Definition of Mutually NonExclusive Events:
Two events A and B are said to be mutually non exclusive events if both the events A and B have atleast one common outcome between them.
The events A and B cannot prevent the occurrence of one another so from here we can say that the events A and B have something common in them.
For example, in the case of rolling a die the event of getting an ‘oddface’ and the event of getting ‘less than 4’ are not mutually exclusive and they are also known as compatible event.
The event of getting an ‘oddface’ and the event of getting ‘less than 4’ occur when we get either 1 or 3.
Let ‘X’ is denoted as event of getting an ‘oddface’ and
‘Y’ is denoted as event of getting ‘less than 4’
The events of getting an odd number (X) = {1, 3, 5}
The events of getting less than 4 (Y) = {1, 2, 3}
Between
the events X and Y the common outcomes are 1 and 3
Therefore, the events X and Y are compatible events/mutually nonexclusive.
Addition Theorem Based on Mutually NonExclusive Events:
If X and Y are two mutually Non Exclusive Events, then the probability of ‘X union Y’ is the difference between the sum of the probability of X and the probability of Y and the probability of ‘X intersection Y’ and represented as,
P(X ∪ Y) = P(X) + P(Y)  P(X ∩ Y)
Proof: The events X  XY, XY and Y  XY are pairwise mutually exclusive events then,
X = (X  XY) + XY,
Y = XY + (Y  XY)
Now, P(X) = P(X  XY) + P(XY)
or, P(X  XY) = P(X)  P (XY)
Similarly, P(Y  XY) = P(Y)  P(XY)
Again, P(X + Y) = P(X  XY) + P(XY) + P(Y  XY)
⇒ P(X + Y) = P(X)  P(XY) + P(XY) + P(Y)  P(XY)
⇒ P(X + Y) = P(X) + P(Y)  P(XY)
⇒ P(X + Y) = P(X) + P(Y)  P(X) P(Y)
Therefore, P(X ∪ Y) = P(X) + P(Y)  P(X ∩ Y)
Workedout problems on probability of Mutually NonExclusive Events:
1. What is the probability of getting a diamond or a queen from a wellshuffled deck of 52 cards?
Solution:
Let X be the event of ‘getting a diamond’ and,
Y be the event of ‘getting a queen’
We know that, in a wellshuffled deck of 52 cards there are 13 diamonds and 4 queens.
Therefore, probability of getting a diamond from wellshuffled deck of 52 cards = P(X) = 13/52 = 1/4
The probability of getting a queen from wellshuffled deck of 52 cards = P(Y) = 4/52 = 1/13
Similarly, the probability of getting a diamond queen from wellshuffled deck of 52 cards = P(X ∩ Y) = 1/52
According to the definition of mutually nonexclusive we know that, drawing of a wellshuffled deck of 52 cards ‘getting a diamond’ and ‘getting a queen’ are known as mutually nonexclusive events.
We have to find out Probability of X union Y.
So according to the addition theorem for mutually non exclusive events, we get;
P(X ∪ Y) = P(X) + P(Y)  P(X ∩ Y)
Therefore, P(X U Y) 
= 1/4 + 1/13  1/52 = (13 + 4  1)/52
= 16/52 = 4/13 
Hence, probability of getting a diamond or a queen from a wellshuffled deck of 52 cards = 4/13
2. A
lottery box contains 50 lottery tickets numbered 1 to 50. If a lottery ticket
is drawn at random, what is the probability that the number drawn is a multiple
of 3 or 5?
Solution:
Let X be the event of ‘getting a multiple of 3’ and,
Y be the event of ‘getting a multiple of 5’
The events of getting a multiple of 3 (X) = {3,6,9,12,15,18,21,24,27,30,
33,36,39,42,45,48}
Total number of multiple of 3 = 16
P(X) = 16/50 = 8/25
The events of getting a multiple of 5 (Y) = {5, 10, 15, 20, 25, 30, 35, 40, 45, 50}
Total number of multiple of 3 = 16
P(X) = 10/50 = 1/5
Between the events X and Y the favorable outcomes are 15, 30 and 45.
Total number of common multiple of both the number 3 and 5 = 3
The probability of getting a ‘multiple of 3’ and a ‘multiple of 5’ from the numbered 1 to 50 = P(X ∩ Y) = 3/50
Therefore, X and Y are non mutually exclusive events.
We have to find out Probability of X union Y.
So according to the
addition theorem for mutually non exclusive events, we get;
P(X ∪ Y) = P(X) + P(Y)  P(X ∩ Y)
Therefore, P(X U Y) 
= 8/25 + 1/5  3/50 = (16 + 10 3)/50
= 23/50 
Hence, probability of getting multiple of 3 or 5 = 23/50
`Probability
Probability of Tossing Two Coins
Probability of Tossing Three Coins
Probability for Rolling Two Dice
Probability for Rolling Three Dice
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