3. independent events and mutually exclusive and exhaustive events

Independent events and mutually exclusive and exhaustive events

Definition 2.

Two events E and F are said to be independent,

 if P(F|E) = P (F) provided P (E) ≠ 0 and P (E|F) = P (E) provided P (F) ≠ 0 Thus, in this definition we need to have P (E) ≠ 0 and P(F) ≠ 0

 Now, by the multiplication rule of probability, we have P(E ∩ F) = P(E) . P (F|E) ……………. (1)

If E and F are independent, then (1)

becomes P(E ∩ F) = P(E) . P(F) ………………………….... (2)

Thus, using (2), the independence of two events is also defined as follows:

Mutually exclusive:

Two events E and F are said to be mutually exclusive if E ∩ F = ɸ 

i.e.,  P(E ∩ F)=0

Exhaustive events

Two events E and F are said to be exhaustive events  if E U F = S

i.e.,  P(E U F)=1

Definition 3.  Let E and F be two events associated with the same random experiment, then E and F are said to be independent if P(E ∩ F) = P(E) . P (F)

Similarly for Three  events.

Three events A, B and C are said to be mutually independent,

 if P(A ∩ B) = P (A) P (B)

 P(A ∩ C) = P (A) P (C)

 P(B ∩ C) = P (B) P(C)

and P(A ∩ B ∩ C) = P (A) P (B) P (C)

Properties of Independent:

1. If E and F are independent events, then the events E and F′ are independent.
2. If E and F are independent events, then the events E’ and F are independent.
3. If E and F are independent events, then the events E’ and F′ are independent.

Example .  Prove that if E and F are independent events, then so are the events E and F′.

Solution :  Since E and F are independent, we have P(E ∩ F) = P(E) . P(F) ....(1)

it is clear that E ∩ F and E ∩ F′ are mutually exclusive events

and also E =(E ∩ F) ∪ (E ∩ F′).

Therefore, P(E) = P(E ∩ F) + P(E ∩ F′)

• P(E ∩ F′) = P(E) − P(E ∩ F) = P(E) − P(E) . P(F)      (by (1))

                              = P(E) (1−P(F)) = P(E). P(F′)

Hence, E and F′ are independent

Example .  If A and B are two independent events, then the probability of occurrence of at least one of A and B is given by 1– P(A′) P(B′)

Solution:  We have P(at least one of A and B) = P(A ∪ B)

 P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = P(A) + P(B) − P(A) P(B)

= P(A) + P(B) [1−P(A)] = P(A) + P(B). P(A′)

 = 1− P(A′) + P(B) P(A′)

= 1− P(A′) [1− P(B)]

= 1− P(A′) P (B’)

Example .  Prove that if E and F are independent events, then so are the events E’ and F.

 Solution .  Since E and F are independent, we have P(E ∩ F) = P(E) . P(F) ....(1)

it is clear that E ∩ F and E ‘∩ F are mutually exclusive events

and also E =(E ∩ F) ∪ (E’ ∩ F).

Therefore, P(E) = P(E ∩ F) + P(E’ ∩ F)

• P(E’ ∩ F) = P(F) − P(E ∩ F) = P(F) − P(E) . P(F)      (by (1))

                              = P(F) (1−P(E)) = P(E’). P(F)

Hence, E’ and F are independent