[Maths Class Notes] on Bayes’ Theorem Formula Pdf for Exam

Bayes’ theorem relates to the concept of probability in mathematics. It is one of the formulas that have their practical application in almost all the areas that derive their results from predictions and probability. One can think of weather predictions where the meteorologists use the Bayes’ Theorem to predict the day-to-day weather forecast. The scientist can even forewarn the people about prevailing possibilities of bad weather so they can make the required arrangements. Similarly, there are many other fields where Bayes’ Theorem is applied. Let’s see more details about Bayes’ theorem.

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Components of Bayes’ Theorem 

Bayes’ theorem is a tool to calculate the probability of an event taking place. This could be done using Conditional probability. There are two types of probabilities namely, dependent and independent ones.  This can be explained using the example of coins. If a person tosses a single coin the probability of getting heads depends upon the probability of getting tales. This is called a dependent event. On the other hand, if you toss two different coins the probability of getting heads or tales on two coins doesn’t depend on each other. This is called an independent event. The concept of conditional probability under Bayes’ theorem calculates dependent events.

Bayes’ Theorem of Probability 

We can revise probabilities when new or additional information is supplied by a random experiment or ds. 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 conditional probability. 

What is a Conditional Probability? 

Sometimes an event or an outcome occurs based on 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.

Understanding Conditional Probability 

The concept of the conditional probability of Bayes’ theorem could be better explained using the marble example. Following is the example:

Example: 

There is a bag of five marbles. Three marbles are red and two are black in colour. A student is asked to pick out the black marbles. Here the probability of picking out black marble is dependent upon the other marbles that give us equation ⅖.

 

Now a student is asked to pick out another black marble from the bag. The probability of picking out this black marble is ¼. The possibility of picking black marble now depends upon the event of taking out black marble that was taken out before. This is called conditional probability which is a dependent event. 

Bayes’ Theorem Solved Examples 

Given below are a few Bayes’ Theorem examples that will help you to solve problems easily.

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, 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} times 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).== 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 occur 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 2000, 2500, and 4000 bolts every day. Of their output 3%, 4%, and 2.5% are defective bolts. One of the bolts 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 is 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, 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}]