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2. 2 LECTURE 6 OF 6 CONDITIONAL PROBABILITY & BAYES’ THEOREM

3. 3 At the end of the lesson, students should be able to: (a) understand and use Bayes’ Theorem to solve probability problems

4. 4 REMEMBER – THE PREVIOUS LECTURE the conditional probability of A given B is written as P(A | B) The event that has already occurred The event whose probability is to be determined

5. 5 P(B | A) = 0.2 P(A | B) = ? When the condition is reversed, Bayes’ Theorem is used to solve such problems. If you are given How do you find ?

6. 6 A B B B’ A’ P( B | A’ ) P( A) P( B | A ) P( A’) P( B )=P(A) x P( B | A ) +P(A’) x P( B | A’ ) TOTAL PROBABILITY OF EVENT B = P(B)

7. 7 In general, if events A 1,A 2,.……., A n are mutually exclusive and exhaustive events, then the probability of event B is given by : P( B )=P(A 1 ) x P( B | A 1 ) +P(A 2 ) x P( B | A 2 ) + P(A 3 ) x P( B | A 3 )+…..+ P(A n ) x P( B | A n ) TOTAL PROBABILITY OF EVENT B = P(B) THE TOTAL PROBABILITY THEOREM

8. 8 P (A|B) : “the probability of A given B” P (B|A) : “the probability of B given A”

9. 9 BAYES’ THEOREM (2) (1) Substitute (2) into (1), we get :

10. 10 BAYES’ THEOREM where A 1, A 2, ….., A n are n mutually exclusive and exhaustive events so that A 1 A 2 ……. A n = S, the possibility space, and B is an arbitrary event of S ( i = 1,2,3,…..,n ). P(B) is the total probability of event B. Bayes’ Theorem is useful when we have to ‘ reverse the conditions ’ in a problem.

11. 11 Example 1 There are 12 red balls and 8 green balls in a bucket. Two balls are taken out in sequence without replacement. By using a tree diagram, find the probability that (a) the first ball is red (b) the second one is red if the first is red (c) the second one is red if the first is green (d) the second one is red (e) the first one is red if the second is red

12. 12 Solution: 1 st draw R1R1 G1G1 2 nd draw G2G2 R2R2 G2G2 R2R2 R ~ red ball G ~ green ball

13. 13 (a) P( first ball is red) = P(R 1 ) (b) P( R 2 | R 1 ) Direct from the tree diagram Or using the formula of conditional probability P( R 2 | R 1 )

14. 14 (c) P( R 2 | G 1 ) Direct from the tree diagram (d) P( R 2 )= P( R 1 ∩ R 2 ) + P(G 1 ∩ R 2 )

15. 15 (e) P(R 1 | R 2 ) = ‘Reverse condition’ use Bayes’ Theorem |

16. 16 Example 2 I travel to work by route A or route B. The probability that I choose route A is. The probability that I am late for work if I go via route A is and the corresponding probability if I go via route B is. (a)What is the probability that I am late for work on Monday ? (b)Given that I am late for work, what is the probability that I went via route B ?

17. 17 Solution: ROUTE B AL’ (not late) L (late) L’ (not late) L (late) ARRIVE AT WORK P(A) x P(L|A) P(B) x P(L|B)

18. 18 P(A) x P( L | A )(a) P ( L ) =+ P(B) x P( L | B) (b) BAYES’ THEOREM

19. 19 Example 3 Aishah, Siti and Muna pack biscuits in a factory. Aishah packs 55%, Siti 30% and Muna 15% from the batch allotted to them. The probability that Aishah breaks some biscuits in a packet is 0.7, and the respective probabilities for Siti and Muna are 0.2 and 0.1. What is the probability that a packet with broken biscuits found by the checker was packed by Aishah ?

20. 20 Solution: A B B B’ S M 0.3 0.55 0.7 B B’ 0.15 A – Aishah, S – Siti, M - Muna B – Broken Biscuits 0.3 0.2 0.8 0.1 0.9

21. 21 BAYES’ THEOREM

22. 22 Example 4 According to a firm’s internal survey, of those employees living more than 2 miles from work, 90% travel to work by car. Of the remaining employees, only 50% travel to work by car. It is known that 75% of employees live more than 2 miles from work. Determine : (i) the overall proportion of employees who travel to work by car. (ii) the probability that an employee who travels to work by car lives more than 2 miles from work.

23. 23 Define the events C, B 1, B 2 as follows : C : Travels to work by car B 1 : Lives more than 2 miles from work B 2 : Lives not more than 2 miles from work The events B 1 and B 2 are mutually exclusive and exhaustive. P(B 1 ) = 0.75, P(B 2 ) = 0.25 P( C | B 1 ) = 0.9and P( C | B 2 ) = 0.5 Solution:

24. 24 B2B2 B1B1 C’ C C (i) P(C) = P(B 1 ) x P( C | B 1 ) + P(B 2 ) x P( C | B 2 ) = ( 0.75 x 0.9 ) + ( 0.25 x 0.5 ) = 0.8 P( C | B 1 ) = 0.9P( C | B 2 ) = 0.5 P(B1) = 0.75,P(B2) = 0.25 Solution: 80% of employees travel to work by car.

25. 25 BAYES’ THEOREM (ii)

26. 26 BAYES’ THEOREM THE TOTAL PROBABILITY THEOREM P( B )=P(A 1 ) x P( B | A 1 ) +P(A 2 ) x P( B | A 2 ) + P(A 3 ) x P( B | A 3 )+…..+ P(A n ) x P( B | A n ) TOTAL PROBABILITY OF EVENT B = P(B)

27. 27 Exercise : 1.Three children, Azman, Mariam and Nasir, have equal plots in a circular patch of garden. The boundaries are marked out by pebbles, Azman has 80 red and 20 white flowers in her patch, Mariam has 30 red and 40 white flowers and Nasir has 10 red and 60 white flowers. Their young sister, Mumtaz, wants to pick a flower for her teacher. (a) Find the probability that she picks a red flower if she chooses a flower at random from the garden ignoring the boundaries. (b) Find the probability that she picks a red flower if she first chooses a plot at random.

28. 28 (c) If she picks a red flower by the method described in (b), find the probability that it comes from Mariam’s plot. Answer : (a) 0.5(b) (c)