Here, you will learn the definition of bayes theorem and the formula for bayes theorem with example.
Let’s begin –
Formula for Bayes Theorem
Let an event A of an experiment occurs with its n mutually exclusive & exhaustive events \(B_1\), , \(B_2\), \(B_3\),………\(B_n\) & the probabilities P(A/\(B_1\)), P(A/\(B_2\))……..P(A/\(B_n\)) are known, then
P(\(B_i\)/A) = \(P(B_i).P(A/B_i)\over {\sum_{n=1}^{\infty}P(B_i).P(A/B_i)}\)
Explanation :
A = event what we have;
\(B_i\) = event what we want & remaining are alternative events.
Now, P(A\(B_i\)) = P(A).P(\(B_i\)/A) = \(P(B_i).P(A/B_i)\)
P(\(B_i\)/A) = \(P(B_i).P(A/B_i)\over {P(A)}\) = \(P(B_i).P(A/B_i)\over {\sum_{n=1}^{n}P(AB_i)}\)
P(\(B_i\)/A) = \(P(B_i).P(A/B_i)\over {\sum_{n=1}^{\infty}P(B_i).P(A/B_i)}\)
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 box is also of gold?
Solution : Let \(E_1\), \(E_2\), \(E_3\) be the events that boxes I,II and III are chosen, respectively.
Then P(\(E_1\)) = P(\(E_2\)) = P(\(E_3\)) = \(1\over 3\)
Also, Let A be the event that ‘the coin drawn is of gold’
Then P(A|\(E_1\)) = P(a gold coin from box I) = \(2\over 2\) = 1
P(A|\(E_2\)) = P(a gold coin from box II) = 0
P(A|\(E_3\)) = P(a gold coin from box III) = \(1\over 2\)
Now, the probability that the other coin in the box is of gold
= the probability that gold coin is drawn from the box I.
= P(\(E_1\)|A)
By Baye’s theorem, we know that
P(\(E_1\)|A) = \(P(E_1).P(A/E_1)\over {P(E_1).P(A/E_1) + P(E_2).P(A/E_2) + P(E_3).P(A/E_3)}\)
P(\(E_1\)|A) = \({{1\over 3}\times1}\over {{1\over 3}\times 1 + {1\over 3}\times 0 + {1\over 3}\times{1\over 2}}\) = \(2\over 3\)
Hope you learnt the formula for bayes theorem in probability. To learn more practice more questions and get ahead in competition. Good Luck!