Bayes Theorem Examples
Example 1: A person has undertaken a job. The probabilities of completion of the job on time with and without rain are 0.44 and 0.95 respectively. If the probability that it will rain is 0.45, then determine the probability that the job will be completed on time.
Solution:
Let E1 be the event that the mining job will be completed on time and E2 be the event that it rains. We have,
P(A) = 0.45,
P(no rain) = P(B) = 1 − P(A) = 1 − 0.45 = 0.55
By multiplication law of probability,
P(E1) = 0.44, and P(E2) = 0.95
Since, events A and B form partitions of the sample space S, by total probability theorem, we have
P(E) = P(A) P(E1) + P(B) P(E2)
⇒ P(E) = 0.45 × 0.44 + 0.55 × 0.95
⇒ P(E) = 0.198 + 0.5225 = 0.7205
So, the probability that the job will be completed on time is 0.7205
Example 2: There are three urns containing 3 white and 2 black balls; 2 white and 3 black balls; 1 black and 4 white balls respectively. There is an equal probability of each urn being chosen. One ball is equal probability chosen at random. what is the probability that a white ball is drawn?
Solution:
Let E1, E2, and E3 be the events of choosing the first, second, and third urn respectively. Then,
P(E1) = P(E2) = P(E3) =1/3
Let E be the event that a white ball is drawn. Then,
P(E/E1) = 3/5, P(E/E2) = 2/5, P(E/E3) = 4/5
By theorem of total probability, we have
P(E) = P(E/E1) . P(E1) + P(E/E2) . P(E2) + P(E/E3) . P(E3)
⇒ P(E) = (3/5 × 1/3) + (2/5 × 1/3) + (4/5 × 1/3)
⇒ P(E) = 9/15 = 3/5
Example 3: A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both hearts. find the probability of the lost card being a heart.
Solution:
Let E1, E2, E3, and E4 be the events of losing a card of hearts, clubs, spades, and diamonds respectively.
Then P(E1) = P(E2) = P(E3) = P(E4) = 13/52 = 1/4.
Let E be the event of drawing 2 hearts from the remaining 51 cards. Then,
P(E|E1) = probability of drawing 2 hearts, given that a card of hearts is missing
⇒ P(E|E1) = 12C2 / 51C2 = (12 × 11)/2! × 2!/(51 × 50) = 22/425
P(E|E2) = probability of drawing 2 clubs ,given that a card of clubs is missing
⇒ P(E|E2) = 13C2 / 51C2 = (13 × 12)/2! × 2!/(51 × 50) = 26/425
P(E|E3) = probability of drawing 2 spades ,given that a card of hearts is missing
⇒ P(E|E3) = 13C2 / 51C2 = 26/425
P(E|E4) = probability of drawing 2 diamonds ,given that a card of diamonds is missing
⇒ P(E|E4) = 13C2 / 51C2 = 26/425
Therefore,
P(E1|E) = probability of the lost card is being a heart, given the 2 hearts are drawn from the remaining 51 cards
⇒ P(E1|E) = P(E1) . P(E|E1)/P(E1) . P(E|E1) + P(E2) . P(E|E2) + P(E3) . P(E|E3) + P(E4) . P(E|E4)
⇒ P(E1|E) = (1/4 × 22/425) / {(1/4 × 22/425) + (1/4 × 26/425) + (1/4 × 26/425) + (1/4 × 26/425)}
⇒ P(E1|E) = 22/100 = 0.22
Hence, The required probability is 0.22.
Example 4: Suppose 15 men out of 300 men and 25 women out of 1000 are good orators. An orator is chosen at random. Find the probability that a male person is selected. Assume that there are equal numbers of men and women.
Solution:
Gievn,
- Total Men = 300
- Total Women = 1000
- Good orators among Men = 15
- Good orators among Women = 25
Total number of good orators = 15 (from men) + 25 (from women) = 40
Probability of selecting a male orator:
P(Male Orator) = Numbers of male orators / total no of orators = 15/40
Example 5: A man is known to speak the lies 1 out of 4 times. He throws a die and reports that it is a six. Find the probability that is actually a six.
Solution:
In a throw of a die, let
E1 = event of getting a six,
E2 = event of not getting a six and
E = event that the man reports that it is a six.
Then, P(E1) = 1/6, and P(E2) = (1 – 1/6) = 5/6
P(E|E1) = probability that the man reports that six occurs when six has actually occurred
⇒ P(E|E1) = probability that the man speaks the truth
⇒ P(E|E1) = 3/4
P(E|E2) = probability that the man reports that six occurs when six has not actually occurred
⇒ P(E|E2) = probability that the man does not speak the truth
⇒ P(E|E2) = (1 – 3/4) = 1/4
Probability of getting a six ,given that the man reports it to be six
P(E1|E) = P(E|E1) × P(E1)/P(E|E1) × P(E1) + P(E|E2) × P(E2) [by Bayes’ theorem]
⇒ P(E1|E) = (3/4 × 1/6)/{(3/4 × 1/6) + (1/4 × 5/6)}
⇒ P(E1|E) = (1/8 × 3) = 3/8
Hence the probability required is 3/8.
Bayes’ Theorem
Bayes’ Theorem is used to determine the conditional probability of an event. It was named after an English statistician, Thomas Bayes who discovered this formula in 1763. Bayes Theorem is a very important theorem in mathematics, that laid the foundation of a unique statistical inference approach called the Bayes’ inference. It is used to find the probability of an event, based on prior knowledge of conditions that might be related to that event.
For example, if we want to find the probability that a white marble drawn at random came from the first bag, given that a white marble has already been drawn, and there are three bags each containing some white and black marbles, then we can use Bayes’ Theorem.
This article explores the Bayes theorem including its statement, proof, derivation, and formula of the theorem, as well as its applications with various examples.