Table of Contents
Conditional Probability
To understand Conditional Probability let us have examples from the real life situations. In the experiment of throwing a dice one might be interested in the probability of the event ‘multiple of 3’, given that the event ‘even number’ occurs in a particular throwing. In the experiment of finding the life of a light bulb, we might be interested in the probability that the bulb will last 75 hours, given that it has already lasted 30 hours. Probability questions of the above type are considered in the framework of conditional probability.
Definition
Let E be a given random experiment and X, Y be two events of E where P(Y) is not zero. The conditional probability of the event X on the hypothesis that the event Y has happened, denoted by P(X | Y), is defined by
\[P(X|Y)=\underset{N\to \infty }{\mathop{\lim }}\,\frac{N(X\cap Y)}{N(Y)}\]
assuming that the limit exists.
If X, Y are two events of a given random experiment, then
\[P(X\cap Y)=P(X|Y)P(Y),P(Y)\ne 0\]
\[P(X\cap Y)=P(Y|X)P(X),P(X)\ne 0\]
Proof:
Let E be the given random experiment and let E be repeated under identical conditions N times. If N(XY), N(X), N(Y) be respectively the number of occurrences of the events XY, X, Y then
\[P(X|Y)=\underset{N\to \infty }{\mathop{\lim }}\,\frac{N(X\cap Y)}{N(Y)}\]
\[=\underset{N\to \infty }{\mathop{\lim }}\,\frac{\frac{N(X\cap Y)}{N}}{\frac{N(Y)}{N}}\]
\[=\frac{\underset{N\to \infty }{\mathop{\lim }}\,\frac{N(X\cap Y)}{N}}{\underset{N\to \infty }{\mathop{\lim }}\,\frac{N(Y)}{N}}\]
\[=\frac{P(X\cap Y)}{P(Y)}\]
Hence we get,
\[P(X\cap Y)=P(X|Y)P(Y),P(Y)\ne 0\]
It can be proved similarly that,
\[P(X\cap Y)=P(Y|X)P(X),P(X)\ne 0\]
Let S be the sample space of a random experiment with the events E1, E2, E3, …. , En which are n number of mutually exclusive and exhaustive events and if A be an event which occurs with one or more events of E1 or E2 … En then
\[P(A)=P({{E}_{1}}).P(A|{{E}_{1}})+P({{E}_{2}}).P(A|{{E}_{2}})+…+P({{E}_{n}}).P(A|{{E}_{n}})\], \[P({{E}_{i}})>0\]
\[P(A)=\sum\limits_{i=1}^{n}{P({{E}_{i}}}).P(A|{{E}_{i}})\]
A bag contains 4 red and 3 black color balls and another bag contains 2 red and 4 black color balls. Now randomly we have to choose one bag and choose a ball from that bag. Now what is the probability that the ball is red?
We can have red ball in two ways:
1. We have choose the first bag and the ball is red
2. We have choose the second bag and the ball is red
The events are X = ‘first bag chosen’ , Y = ‘Second bag chosen’ , and Z = ‘red ball chosen’
\[\therefore P(X)=\frac{1}{2},P(Y)=\frac{1}{2}\]
Now, the probability of chosen red ball when already first bag is chosen
\[P(Z|X)=\frac{4}{7}\]
[As the first bag contains 4 red balls and 3 black balls]
The probability of chosen red ball when already second bag is chosen
\[P(Z|Y)=\frac{2}{6}\]
[As the second bag contains 2 red balls and 4 black balls]
So, the probability of getting red ball is
\[P(A)=P(X).P(Z|X)+P(Y).P(A|Y)\]
\[=\frac{1}{2}.\frac{4}{7}+\frac{1}{2}.\frac{2}{6}\]
\[=\frac{19}{42}\]
A bag contains 6 red balls and 5 blue balls, another bag also contains 5 red balls and 8 blue balls. A ball is picked from first bag randomly and put it in second bag. Then a ball is randomly picked from second bag. Find the probability that the ball is blue.
Solution:
Let, the events E1 = ‘the ball picked from the first bag is blue’, E2 = ‘the ball picked from the first bag is red’ and E = ‘the ball picked from the second bag is blue’.
Since, first bag has 6 red and 5 blue balls,
\[\therefore P({{E}_{1}})=\frac{5}{11};P({{E}_{2}})=\frac{6}{11}\]
If E1 already happened i.e., if blue ball is picked from first bag and put in the second bag then second bag will have 5 red balls and 9 blue balls. Then the probability of picking the blue ball from the second bag is,
\[P(E|{{E}_{1}})=\frac{9}{14}\]
Similarly,
\[P(E|{{E}_{2}})=\frac{8}{14}\]
Now using the rule of total Probability we get,
\[P(E)=P({{E}_{1}}).P(E|{{E}_{1}})+P({{E}_{2}}).P(E|{{E}_{2}})\]
\[=\frac{5}{11}.\frac{9}{14}+\frac{6}{11}.\frac{8}{14}\]
\[=\frac{93}{154}\]
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