is I prize’ et mmm ay tae wt eno Solution. A can draw a ticket in °C, = 3 way ‘The number of cases in which A can get a prize is clearly 1. the probability of A's success = 9-8 32 Again B can draw a ticket in °C, = 84 ways. ‘The number of ways in which B gets all blanks = °C, = eee the number of ways of getting a prize = 84 — 20 = 64. ‘Thus the probability of B's success = 64/84 = 16/21. Hence A’s probability of success : B's probability of success PROBABILITY AND SET NOTATIONS (1) Random experiment. Experiments which are performed essentially under the same conditions and whose results cannot be predicted are known as random experiments. e.g., Tossing a coin or rolling a die are random experiments. (2) Sample space. The set of all possible outcomes of a random experiment is called sample space for that experiment and is denoted by S. ‘The elements of the sample space S are called the sample points. e.g, On tossing a coin, the possible outcomes are the head (H) and the tail (7). Thus S = (H, T). (8) Event. The outcome of a random experiment is called an event. Thus every subset of a sample space S is an event. The null set is also an event and is called an impossible event. Probability of an impossible event is zero ie, PQ) = (4) Axioms (@ The numerical value of probability lies between 0 and 1. i.e., for any event A of S,0.< P(A) <1. ) The sum of probabilities of all sample events is unity ive., P(S) = I. (iii) Probability of an event made of two or more sample events is the sum of their probabilities. (5) Notations (i) Probability of happening of events A or B is written as P(A + B) or P(A UB). (i) Probability of happening of both the events A and B is written as P (AB) or P(A 4B). (iii) ‘Event A implies (=) event B’is expressed as A B. (iv) ‘Events A and B are mutually exclusive’ is expressed as A > B = 6. (6) For any two events A and B, P(ANB) =P (A)-P (ANB) Proof. From Fig. 26.1, (ANB)U(ANB)=A PUAN B)U(A OB) = P(A) or P(ANB')+P(ANB)=P(A) or P(ANB’)=P(A)-P(AnB) Similarly, P(A’ 9 B) = P(B)~ P(A B)Ho Enameerine Maniatis | ADDITION LAW OF PROBABILITY or THEOREM OF TOTAL PROBABILITY (1) If the probability of an event A happening as a result of « trial is P(A) and the probability ofa mutually exclusive event B happening is P(B), then the probability of either of the events happening as a result ofthe trial is P(A +B) or P(A UB) = PA) + PB). Proof. Let n be the total number of equally likely cases and let m, be favourable to the event A and m, be favourable to the event B. Then the number of cases favourable to A or B is m, + mo. Hence the probability of A or B happening as a result of the trial = Mit, Te = P(A) + PIB). (2) If A, B, are any two events (not mutually exclusive), then P(A+B)=P(A) +P (B)-P (AB) or P(AUB)=P (A) +P (B)-P (ANB) Ifthe events A and B are any two events then, there are some outcomes which favour both A and B. If m, be their number, then these are included in both m, and m,. Hence the total number of outcomes favouring either A or B or both my + my—my. Thus the probability of occurrence of A or B or both PL n nin 7 Hence ‘PIA + B) = P(A) + P(B) — PAB) or PIA vB) = P\A) + PB) - P(A OB) (8) IFA, B, C are any three events, then P(A+B+C)=P (A) +P (B) +P (C)-P (AB) -P (BC) - P (CA) + P (ABC). P(AUBUC) =P (A) +P (B) +P (C)-P (ANB) -P (BAC)-P (COA) +P ANBOO) Proof. Using the above result for any two events, we have P(AUBUC)=PIAUB)UC) =P(AUB)+P(C)-PtAGB NC) = (IP (A)+ P(B)-P(ANB)] +P (C)-P(ANQC)UBOO) (Distributive Law) = P(A) + P(B) + P(C)~P(ANB)~ (P(A C)+P(BOC)-PIANBOO) (ANOOBAC)=AnBod = P(A) + P(B)+P(C)~ P(A B)~P(BOC)-P(COA)+PIANBOO) Is ANC=COAI Solution. Since it is not possible for all the horses to cover the same distance in the same time (a dead heat), the events are mutually exclusive. If, Pa» Py» Py be the probabilities of winning of the horses H,, H,, H, H,, respectively, then meet I: Odds in favour of H, are 1: 4] 1445 and P=