Pairwise and Mutually Independent Events,

Pairwise and Mutually Independent Events

Pairwise Independent Events

Let A,B and C be three events in a sample space S, then these events are said to be pairwise independent if every pair of two events is independent i.e. if and only if

$$P(A \cap B)=P(A).P(B)$$

$$P(B \cap C)=P(B).P(C)$$

$$P(A \cap C)=P(A).P(C)$$

Difinition:The events \(A_1,A_2,.........,A_n\) are said to be pairwise independent if and only if

$$P(A_i \cap A_j)=P(A_i)P(A_j) \; \; \; \; \; for \; \; all \; \; i \neq j=1,2,.....,n$$

$$provided \; \; that \; \; P(A_i)> 0 \; \; \; \; ;i=1,2,....,n$$

Mutually Independent Events

Difinition: The three A, B and C are said to be mutually independent if they are independent by pairs and by triplets. i.e if and only if

$$P(A \cap B)=P(A).P(B)$$

$$P(B \cap C)=P(B).P(C)$$

$$P(A \cap C)=P(A).P(C)$$

$$ and \; \; P(A \cap B \cap C)=P(A).P(B).P(C)$$

$$Provided \; \; that \; \; P(A)> 0,P(B)> 0 \; and \; P(C)> 0 .$$

This difinition can also be generalized to n events \(A_1,A_2,....A_n\).

Theorem 17.

For any two events A and B so that P(A)> 0 and P(B)> 0,

$$P(A)=P(B)P(\frac {A}{B})+P(\bar B)P(\frac {A}{\bar B})$$

Proof: By multiplication theorem of probability ,we have,

$$P(A \cap B)=P(B).P(\frac{A}{B})$$

$$P(A \cap \bar B)=P(\bar B).P(\frac{A}{\bar B})$$

Again, we have

$$ A=(A \cap \bar B) \cup (A \cap B)$$

$$ \therefore P(A)=P[(A \cap \bar B) \cup (A \cap B)]$$

$$=P(A \cap \bar B)+P(A \cap B)$$

$$=P(\bar B).P(\frac {A}{\bar B})+P(B).P(\frac {A}{B})$$

Theorem 18

If A ,B and C are mutually independent events, then \(A \cup B\) and C are also independent.

Proof:Here,we have to prove

$$P[(A \cup B) \cap C]=P(A \cap B).P(C)$$

$$Now, \; P[(A \cup B) \cap C]=P[(A \cap C) \cup :(B \cap C)] \; \; \; (using \; \; distributive \; \; law)$$

$$=P(A \cap C)+P(B \cap C)-P(A\cap B \cap C)$$

$$=P(A).P(C)+P(B).P(C)-P(A).P(B).P(C) \; \; \; \; [\therefore A,B \; and \; C \; are \; mutually \; independent]$$

$$=P(C).[P(A)+P(B)-P(A \cap B)]$$

$$=P(C).P(A \cup B)$$

Hence \(A \cup B\) and C are independent.

Theorem 19

If an event A is independent of the events B,\(B \cup C\) and \(B \cap C\),then it is also independent of C.

Proof: Given that ,

$$P(A \cap B)=P(A).P(B) \; \; \; \; \; \; \; ................(i)$$

$$P(A \cap B \cap C)=P(A).P(B \cup C)\; \; \; \; \; \; \; ................(ii)$$

$$P(A \cap B \cap C)=P(A).P(B \cap )\; \; \; \; \; \; \; ................(iii)$$

$$Also, \; \; P[A \cap (B \cup C)]=P(A).[P(B)+P(C)-P(B \cap )]$$

$$=P(A).P(B)+P(A).P(C)-P(A)P(B \cap C)$$

$$=P(A \cap B)+P(A)P(C)-P[A \cap (B \cap C)]\; \; \; \; \; \; \; ................(iv) [Using \; \; (i) \; \; and \; \; (ii)$$

$$Again,P[A \cap (B \cup C)]=P[(A \cap B) \cup (A \cap C)]$$

$$=P(A \cup B)+P(A)P(C)-P[(A \cap B) \cap P(A \cap C)$$

$$=P(A \cup B)+p(A \cap C)-P[A \cap (B \cap C)]\; \; \; \; \; \; \; ................(v)$$

$$From \; \; (iv) \; \; and \; \; (V), \; \; we \; \; get $$

$$P(A \cap C )=P(A).P(C)$$

$$ Hence, \; \; A \; \; and \; \; C \; \; are \; \; independent \; \; events.$$

Theorem 20

If A, B and C are random events in a sample space and if A,B and C are pairwise independent and A is independent o \(B \cup C\), then A, B and C are mutually independent.

Proof: Given,

$$P(A \cap B)=P(A).P(B)$$

$$P(B \cap C)=P(B).P(C)$$

$$P(A \cap C)=P(A).P(C)$$

$$ and \; \; P[A \cap (B \cup C)]=P(A).P(B).P(C)$$

$$Now,P[A \cap (B \cup C)]=P[(A \cap B) \cup (A \cap C)]$$

$$=P(A \cap B)+ P(A \cap C)-p[(A \cap B) \cap (A \cap C)]$$

$$=P(A).P(B)+P(A).P(C)P(A \cap B \cap C) \; \; \; from (i) \; \; \; \; .......(ii)$$

$$ and\; \; \; P(A).P(B \cup C)=P(A)[P(B)+P(C)-P(B \cap C)]$$

$$ =P(A).P(B)+P(A).P(C)-P(A)P(B \cap C) \; \; \; \; \; \; ......(iii)$$

$$ Equating \; \; (ii) \; \; and \; \; (iii) \; \; on \; \; using\; \; (i) \; \;, we \; \; get$$

$$P(A).P(B)+P(A).P(C)-P(A \cap B \cap C)=P(A)P(B)+P(A).P(C)-P(A)P(B \cap C)$$

$$ \Rightarrow \; \; \; P(A \cap B \cap C)=P(A).P(B \cap C)=P(A).P(B).P(C)$$

$$Hence, A,B \; \; and \; \; C \; \; are \; \; mutually \; \; independent.$$

Examples:

1) Two fair coins are tossed simultaneously.Find the probability of occurrence of (i) head on the first coin and tail on the second coin. (ii) head and tail alternately.

Solution

In a random experiment of tossiing two coins,the sample space is

$$S={HH,HT,TH,TT} \; \; \; \; \; \therefore n(S)=4$$

Let A be event of occurring head on the first coin and B be the event of occurring tail on the second coin.

$$Then \; A={HH,HT} \; \; \; \; \therefore n(A)=2$$

$$B={HT,TT} \; \; \; \; n(B)=2$$

$$P(A)=\frac{n(A)}{n(S)}=\frac{2}{4}=\frac{1}{2} \; \; \; and \; \; \; P(B)=\frac{2}{4}=\frac{1}{2}$$

Since one outcome HT is common to both A and B, i.e. \(A \cap B\)={HT}, then

$$P(A \cap B)=\frac{1}{4}$$

$$As, P(A).P(B)= \frac {1}{2} \times \frac{1}{2}=\frac {1}{4} \; \; \; we \; \; get$$

$$P(A \cap B)=P(A).P(B)=\frac{1}{4}$$

$$Hence \; \; two \; \; events \;\; A \; \; and \; \; B \; \; are \; \; independent$$

\(\therefore \) The required probability is given by multiplication theorem of probability as

$$P(A \cap B)=P(A).P(B)=\frac {1}{2} \times \frac{1}{2} =\frac{1}{4}$$

ii) Let E=event of getting head and tail alternately .Then E can happen in either of the following two mutually exclusive events:

$$E_1={HT} \Rightarrow P(E_1)=\frac{1}{4} \; \; and\; \; E_2={TH} \Rightarrow P(E_2)=\frac {1}{4}$$

$$ \therefore P(Head \; and \; tail \; alternately ) \; is \; given \; by$$

$$P(E)=P(E_1 \cup E_2)=P(E_1)+P(E_2)=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$$

2) An urn contains 4 White and 3 red marbles.Four marbles are drawn one after another from the urn.What is the probability that first marble will be white ,the second red,the third white and the fourth a red one?

Solution

Let A,B,C and D denote the event that the first marble drawn is white ,the second is red ,the third is white and the fourth is red respectively.

Case (i) : Withe replacement:

In this case, the four events A,B,C and D are independent.Thus,

$$P(A)=\frac {4}{7},\; \; P(B)=\frac{3}{7}, \; \; \; P(C)=\frac{4}{7} \; \; and \; \; P(D)=\frac{3}{7}$$

$$Therefore, \; the \; required \; probability \; is \; given \; by $$

$$P(A \cap B \cap C \cap D)=P(A).P(B).P(C).P(D)$$

$$=\frac{4}{7} \times \frac{3}{7} \times \frac{4}{7} \times \frac{3}{7} =\frac{144}{2401}$$

Case (ii) : without replacement:

In this case, the event B is dependent on event A ,event C is dependent on the previous events A and B and event D is dependent on the previous events A,B and C.Then,

$$P(A)=\frac {4}{7}, \; \; \; P(\frac {B}{A})=\frac{3}{6},\; \; \; P(\frac{C}{A \cap B})=\frac{3}{5} \; \; and \; \; \; P(\frac {D}{A \cap B \cap C})=\frac{2}{4}$$

$$ Therefore, \; the \; required \; probability \; is \; given \; by $$

$$P(A \cap B \cap C \cap D)=P(A).P(\frac {B}{A} ).P(\frac {C}{A \cap B}).P(\frac{D}{A \cap B \cap C})$$

$$=\frac {4}{7} \times\frac {3}{6} \times\frac {3}{5} \times\frac {2}{4}=\frac{3}{35}$$

References

Sukubhattu,Narendra Prasad. Probability and Inference-I. Asmita Books Publishers & Distributors (P)Ltd. 2013