respectively. It is known that 2%, 5% and 8% of the items supplied by the respective
suppliers are defective. On a particular day, a normal shipment arrives from each of the three
suppliers and the contents get mixed. A component is chosen at random from the day's shipment:

(a) What is the probability that it is defective?
(b) If this component is found to be defective, what is the probability that it was supplied by
(i) A, (ii) B, (iii) C ?

Solution.
Let A be the event that the item is supplied by A. Similarly, B and C denote the events that the
item is supplied by B and C respectively. Further, let D be the event that the item is defective. It
is given that :
P(A) = 0.6, P(B) = 0.3, P(C) = 0.1, P(D/A) = 0.02
P(D/B) = 0.05, P(D/C) = 0.08.

(a) We have to find P(D)
From equation (1), we can write
  
P D P A  D  P B  D  P C  D 

  D/A
 P A P  P B P   D/B  P C P   D/C 
= 0.6  0.02 + 0.3  0.05 + 0.1  0.08 = 0.035
(b) (i) We have to find P(A/D)

  D/A
P A P  0.6 0.02


P A/D    0.343
P D 0.035
 

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