Independent Events

Proof:

Since A and B are independent, therefore

P(A∩B) = P(A).P(B)

(a) P(A∩B’) = P(A) - P(A∩B)

= P(A) – P(A).P(B)

= P(A) [1- P(B)]

= P (A). P(B’)

Therefore, A and B’ are also independent

(b) P(A’ ∩ B’) = P(A U B)’             [By De Morgan’s Law]

= 1- P (A U B)

= 1- [P (A) +P (B) – P(A∩B)]

= 1- [P (A) +P (B) – P (A∩B)]

= 1- P (A) – P (B) + P (A).P (B)

= [1-P (A)] – P (B) [1-P (A)]

= [1-P (A)]. [1-P (B)]

= P (A’). P (B’)

Therefore, A’ and B’ are also independent

Formula of Independent Events

Problems

Q1.  Two cards are drawn at random one after the other. Given that first card drawn is non-face red card, what is the probability that second card is face card, if the cards are drawn

i) Without Replacement? ii) With Replacement?

Sol. Let event A: first card is a non-face red card and event B : second card drawn is face card

P (A) = 20/52 = 5/13 and P(B) =12/52 = 3/13

Required probability = P (second card drawn is face card given that it is a red card)

= P(B/A)

i)  Without Replacement: Since first non-face red card is not replaced, therefore now we have 51 cards containing 12 face cards.

P(B/A) = 12/51 ≠ P(B). In this case A and B are not independent

ii)  With Replacement: Since first non-face red card is replaced, therefore now again we have 52 cards containing 12 face cards.

P (B/A) = 12/52 = 3/13 = P(B)

In this case A and B are independent
 

Q2.  A drawer contains a pair of socks each one with the following colours: blue, brown, red, white and black. Each pair is kept together folded in a matching set. You reach into the socks drawer and choose a pair of socks without looking. You replace this pair and then choose another pair of socks. What is the probability that you will choose the blue pair of socks both times?

Sol. P(blue) = 1/5

Difference between Independent and Mutually exclusive( or disjoint ) events

If two events A and B are independent, then

P [A and B] = P [A]. P [B]

that is, the probability that both A and B occur is equal to the probability that A occurs times the probability that B occurs.

If A and B are mutually exclusive, then

P[A and B] = 0

that is, the probability that both A and B occur is zero.

Are all independent events disjoint events ( or mutually exclusive events )?

Clearly, if

A and B are nontrivial events (P [A] and P[B] are nonzero), that is,

P[A] > 0

P[B] >0,

then they cannot be both independent and mutually exclusive.

Example

Consider a fair coin and a fair six-sided die.

Let event A be obtaining heads, and event B be rolling a 6.  Here, the events A and B are independent, because the outcome of one does not affect the outcome of the other.  The probability that both A and B occur is

P[A and B] = P[A]P[B] = (1/2)(1/6) = 1/12.

Since this value is not zero, then events A and B cannot be mutually exclusive.

Basis for comparison	Mutually exclusive events	Independent events
Meaning	Two events are said to be mutually exclusive, when their occurrence is not simultaneous.	Two events are said to be independent, when the occurrence of one event cannot control the occurrence of other.
Influence	Occurrence of one event will result in the non-occurrence of the other.	Occurrence of one event will have no influence on the occurrence of the other.
Mathematical formula	P(A and B) = 0	P(A and B) = P(A) P(B)
Sets in Venn diagram	Does not overlap	Overlaps

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