Mutually Non-Exclusive Events
Definition
of Mutually Non-Exclusive 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 ‘odd-face’ 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 ‘odd-face’ 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 ‘odd-face’ 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
non-exclusive.
Addition Theorem Based on Mutually Non-Exclusive 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 pair-wise 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)
Worked-out problems on probability of Mutually Non-Exclusive Events:
1. What is the probability of getting a diamond or a queen from a well-shuffled 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 well-shuffled deck of 52 cards there are 13 diamonds and 4 queens.
Therefore, probability of getting a diamond from well-shuffled deck of 52 cards = P(X) = 13/52 = 1/4
The probability of getting a queen from well-shuffled deck of 52 cards = P(Y) = 4/52 = 1/13
Similarly, the probability of getting a diamond queen from well-shuffled deck of 52 cards = P(X ∩ Y) = 1/52
According to the definition of mutually non-exclusive we know that, drawing of a well-shuffled deck of 52 cards ‘getting a diamond’ and ‘getting a queen’ are known as mutually non-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) |
= 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 well-shuffled 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 = 23/50 |
Hence, probability of
getting multiple of 3 or 5 = 23/50