Page 119 - DMTH404_STATISTICS

Unit 7: Modern Approach to Probability

The conditional probability of an event A given that D has already occurred, is given by Notes
k

P (A  D ) P ( ) ( /A .P D A )
P (A k /D )  k  k k .... (2)
( )
( )
P D P D
Substituting the value of P(D) from (1), we get
P ( ) ( /A .P D A )
P (A k /D )  n k k .... (3)
å P ( ) ( /A i .P D A i )
i 1

Example 54: 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 b g b  Dg b  Dg b  Dg
P C

P B
P A
+
+
a
a
= P A a fP D/Af+P B a fP D/Bf+P C a fP D/Cf
a
= 0.6 ´ 0.02 + 0.3 ´ 0.05 + 0.1 ´ 0.08 = 0.035
(b) (i) We have to find P(A/D)
P ( ) ( /A P D A ) 0.6 0.02
´
)
( /P A D    0.343
( )
P D 0.035
´
Similarly, (ii) P ( /B D  P ( ) ( /B P D B )  0.3 0.05  0.429
)
( )
P D 0.035
P ( ) ( /C P D C ) 0.1 0.08
´
)
and (iii) ( /P C D    0.228
( )
P D 0.035

LOVELY PROFESSIONAL UNIVERSITY 111