Let us assume that a number is selected at random from $1, 2, 3$. We define
$$A = \{1, 2\},\quad B = \{2, 3\},\quad C = \{1, 3\}$$
Then are $A$, $B$ and $C$ mutually independent or pairwise independent or both?
I am confused between mutually vs pairwise independent.
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$\begingroup$Mutual independence: Every event is independent of any intersection of the other events.
Pairwise independence: Any two events are independent.
$A, B, C$ are mutually independent if $$P(A\cap B\cap C)=P(A)P(B)P(C)$$ $$P(A\cap B)=P(A)P(B)$$ $$P(A\cap C)=P(A)P(C)$$ $$P(B\cap C)=P(B)P(C)$$
On the other hand, $A, B, C$ are pairwise independent if $$P(A\cap B)=P(A)P(B)$$ $$P(A\cap C)=P(A)P(C)$$ $$P(B\cap C)=P(B)P(C)$$
I'm sure you can solve your problem now.
$\endgroup$ 8 $\begingroup$Pairwise Independent Conditions:
- P(AB) = P(A)P(B)
- P(BC) = P(B)P(C)
- P(AC) = P(A)P(C)
Mutual Independent Conditions:
- P(AB) = P(A)P(B)
- P(BC) = P(B)P(C)
- P(AC) = P(A)P(C)
- P(ABC) = P(A)P(B)P(C)
Let us take an example to understand the situation clearly.
A lot contains 50 defective and 50 non-defective pens. Two pens are drawn at random, one at a time, with replacement. The events A, B, C are defined as :
- A = ( the first pen is defective)
- B = (the second pen is non-defective)
- C = (the two pens are both defective or both non-defective).
Determine whether:
- (i) A, B, C are pairwise independent.
- (ii) A, B, C are independent. [IIT 1992]
Solution :
- D := Defective , N := Not Defective
- P(A) = P{the first pen is defective} = P($[D_{1}D_{2},D_{1}N_{2}] ) = (\frac{1}{2})^2 + (\frac{1}{2})^2 = \frac{1}{2}$
- P(B) = P{the second pen is non-defective} = P($D_{1}N_{2},N_{1}N_{2}$) = $\frac{1}{2}$
- P(C) = P{the two pens are both defective OR not defective} = P($D_{1}D_{2},N_{1}N_{2}$) = $\frac{1}{2}$
Then P($A\bigcap B$) = P(AB) = P($D_{1}N_{2}$) = $\frac{1}{2} \cdot \frac{1}{2}$=$\frac{1}{4}$, P(AC)=P($D_{1}D_{2}$) = $\frac{1}{4}$ , P(BC)= P($N_{1}N_{2}$) = $\frac{1}{4}$.
But, P($A\bigcap B\bigcap C) = P(ABC) = \varnothing = null$ = no common elements = $0$.
Therefore, $P(AB) = P(BC) = P(AC) = \frac{1}{4}$. But $P(ABC) = 0 \neq P(A) \cdot P(B) \cdot P(C)$.
Hence, A, B, C are pairwise independent BUT Not Mutually Independent.
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