by Addition Theorem on probability:
If A and B are any two events then the probability of happening of at least one of the events is defined as P(AUB) = P(A) + P(B)- P(A∩B).
Since events are nothing but sets,
From set theory, we have
n(AUB) = n(A) + n(B)- n(A∩B).
Dividing the above equation by n(S), (where S is the sample space)
n(AUB)/ n(S) = n(A)/ n(S) + n(B)/ n(S)- n(A∩B)/ n(S)
Then by the definition of probability,
P(AUB) = P(A) + P(B)- P(A∩B).
Example:
If the probability of solving a problem by two students George and James are 1/2 and 1/3 respectively then what is the probability of the problem to be solved.
Solution:
Let A and B be the probabilities of solving the problem by George and James respectively.
Then P(A)=1/2 and P(B)=1/3.
The problem will be solved if it is solved at least by one of them also.
So, we need to find P(AUB).
By addition theorem on probability, we have
P(AUB) = P(A) + P(B)- P(A∩B).
P(AUB) = 1/2 +.1/3 – 1/2 * 1/3 = 1/2 +1/3-1/6 = (3+2-1)/6 = 4/6 = 2/3
Note:
If A and B are any two mutually exclusive events then P(A∩B)=0.
Then P(AUB) = P(A)+P(B).
Multiplication theorem on probability:
If A and B are any two events of a sample space such that P(A) ≠0 and P(B)≠0, then
P(A∩B) = P(A) * P(B|A) = P(B) *P(A|B).
Example: If P(A) = 1/5 P(B|A) = 1/3 then what is P(A∩B)?
Solution: P(A∩B) = P(A) * P(B|A) = 1/5 * 1/3 = 1/15
INDEPENDENT EVENTS:
Two events A and B are said to be independent if there is no change in the happening of an event with the happening of the other event.
i.e. Two events A and B are said to be independent if
P(A|B) = P(A) where P(B)≠0.
P(B|A) = P(B) where P(A)≠0.
i.e. Two events A and B are said to be independent if
P(A∩B) = P(A) * P(B).
Example:
While laying the pack of cards, let A be the event of drawing a diamond and B be the event of drawing an ace.
Then P(A) = 13/52 = 1/4 and P(B) = 4/52=1/13
Now, A∩B = drawing a king card from hearts.
Then P(A∩B) = 1/52
Now, P(A/B) = P(A∩B)/P(B) = (1/52)/(1/13) = 1/4 = P(A).
So, A and B are independent.
[Here, P(A∩B) = = = P(A) * P(B)]
Note:
(1) If 3 events A,B and C are independent the
P(A∩B∩C) = P(A)*P(B)*P(C).
(2) If A and B are any two events, then P(AUB) = 1-P(A’)P(B’).
