2. Introduction  Probability Theory was first used to solve problems in gambling  Blaise Pascal (1623-1662) - laid the foundation for the Theory of Probability  Now this theory is used in business, Science and industry  Next we define-Experiment and Event

3. Simple Probability  An Experiment is an operation or a process with a result or an outcome which is determined by, or depends on, Chance.  Examples: (1) Tossing a coin.  (2) Tossing a die  An Event is the outcome of an experiment  Example: Tossing a HEAD, getting a SIX are events.

4. Definition of Probability  In an experiment resulting in n equally likely outcomes, if m of these outcomes are favour the occurrence of an event E Then the Probability of event E happening, written as P(E), is defined as No. of outcomes favourable to the occurance of E Total number of equally likely outcomes P(E) = = m n

5. A two-digit number is written down at random. Find the probability that the number will be (i) smaller than 20 (ii) even (iii) a multiple of 5 Example 1:

6. How many two digit numbers are there? Is it 100 or 90 or 91 The correct answer is 90 Now we will find the probabilities. How many numbers are less than 20 ? How many are even numbers ? How many are multiples of 5 ? 10 45 18 Example 1:

7. Probability The possible outcomes is called the Sample space (S) Hence, the probability of an Event E, P(E) = n(E) n(S) If P(E) = 0, then the event cannot possibly occur If P(E) = 1, then the event will certainly occur. In the Probability Theory, an event is any subset of a Sample Space

8. Possible outcomes of the following experiments 1. Tossing a Coin: S = [ H, T ] 2. Tossing a die : S = [ 1, 2, 3, 4, 5, 6 ] 3. Tossing two coins: S = [ HH, HT, TH, TT ] 4. Tossing two dice : S = [ (1,1),(1,2) ….. (6,5),(6,6)] We can draw sample space for the above experiments

9. Example 2 : Two dice are thrown together. Find the probability that the sum of the resulting numbers is (a) odd (b) even (c) a prime number (d) a multiple of 4 (e) at least 7 First we draw the sample space, then using that we can find the probability.

10. All the possible sums are displayed in the above diagram This is called the sample space of this experiment + First die 12345 6 Second die 1 2 3 4 5 6 12 7 891011 234567 34567 84567 8 9 98 10 76 9 8 5 7 6

11. We define the following: A: the sum is odd B: the sum is even C: the sum is a prime number D: the sum is a multiple of 4 E: the sum is at least 7 Total possible outcomes are 36. Hence n(S) = 36 n(A) = 18 A and B are complementary events. n(B ) = 36 - n(A) n(C) = 15 n(D) = 9 n(E) = 21 Now, it is very easy to calculate the probabilities. Example 2 :

12. Answers: Hence P(E’) = 1 - P(E), where E’ is the complement of E

13. Box A contains 4 pieces of paper numbered 1,2,3,4 Box B contains 2 pieces of paper numbered 1,2. One piece of paper is removed at random from each box The sample space is as follows 1234 1 2 Box A Box B Another way to illustrate the possible outcome Example 3 :

14. TREE DIAGRAM Box A Box B Outcome 1 1 2 2 1 2 3 1 2 4 1 2 (1,1) (1,2) (2,1) (2,2) (3,1) (3,2) (4,1) (4,2) Hence n (S) = 8

15. A coin is tossed three times. Display all the outcomes using a tree diagram find the probability of getting (i) three heads (ii) exactly two heads (iii) at least two heads T H H [T,H,H] T H [H,T,H] H H H[H,H,H] T [H,H,T] T [H,T,T] T[T,H,T] T H[T,T,H] T [T,T,T] n (S) = 8 Now it is very easy to find the probabilities.

16. Adding Probabilities - Mutually Exclusive Events Two events A and B are said to be Mutually Exclusive(ME) if the occurance of one event will not affect the occurance of the other event. Set theoritically Hence these two events A and B cannot occur simultaneously. If you want to calculate the probability of A or B, then P(A or B )= P(AUB) = P(A) + P(B) Example: A - getting an odd number B - getting an even number, while tossing a die once Also, P(AUBUC) = P(A) + P(B) + P(C)

17. The probabilities of three teams L, M and N, winning a football competition are 1/4, 1/8 and 1/10 respectively. Calculate the probability that (i) either L or M wins, (ii) neither L nor N wins. Example 4 :

18.  We assume that only one team can win, so the events are mutually exclusive. (i) P( L or M wins) = P(Lwins) + P(M wins) = 1/4 + 1/8 = 3/8 ( ii) P(L or N wins) = 1/4 + 1/10 = 7/20 P(neither L nor N wins ) = 1 - P(L or N wins) = 1 - 7/20 = 13/20 Note: “Branches” of a “Probability Tree” represent outcomes which are mutually exclusive Example 4 :

19. Example 5 Consider the experiment, Tossing a die once let A - getting an odd number [ 1,3,5] B - getting a prime number [2,3,5] Here, A intersection B is not empty P(A) = 3/6 = 1/2, P(B) = 3/6 = 1/2 A B = [3,5], P(A B) = 2/6 = 1/3 P(AUB) = P(A) + P(B) - P(A B) = 3/6 + 3/6 - 2/6 = 4/6 = 2/3

20. TREE DIAGRAM Box A Box B Outcomes 3 1 2 2 1 2 1 1 2 4 1 2 (1,1) (1,2) (2,1) (2,2) (3,1) (3,2) (4,1) (4,2) Hence n (S) = 8

21. Now, we go back to the same example If we replace the numbers 1,2,3.. by the corresponding probabilities, we get Box A contains 4 pieces of paper numbered 1,2,3,4 Box B contains 2 pieces of paper numbered 1,2. One piece of paper is removed at random from each box

22. TREE DIAGRAM Box A Box B Probability P(1,1)= 1/8 1 1 P(3,2)=1/8 P(2,1)=1/8 P(3,1)=1/8 P(2,2)=1/8 P(4,2)=1/8 P(1,2)=1/8 P(4,1)=1/8 2 2 1 3 1 4 1 2 2 2 To find P(1,2), we multiply along the branches

23. Multiplying Probabilities - Probability Tree The probability that two events, A and B, will both occur, written as P(A occurs and B occurs) or simply P(A and B), is given by P(A and B ) = P(A) x P(B) Note: we have to multiply the probabilities along the branch,

24. Consider the following example: Suppose in a bag, there are 5 blue and 3 yellow marbles. A marble is drawn at random from the bag, the colour in noted and the marble is replaced. A second marble is then drawn. Y B Y Y B B 1 st draw 2 st draw This shows the Probability Tree Since the first marble drawn is replaced, the total number of marbles in the bag remains the same for the second draw.

25. Hence, we can say that the results of the two draws are independent and the two outcomes from each of the two draws are independent events. If the marble is not replaced, then the probability of selecting a marble from the second draw is affected. This kind of events are called dependent events Probability Tree diagram for dependent events B Y Y Y B B

26. A garden has three flower beds. The first bed has 20 daffodils and 20 tulips, the second has 30 daffodils and 10 tulips and the third has 10 daffodils and 20 tulips. A flower bed is to be chosen by throwing a die which has its six faces numbered 1,1,1,2,2,3. If the die shows a ‘1’, the first flower bed is chosen, if it shows a ‘2’ the second bed is chosen and so on. A flower is then to be picked at random from the chosen bed. Copy and complete the Probability Tree:

27. Bed 1 Bed 2 Bed 3 daffodil Tulip ( ) 2 2 2 4 6 1 1 1 1 Sample Space S = [ 1,1,1,2,2,3] [20] [30] [10] [20] [40] [30] 11 3 3 4 1 3 2 3 Prob. of picking a daffodil =

28. SUMMARY  In an experiment which results in n equally likely outcomes, if m of these outcomes favour the occurrence of an event E, then probability of the event E is given by P(E) = m / n, also 0  P(E)  1  The Sample Space refers to the set of all the possible outcomes of an experiment.  An event is any subset of the sample space

29.  If A and B are Mutually Exclusive, then P(A or B) = P(A) + P(B)  If A and B are independent, then P(A and B) = P(A) x P(B) SUMMARY