MCom I Semester Statistical Analysis Probability Study Material notes ( Part 3 )

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MCom I Semester Statistical Analysis Probability Study Material notes ( Part 3 )

MCom I Semester Statistical Analysis Probability Study Material notes ( Part 3 ) : Calculation of Probability in Cas of Odds in Known Bernoulli Theorem Bayes Theorem Inverse Probability  Mathematical Expectation or Expected Value Advanced Miscellaneous Illustrations ( Most Important Chapter Wise  Notes for MCom I Semester Students )

 Probability Study Material
Probability Study Material

MCom I Semester Business Environment Collaboration Light Recent Changes Study Material Notes

Calculation of Probability in Case of Odds is Known

Sometimes, the possibility in favor of and against to is expressed in terms of odd. Hence, we must understand the concept of odds. In simple words, odds relate the chances in favor of an event to the chances against it. For instance, the odds are 3 : 1 that X will get a job, which means that there are 3 chances that he will get the job and 1 chance against his getting the job. This can also be converted into probability as getting the job = 3/4. Therefore, if the odds are a :b in favor of an

Further, it may be noted that the odds are a:b in favor of an event is the same as to say that the odds are b: an against the event.

If the probability of an event is p, then the odds in favor of its occurrence are p to 1-p and the odds against its occurrence are 1 – p to p.

Probability Study Material

Illustration 35.

(i) The odds in favor of A solving a question is 7 to 6, and the odds against B solving the same question is 11 to 8. What is the probability that if both of them try, the question will be solved.

 (ii) Only one out of three events A, B and C can occur. If the odds against A are 7 to 4 and against B are 6 : 3. Find out odds against C.

(iii) In a race, odds in favor of 4 horses, A, B, C and D are 1:3, 1:4. 1:5 and 1 : 6 respectively. Assuming that dead heat is impossible that is, two or more horses can not win simultaneously, find the probability that one of these horses will win the race.

Solution.

(i) The question would be solved if (a) A solves and B fails to solve it or (b) B solves it and A fails to solve it or (C) both A and B solve it. In other words, at least one of these two should solve the question. The probability of such situation can be computed as follows:

Bayes’ Theorem-Inverse Probability

Up till now we have been concerned with such problems in which our knowledge of various causes which may produce an event was enough to enable us to determine the chances of the happening of the event. We shall now discuss some problems of a reverse type. For example, we may know that an event has happened as a result of some one of a certain number of causes, and we may have to find out the probability of a particular cause being the true one. Such problems are known as problems of inverse probability. These probabilities are computed by Bayes’ rule.

Suppose a bag contains four black balls and five white ones. Another bag contains five black balls and six white ones. A black ball has been drawn from one of the bags and we have to find out the probability that it came from the first bag. It is clearly a question of inverse probability. Here we know that a black ball must have come either from the first bag or from the second one and we have to find out the probability of the first case.

Infact, one of the important applications of the conditional probability is in the computation of unknown probabilities, on the basis of the information supplied by the experiment or past records. That is, the applications of the results of probability theory involves estimating unknown probabilities and making decisions on the basis of new sample information. This concept is reffered to as Bayes’ Theorem.

The theorem owes its name to the British mathematician Thomas Bayes who propounded it through a paper published posthumously in 1763. The popularity of the theorem has been mainly because of its usefulness in revising a set of old probabilities called prior probabilities (derived subjectively or objectively) in the light of additional information made available and derive a set of new probabilities called the posterior probabilities. The revision of old (given) probabilities in the light of the additional information supplied by the experiment or past records is of extreme help to business and management executives in arriving at valid decisions in the face of uncertainties.

In short, Bays’ Theorem may be stated as follows:

“An event is known to have proceeded from one of n mutually exclusive causes whose probabilities are P1, P2,…,P Further more let P1P2…., be the respective probabilities that when one of the n causes exists the event will then have followed. The probability that event proceeded from the mth cause is then :

P.P. + P P. + P3P3 + … + P.P.

The following example illustrates the application of Bays’ Theorem.

Probability Study Material

Illustration 38.

A bag contains 8 white and 4 black balls. Another bag contains 5 white and 3 black balls. A black ball was drawn from one of the two bags. What is the probability that it was drawn (i) from the first bag and (ii) from the second bag ?

 

(1) Expected Number : If p happens to be the probability of the happening of an event in a single trial, then the expected number of occurrences of that event in n trials is given by n.p. (where n means the number of trials and p means the probability of happening of an event) thus, the expectation may be regarded as the likely number of success in n trials.

(2) Expected Value : The expected value of a random variable is obtained by considering the values that the variable can take multiplying these by their corresponding probabilities and then summing these products.

If X denotes a discrete random variable which can assume the values X1, X2, X3, …, X, with respective probabilities P.P.P. ….P k where P. + P2 + P3 + … + P k = 1, the mathematical expectation of X denoted by E(X) is defined as

E(X) = p X, +p X, + P3X3 + … +PX Thus, the expected value is the sum of the product of the value of X and the probability attached with each event.

In calculating the mathematical expectation, following points should be kept in mind :

Probability Study Material

(1) If various possibilities of the events are positive then we calculate the sum of every expectation as E(X) = P1X1 + X: +…

(2) In case the expected possibility carries both positive and negative expectation then we get the net difference of gain and loss as

E(X) = p.x (positive)-pox (negative)

The mathematical expectations are used to compare different possibilities under risks and un-certainties and to choose comparatively better expectations.

The significance of this expectation lies in the fact that if a player pays more than this, by way of fair price, per game then he is sure to lose but if he plays long enough and if he pays less than his expectation per game he is certain to win in the long run. It is on this principle that speculators and businessmen take decisions in real life situations.

Illustration 44.

(i) The probability of a businessman’s earning a profit of Rs. 5,000 in grain business is 45%, the probability of his earning a profit of Rs. 3,000 in sugar business is 35% and the probability of his earning a profit of Rs. 2,000 in cement business is 20%. In case he starts all the three businesses, what is the amount of profit expected to be received by him ?

(ii) If the probability of the above businessman’s earning a profit of Rs. 5,000 in grain business is 50% and the probability of his incurring a loss of Rs. 1,000 in the same business is 50%, calculate his expected profit from it.

Solution.

= Rs. 2,250

(i) Expected profit in grain business = Rs. 5,00

Expected profit in sugar business = Rs. 3,00 350 = Rs. 1,050

Expected profit in cement business = Rs. 2,000 x 100 = Rs. 400

 

Probability Study Material

 

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