 # Bayes' Theorem Formula  View Notes

Bayes’ Theorem of Probability

We can revise probabilities when new or additional information is supplied by a random experiment or past records. The revision of old (given) probabilities in the light of additional information is of immense help to business and management executives in arriving at a valid decision in the face of uncertainties. The procedure for revising probabilities due to a specific cause is known as Bayes’ theorem and it was originally developed by Rev. Thomas Bayes. It gives a probability law relating a posteriori probability to a priori probability. Bayes’ theorem formula is actually of great help if we want to calculate the conditional probability.

What is a Conditional Probability?

Sometimes an event or an outcome occurs on the basis of previous occurrences of events or outcomes, this is known as conditional probability. We can calculate the conditional probability if we multiply the probability of the preceding event by the updated probability of the succeeding, or conditional, event.

Bayes Theorem Examples

Given below are a few Bayes theorem examples that will help you to solve problems easily. For more practice, you can also go through a lot of Bayes theorem examples present on the internet.

Example 1) Three identical boxes contain red and white balls. The first box contains 3 red and 2 white balls, the second box has 4 red and 5 white balls, and the third box has 2 red and 4 white balls. A box is chosen very randomly and a ball is drawn from it. If the ball that is drawn out is red, what will be the probability that the second box is chosen?

Solution 1) Let A1, A2, and A3 represent the events of choosing the first, second, and third box respectively, and let X be the event of drawing a red ball from the chosen box.

Then we are to find the value of P(A2/X)

Since the boxes are identical, hence

P(A1) = P(A2) = P(A3) 1/3

Again, by the problem

P(X/A1) = 3/3+2 = 3/5;          P(X/A2) = 4/4+5 = 4/9; and           P(X/A3) = 2/2+4 = ⅓

Now, the event X occurs if one of the mutually exclusive and exhaustive events A1, A2, and A3 occurs. Therefore, using Bayes’ theorem formula we get,

P($\frac{A2}{X}$) = $\frac{P(A2).P(\frac{X}{A2})}{P(A1).P(\frac{X}{A1})+P(A2).P(\frac{X}{A2})+P(A3).P(\frac{X}{A3})}$

=$\frac{\frac{1}{3}.\frac{4}{9}}{\frac{1}{3}.\frac{3}{5}+\frac{1}{3}.\frac{4}{9}+\frac{1}{3}.\frac{1}{3}}$ = $\frac{\frac{4}{9}}{\frac{62}{45}}$ = $\frac{4}{9}$ x $\frac{45}{62}$ = $\frac{10}{31}$

Example 2) Two urns contain respectively 2 red, 3 white, and 3 red, 5 white balls. One ball is drawn at random from the first urn and transferred into the second one. A ball is then drawn from the second urn and it turns out that the ball is red. What will be the probability that the transferred ball was white?

Solution 2) Let A1 and A2 denote the events that the transferred ball from the first urn to the second is white and red respectively. Then,

P(A1) = 3/3+2 = 3/2 and P(A2) = 2/3+2 = 2/5

Again, let X denote the event of drawing a red ball from the second urn after the occurrence of A1 or A2. then we are to find the value of P(A1/X).

Clearly, P(X/A1) = 3/3+5+1 = 3/9 = ⅓ and P(X/A2) = 3+⅓+5+1 = 4/9

Now the event X occurs if any of the mutually exclusive and exhaustive events A1 and A2 occurs. Therefore, using the Bayes' theorem formula we get,

P($\frac{A1}{X}$) = $\frac{P(A1).P(\frac{X}{A1})}{P(A1).P(\frac{X}{A1})+P(A2).P(\frac{X}{A2})}$

=$\frac{\frac{3}{5}.\frac{1}{3}}{\frac{3}{5}.\frac{1}{3}+\frac{2}{5}.\frac{4}{9}}$ = $\frac{\frac{1}{5}}{\frac{17}{45}}$ = $\frac{9}{17}$

Example 3) In a bolt factory, three machines M1, M2, and M3 manufacture respectively 2000, 2500, and 4000 bolts every day. Of their output 3%, 4%, and 2.5% are defective bolts. One of the bolt is drawn very randomly from a day’s production and is found to be defective. What is the probability that it was produced by machine M2?

Solution 3) let A1, A2, and A3 denote the events that the randomly drawn bolt from a day’s production was manufactured by machines M1, M2, and M3 respectively. If X be the event that the drawn bolt is defective, then we are to find the value of P)A2/X)

Now by the question we have,

P(A1) = $\frac{2000}{2000+2500+4000}$ =  $\frac{2000}{8500}$ = $\frac{4}{17}$

P(A2) = $\frac{2500}{2000+2500+4000}$ =  $\frac{2500}{8500}$ = $\frac{5}{17}$

P(A3) = $\frac{4000}{2000+2500+4000}$ =  $\frac{4000}{8500}$ = $\frac{8}{17}$

Again, P(X/A1) = 3/100, P(X/A2) = 4/100 and P(X/A3) = 2.5/100

Now, the event X occurs if one of the mutually exclusive and exhaustive events A1, A2, and A3 occurs. Therefore, using Bayes' theorem formula we get,

P($\frac{A2}{X}$) = $\frac{P(A2).P(\frac{X}{A2})}{P(A1).P(\frac{X}{A1})+P(A2).P(\frac{X}{A2})+P(A3).P(\frac{X}{A3})}$

=$\frac{\frac{5}{17}.\frac{4}{100}}{\frac{4}{17}.\frac{3}{100}+\frac{5}{17}.\frac{4}{100}+\frac{8}{17}.\frac{25}{100}}$ = $\frac{20}{12+20+20}$ = $\frac{20}{52}$ =  $\frac{5}{13}$

Question 1: What is a Mutually Exclusive Event?

Answer: two events A and B connected with a random experiment E, are said to be mutually exclusive if they cannot occur simultaneously. Symbolically, events A and B are mutually exclusive when A⋂B = Ø or P(A⋂B) = 0, Where Ø is the impossible event. Two simple events connected with a random experiment are always mutually exclusive but two compound events may be or may not be so. Let A, B, and C be the events ‘even face’, ‘odd face’, and ‘a multiple of three’ respectively in the random experiment of throwing an unbiased die. Clearly the events A and B cannot occur simultaneously and hence, they are mutually exclusive, but the events B and C occur simultaneously if the result of the experiment is three and hence, they are not mutually exclusive.

Question 2: What is an Exhaustive Event?

Answer: A set of events connected with a random experiment, is said to be exhaustive id at least one of the set is sure to occur at every performance of the experiment. Simple events connected with a random experiment always constitute an exhaustive set of events. Consider the random experiment of throwing an unbiased die from the box. Let A1, A2,......A6 be the events ‘one’, ‘two’,..... ‘Six’, respectively. Clearly, at least one of these events will occur at every performance of the experiment and hence, they form an exhaustive set of events. In the same experiment, let A, B, and Cbe the events ‘even face’, ‘multiple of three’, and ‘five’ respectively.  Obviously none of these events A, B, and C occur when the outcome of the experiment is ‘one’ and hence, the set of events A, B, and C are not an exhaustive set where D denotes at least one of these four events must necessarily occur at every performance of the experiment.

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