Consider that there are two bags I and II. Bag I contains 2 white and 3 red balls and Bag II contains 4 white and 5 red balls. One ball is drawn at random from one of the bags. We can find the probability of selecting any of the bags (i.e. 1 /2 ) or probability of drawing a ball of a particular colour (say white) from a particular bag (say Bag I). In other words, we can find the probability that the ball drawn is of a particular colour, if we are given the bag from which the ball is drawn. But, can we find the probability that the ball drawn is from a particular bag (say Bag II), if the colour of the ball drawn is given? Here, we have to find the reverse probability of Bag II to be selected when an event occurred after it is known. Famous mathematician, John Bayes’ solved the problem of finding reverse probability by using conditional probability. The formula developed by him is known as ‘Bayes theorem’ which was published posthumously in 1763.
We shall now state and prove the Bayes’ theorem.
Bayes’ Theorem If E1, E2 ,…, En are n non empty events which constitute a partition of sample space S, i.e. E1, E2 ,…, En are pairwise disjoint and E1∪ E2∪ … ∪ En = S and A is any event of nonzero probability, then
Proof By formula of conditional probability, we know that
Remark The following terminology is generally used when Bayes’ theorem is applied.
The events E1, E2, …, En are called hypotheses.
The probability P(Ei) is called the priori probability of the hypothesis Ei
The conditional probability P(Ei |A) is called a posteriori probability of the hypothesis Ei.
Bayes’ theorem is also called the formula for the probability of “causes”. Since the Ei‘s are a partition of the sample space S, one and only one of the events Ei occurs (i.e. one of the events Ei must occur and only one can occur). Hence, the above formula gives us the probability of a particular Ei (i.e. a “Cause”), given that the event A has occurred.
The Bayes’ theorem has its applications in variety of situations, few of which are illustrated in following examples.
Example Bag I contains 3 red and 4 black balls while another Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it was drawn from Bag II.
Solution Let E1 be the event of choosing the bag I, E2 the event of choosing the bag II and A be the event of drawing a red ball.
Now, the probability of drawing a ball from Bag II, being given that it is red, is P(E2|A)
By using Bayes’ theorem, we have
Example Given three identical boxes I, II and III, each containing two coins. In box I, both coins are gold coins, in box II, both are silver coins and in the box III, there is one gold and one silver coin. A person chooses a box at random and takes out a coin. If the coin is of gold, what is the probability that the other coin in the box is also of gold?
Solution Let E1, E2 and E3 be the events that boxes I, II and III are chosen, respectively.
Also, let A be the event that ‘the coin drawn is of gold’
Now, the probability that the other coin in the box is of gold
By Bayes’ theorem, we know that