Independent Events

Events are said to be independent when the happening of any one of them does not affect the happening of any of the others

(i) A and B are independent if

P (B/A) = P(B) and P(A/B) = P(A).

(ii) The probability of the occurrence of several independent events is the product of their individual probabilities.

i.e. If E

_{1}, E_{2}, ......... E_{n}are n independent events then P (E_{1}∩ E_{2 }∩ ... ∩ E_{n})= P (E

_{1}) P (E_{2}) P (E_{3})...P (E_{n}).

Remark:The converse of this is also true i.e. if n given events satisfy the above condition then they will be independent.

Pairwise Independent EventsThree events E

_{1}, E_{2}and E_{3}are said to pairwise independent ifP(E

_{1 }∩ E_{2}) = P(E_{1})P(E_{2}), P(E_{2}∩ E_{3}) = P(E_{2}) P(E_{3}) and P(E_{3 }∩ E_{1}) = P(E_{3})P(E_{1})i.e. Events E

_{1}, E_{2}, E_{3}, ......... E_{n}will be pairwise independent ifP (A

_{i }∩ A_{j}) = P(A_{i}) P(A_{j}) i ≠ j.Three events are said to be mutually independent if

P (E

_{1}∩ E_{2}) = P(E_{1}) P(E_{2}), P(E_{2 }∩ E_{3}) = P(E_{2}) P(E_{3}) and P(E_{3}∩ E_{1}) = P(E_{3}) P(E_{1})and P(E

_{1}∩ E_{2}∩ E_{3}) = P(E_{1}) P(E_{2}) P(E_{3}).If two events A and B are mutually exclusive,

P (A ∩ B) = 0 but P(A) P(B) ≠ 0 (In general).

=> P(A ∩ B) ≠ P(A) P(B)

=> Mutually exclusive events will not be independent.

So if two events are independent, they have to have some common element between them i.e. they cannot be mutually exclusive. Mutually exclusiveness is used when the events are taken from the same experiment and independence is used when the evens are taken from different experiments.

For Example(i) Two fair die are thrown. Let two events be 'first die shows an even number' and second die shows an odd number'. These two events are independent events since the result of the first die does not depend upon the result of 2

^{nd}die. But these events are not mutually exclusive since both the events may simultaneously occur.(ii) From a pack of cards, (a) Two cards are drawn in succession without replacement. Let an event be 'card drawn is Ace'. Then the two events (i.e. two cards being aces) are dependent since drawing of 2

^{nd}card very much depends upon which is the first drawn card. (b) If the cards are drawn with replacement, the drawing of second card will not depend upon first card. So, two events will be independent.

Illustration 1:An event A

_{1}can happen with probability p_{1}and event A_{2}can happen with probability p_{2}. What is the probability that(i) Exactly one of them happens.

(ii) at least one of them happens (Given A

_{1}and A_{2}are independent events).

Solution:(i) The probability that A

_{1}happens is p_{1}The probability that event A

_{1}fails 1-p_{1}. Also the probability that A_{2}happens is p_{2}.Now, the chance that A

_{1}happens and A_{2}fails is p_{1}(1 - p_{2}) and the chance that A_{1}fails and A_{2}happens is p_{2}(1 - p_{1}).=> The probability that one and only one of them happens is

p

_{1}(1 - p_{2}) + p_{2}(1 - p_{1}) = p_{1}+ p_{2}+ 2p_{1}p_{2}.(ii) The probability that both of them fail = (1 - p

_{1}) (1 - p_{2})=> Probability that at least one happens = 1- (1 - p

_{1}) (1 - p_{2})

= p_{1}+ p_{2}+ p_{1}p_{2}.

Illustration 2:

A person draws a card from a pack of 52, replaces it and shuffles it. He continues doing it until he draws a spade. What is the chance that he has to make?

(i) At least 3 trials

(ii) Exactly 3 trials

Solution:

(i) For at least 3 trials, he has to fail at the first 2 attempts and then after that it doesn’t make a difference if he fails or wins at the 3rd or the subsequent attempts.

Chance of success at any attempt = 1/4

=> chance of failure = 3/4

=> chance of failing in first 2 attempt = (3/4)^{2 }= Chance of success at any attempt = 1/4

=> chance of failure = 3/4=> chance of failing in first 2 attempt = (3/4)

^{2}= 9/16.

(ii) For exactly 3 attempts, he has to fail in the first two attempts and succeed in the 3rd attempt. The required probability = (3/4)^{2}.1/4 = 9/64.

Illustration 3:

A, B and C in order, toss a coin. The one who gets a head first wins. Find their respective probabilities of winning.

Solution:

P (Head) = 1/2 = P (Tail) in any toss of the coin. A wins if he gets a head in the first trial or if he fails, then B & C fail and then he gets a head and so on.

Therefore, P (A) = 1/2 + (1/2)^{3}.1/2 + (1/2)^{6}.1/2 + ....;

= 1/2 [1 + 1/8 + 1/64 + ....] = 1/2.1/ (1-1/8) = 4/7.

Similarly P (B) = 1/2.1/2 + (1/2)^{4}.1/2 + (1/2)^{7}.1/2+..... = 2/7

and P(C) = (1/2)^{2}.1/2 + (1/2)^{5}.1/2 +..... = 1/7.

Total Probability TheoremLet A

_{1}, A_{2}, .... A_{n}be a the set of mutually exclusive and exhaustive events and E be some event which is associated with A_{1}, A_{2}, ...., An. Then probability that E occurs is given by

P (E) = ∑^{n}_{(i=1)}P(A_{i})P(E/A_{i}).

Illustration:

A bag contains 3 white balls and 2 black balls another contains 5 white and 3 black balls. If a bag is chosen at random and a ball is drawn from it, what is the probability that is white?

Solution:

The probability that the first bag is chosen is 1/2 and the chance of drawing a white ball from it is 3.5.

=> Chance of choosing the first bag and drawing a white ball is 1/2.3/5.

Similarly, the chance of choosing the second bag and drawing a white ball is 1/2.5.8.

Hence the chance of randomly choosing a bag and drawing a white ball

is = 1/2.3/5+1/2.5/8 (Mutually exclusive cases)

= 49/80.

Illustration:

Find the probability that a year chosen at random has 53 Sundays.

Solution:

Let P(L) be the probability that a year chosen at random is leap

P(L) = 1/4

=> P(L^{c}) = 3/4.

Let P(S) be the probability that an year chosen at random has 53 Sundays

Now P(S/L) = Probability that a leap year has 53 Sundays.

A leap year has 366 days, 52 weeks + 2 days; the remaining 2 days may be Sunday-Monday, M–T, T–W.

W–Th, Th–F, F-Sat or Sat–S.

Out of the 7 possibilities, 2 are favourable.

=> P(S/L) = 2/7. Similarly P(S/L^{c}) = 1/7.

=> P(S) = 1/4.2/7 + 3/4.1/7 = 5/28.

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