TOPICS

Miscellaneous

Probability

**Example-1 :-** On her vacations Veena visits four cities (A, B, C and D) in a random order. What is the probability that she visits

(i) A before B?

(ii) A before B and B before C?

(iii) A first and B last?

(iv) A either first or second?

(v) A just before B?

The number of arrangements (orders) in which Veena can visit four cities A, B, C, or D is 4! i.e., 24.Therefore, n (S) = 24. Since the number of elements in the sample space of the experiment is 24 all of these outcomes are considered to be equally likely. A sample space for the experiment is S = {ABCD, ABDC, ACBD, ACDB, ADBC, ADCB, BACD, BADC, BDAC, BDCA, BCAD, BCDA CABD, CADB, CBDA, CBAD, CDAB, CDBA, DABC, DACB, DBCA, DBAC, DCAB, DCBA}

(i) Let the event ‘she visits A before B’ be denoted by E Therefore, E = {ABCD, CABD, DABC, ABDC, CADB, DACB ACBD, ACDB, ADBC, CDAB, DCAB, ADCB} = 12 Hence, P(E) = n(E)/n(S) = 12/24 = 1/2

(ii) Let the event ‘Veena visits A before B and B before C’ be denoted by F. Therefore F = {ABCD, DABC, ABDC, ADBC} = 4 Hence, P(F) = n(F)/n(S) = 4/24 = 1/6

(iii) Let the event ‘Veena visits A first and B last’ be denoted by G. Therefore F = {ACDB, ADCB} = 2 Hence, P(G) = n(G)/n(S) = 2/24 = 1/12

(iv) Let the event ‘Veena visits A either first or second’ be denoted by H. Therefore H = {ABCD, ABDC, ACBD, ACDB, ADBC, ADCB, BACD, BADC, CABD, CADB, DABC, DACB} = 12 Hence, P(H) = n(H)/n(S) = 12/24 = 1/2

(v) Let the event ‘Veena visits A either first or second’ be denoted by I. Therefore I = {ABCD, ABDC, CABD, CDAB, DABC, DCAB} = 6 Hence, P(I) = n(I)/n(S) = 6/24 = 1/4

**Example-2 :-** Find the probability that when a hand of 7 cards is drawn from a well shuffled deck of 52 cards, it contains

(i) all Kings

(ii) 3 Kings

(iii) atleast 3 Kings.

Total number of possible hands = ⁵²C₇ (i) Number of hands with 4 Kings = ⁴C₄ x ⁴⁸C₃ (other 3 cards must be chosen from the rest 48 cards) Hence P (a hand will have 4 Kings) = (⁴C₄ x ⁴⁸C₃)/⁵²C₇ = 1/7735 (ii) Number of hands with 3 Kings and 4 non-King cards = ⁴C₃ x ⁴⁸C₄ Therefore P (3 Kings) = (⁴C₃ x ⁴⁸C₄)/⁵²C₇ = 9/1547 (iii) P(atleast 3 King) = P(3 Kings or 4 Kings) = P(3 Kings) + P(4 Kings) = 9/1547 + 1/7735 = 46/7735

**Example-3 :-** If A, B, C are three events associated with a random experiment, prove that P(A ∪ B ∪ C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(A ∩ C) - P(B ∩ C) + P ( A ∩ B ∩ C)

Consider E = B ∪ C so that P(A ∪ B ∪ C ) = P(A ∪ E ) = P(A) + P(E) - P(A ∩ E) .......(i) Now, P(E) = P(B ∪ C) = P(B) + P(C) - P(B ∩ C) .......(ii) Also A ∩ E = A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) [using distribution property of intersection of sets over the union]. Thus, P(A ∩ E) = P(A ∩ B) + P(A ∩ C) - P[(A ∩ B) ∩ P(A ∩ C)] = P(A ∩ B) + P(A ∩ C) - P(A ∩ B ∩ C) .....(iii) Using (2) and (3) in (1), we get P(A ∪ B ∪ C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(A ∩ C) - P(B ∩ C) + P ( A ∩ B ∩ C)

**Example-4 :-** In a relay race there are five teams A, B, C, D and E.

(a) What is the probability that A, B and C finish first, second and third, respectively.

(b) What is the probability that A, B and C are first three to finish (in any order) (Assume that all finishing orders are equally likely)

If we consider the sample space consisting of all finishing orders in the first three places, we will have ⁵P₃, i.e., (5!)/(5 - 3)! = 5 × 4 × 3 = 60 sample points, each with a probability of 1/60. (a) A, B and C finish first, second and third, respectively. There is only one finishing order for this, i.e., ABC. Thus, P(A, B and C finish first, second and third respectively) = 1/60 (b) A, B and C are the first three finishers. There will be 3! arrangements for A, B and C. Therefore, the sample points corresponding to this event will be 3! in number. So, P(A, B and C are first three to finish) = 3!/60 = 6/60 = 1/10

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