Page 43 - DMTH404_STATISTICS

Unit 3: Conditional Probability and Independence Baye’s Theorem

Solution. Notes

Let A be the event that first bag is selected, B be the event that second bag is selected and D be the
event of drawing a red ball.
Then, we can write

1 2 6 3


P
  
P B
P
P A ,    , D/A  , D/B 
3 3 10 8
Further, P D 1  6  2  3  9 .
  
3 10 3 8 20
P A  D  1 6 20 4

 P A/D     
 
P D 3 10 9 9
Example 4: In a certain recruitment test there are multiple-choice questions. There are 4
possible answers to each questio n out of which only one is correct. An intelligent student knows
90% of the answers while a weak student knows only 20% of the answers.
(i) An intelligent student gets the correct answer, what is the probability that he was guessing?
(ii) A weak student gets the correct answer, what is the probability that he was guessing?
Solution.
Let A be the event that an intelligent student knows the answer, B be the event that the weak
student knows the answer and C be the event that the student gets a correct answer.

P
(i) We have to find  A/C . We can write

  C/A
P  A  C  P A P 

P  A/C   .... (1)
 
P C P A P  P A P   C/A 
   C/A


P
P
It is given that P(A) = 0.90,  C/A  1  0.25 and C/A  1.0
4
P A
From the above, we can also write    0.10
Substituting these values, we get


P  A/C  0.10 0.25  0.025  0.027
0.10 0.25 0.90 1.0 0.925




P
(ii) We have to find  B/C . Replacing A by B , in equation (1), we can get this probability.

P
It is given that P(B) = 0.20,  C/B  0.25 and C/B  1.0
P
P B
From the above, we can also write    0.80
0.80 0.25 0.20


Thus, we get B/C    0.50
P

0.80 0.25 0.20 1.0 0.40


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