TOPICS
Unit-16(Examples)

Example-1 :-  Two coins (a one rupee coin and a two rupee coin) are tossed once. Find a sample space.

Solution :-
  The possible outcomes may be 
  Heads on both coins = (H,H) = HH 
  Head on first coin and Tail on the other = (H,T) = HT 
  Tail on first coin and Head on the other = (T,H) = TH 
  Tail on both coins = (T,T) = TT 
  Thus, the sample space is S = {HH, HT, TH, TT} 
   

Example-2 :-  Find the sample space associated with the experiment of rolling a pair of dice (one is blue and the other red) once. Also, find the number of elements of this sample space.

Solution :-
  S = {(x, y): x is the number on the blue die and y is the number on the red die}. 
  The number of elements of this sample space is 6 × 6 = 36 and the sample space is given below: 
  {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6) (3,1), (3,2), 
   (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6) (5,1), (5,2), (5,3), (5,4), 
   (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)} 
   

Example-3 :-  In each of the following experiments specify appropriate sample space
(i) A boy has a 1 rupee coin, a 2 rupee coin and a 5 rupee coin in his pocket. He takes out two coins out of his pocket, one after the other.
(ii) A person is noting down the number of accidents along a busy highway during a year.

Solution :-
(i) Let Q denote a 1 rupee coin, H denotes a 2 rupee coin and R denotes a 5 rupee coin. 
  Thus, the sample space is S={QH, QR, HQ, HR, RH, RQ} 

(ii) The number of accidents along a busy highway during the year of observation can be either 0 (for no accident )
  or 1 or 2, or some other positive integer. 
  Thus, a sample space associated with this experiment is S= {0,1,2,...} 
         

Example-4 :-  A coin is tossed. If it shows head, we draw a ball from a bag consisting of 3 blue and 4 white balls; if it shows tail we throw a die. Describe the sample space of this experiment.

Solution :-
  Let us denote blue balls by B₁, B₂, B₃ and the white balls by W₁, W₂, W₃, W₄. 
  Then a sample space of the experiment is 
  S = { HB₁, HB₂, HB₃, HW₁, HW₂, HW₃, HW₄, T₁, T₂, T₃, T₄, T₅, T₆}.
   

Example-5 :-  Consider the experiment in which a coin is tossed repeatedly until a head comes up. Describe the sample space.

Solution :-
  In the experiment head may come up on the first toss, or the 2nd toss, or the 3rd toss and so on till head is obtained.
  Hence, the desired sample space is S= {H, TH, TTH, TTTH, TTTTH,...}
   

Example-6 :-  Consider the experiment of rolling a die. Let A be the event ‘getting a prime number’, B be the event ‘getting an odd number’. Write the sets representing the events
(i) A or B
(ii) A and B
(iii) A but not B
(iv) ‘not A’.

Solution :-
  Here S = {1, 2, 3, 4, 5, 6}, A = {2, 3, 5} and B = {1, 3, 5} 
  (i) ‘A or B’ = A ∪ B = {1, 2, 3, 5} 
  (ii) ‘A and B’ = A ∩ B = {3,5} 
  (iii) ‘A but not B’ =  A – B = {2} 
  (iv) ‘not  A’ = A′ = {1,4,6}
    

Example-7 :-  Two dice are thrown and the sum of the numbers which come up on the dice is noted. Let us consider the following events associated with this experiment
A: ‘the sum is even’.
B: ‘the sum is a multiple of 3’.
C: ‘the sum is less than 4’.
D: ‘the sum is greater than 11’.
Which pairs of these events are mutually exclusive?

Solution :-
  There are 36 elements in the sample space S = {(x, y):  x, y = 1, 2, 3, 4, 5, 6}. 
  Then A = {(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (4, 2), 
  (4, 4), (4, 6), (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)}
  B = {(1, 2), (2, 1), (1, 5), (5, 1), (3, 3), (2, 4), (4, 2), (3, 6), (6, 3), (4, 5), (5, 4), (6, 6)} 
  C = {(1, 1), (2, 1), (1, 2)} and 
  D = {(6, 6)} 
  Now, A ∩ B = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1), (6, 6)} ≠ φ 
  A ∩ D = {(6, 6)} ≠ φ 
  B ∩ C = {(2, 1), (1, 2)} ≠ φ 
  B ∩ D = {(6, 6)} ≠ φ 
  Therefore, A and B are not mutually exclusive events. 
  Similarly A ∩ C ≠ φ, A ∩ D ≠ φ, B ∩ C ≠ φ and B ∩ D ≠ φ. 
  Thus, the pairs, (A, C), (A, D), (B, C), (B, D) are not mutually exclusive events. 
  Also C ∩ D = φ and so C and D are mutually exclusive events. 
   

Example-8 :-  A coin is tossed three times, consider the following events.
A: ‘No head appears’,
B: ‘Exactly one head appears’ and
C: ‘Atleast two heads appear’.
Do they form a set of mutually exclusive and exhaustive events?

Solution :-
  The sample space of the experiment is S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} and 
  A = {TTT}, B = {HTT, THT, TTH}, C = {HHT, HTH, THH, HHH} 
  Now A ∪ B ∪ C = {TTT, HTT, THT, TTH, HHT, HTH, THH, HHH} = S 
  Therefore, A, B and C are exhaustive events. 
  Also, A ∩ B = φ, A ∩ C = φ and B ∩ C = φ 
  Therefore, the events are pair-wise disjoint, i.e., they are mutually exclusive.
  Hence, A, B and C form a set of mutually exclusive and exhaustive events.
   

Example-9 :-  Let a sample space be S = {ω₁, ω₂,..., ω₆}.Which of the following assignments of probabilities to each outcome are valid? probability

Solution :-
(a) Condition (i): Each of the number p(ωᵢ) is positive and less than one. 
    Condition (ii): Sum of probabilities = 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 1. Therefore, the assignment is valid.

(b) Condition (i): Each of the number p(ωᵢ) is either 0 or 1. 
    Condition (ii) Sum of the probabilities = 1 + 0 + 0 + 0 + 0 + 0 = 1. Therefore, the assignment is valid.

(c) Condition (i) Two of the probabilities p(ω₅) and p(ω₆) are negative, the assignment is not valid.

(d) Since p(ω₆) = 3/2 > 1, the assignment is not valid .

(e) Since, sum of probabilities = 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 = 2.1, the assignment is not valid. 
    

Example-10 :-  One card is drawn from a well shuffled deck of 52 cards. If each outcome is equally likely, calculate the probability that the card will be
(i) a diamond
(ii) not an ace
(iii) a black card (i.e., a club or, a spade)
(iv) not a diamond
(v) not a black card.

Solution :-
  When a card is drawn from a well shuffled deck of 52 cards, the number of possible outcomes is 52. 
(i) Let A be the event 'the card drawn is a diamond' Clearly the number of elements in set A is 13.
  Therefore, P(A) = 13/52 = 1/4
  i.e. Probability of a diamond card = 1/4

(ii) We assume that the event ‘Card drawn is an ace’ is B Therefore  ‘Card drawn is not an ace’ should be B′.
  We know that  P(B′) = 1 – P(B) = 1 - 4/52 = 1 - 1/13 = 12/13 

(iii) Let C denote the event ‘card drawn is black card’ 
  Therefore, number of elements in the set C = 26 
  i.e. P(C) = 26/52 = 1/2
  Thus, Probability of a black card = 1/2.

(iv) We assumed in (i) above that A is the event ‘card drawn is a diamond’, 
  so the event ‘card drawn is not a diamond’ may be denoted as A' or ‘not A’
  Now P(not A) = 1 – P(A) = 1 - 1/4 = 3/4

(v) The event ‘card drawn is not a black card’ may be denoted as C′ or ‘not C’. 
  We know that P(not C) = 1 – P(C) = 1 - 1/2 = 1/2 
  Therefore,  Probability of not a black card = 2
   

Example-11 :-  A bag contains 9 discs of which 4 are red, 3 are blue and 2 are yellow. The discs are similar in shape and size. A disc is drawn at random from the bag. Calculate the probability that it will be
(i) red,
(ii) yellow,
(iii) blue,
(iv) not blue,
(v) either red or blue.

Solution :-
  There are 9 discs in all so the total number of possible outcomes is 9. 
  Let the events A, B, C be defined as 
  A: ‘the disc drawn is red’ 
  B: ‘the disc drawn is yellow’ 
  C: ‘the disc drawn is blue’. 
        
(i) The number of red discs = 4, i.e., n (A) = 4
  Hence P(A) = 4/9 

(ii) The number of yellow discs = 2, i.e., n (B) = 2
  Therefore, P(B) = 2/9 

(iii) The number of blue discs = 3, i.e., n(C) = 3
  Therefore, P(C) = 3/9 = 1/3

(iv) Clearly the event ‘not blue’ is ‘not C’. We know that P(not C) = 1 –  P(C)
  Therefore P(not C) = 1 - 1/3 = 2/3

(v) The event ‘either red or blue’ may be described by the set ‘A or C’ 
  Since,A and C are mutually exclusive events, we have
  P(A or C) =  P (A ∪ C) = P(A) + P(C) = 4/9 + 1/3 = 7/9 
   

Example-12 :-  Two students Anil and Ashima appeared in an examination. The probability that Anil will qualify the examination is 0.05 and that Ashima will qualify the examination is 0.10. The probability that both will qualify the examination is 0.02. Find the probability that
(a) Both Anil and Ashima will not qualify the examination.
(b) Atleast one of them will not qualify the examination and
(c) Only one of them will qualify the examination.

Solution :-
  Let E and F denote the events that Anil and Ashima will qualify the examination, respectively. 
  Given that P(E) = 0.05, P(F) = 0.10 and P(E ∩ F) = 0.02. Then 

(a) The event ‘both Anil and Ashima will not qualify the examination’ may be expressed as  E´ ∩ F´. 
  Since,E´ is ‘not E’, 
  i.e., Anil will not qualify the examination and F´ is ‘not F’, 
  i.e., Ashima will not qualify the examination. 
  Also E´ ∩ F´ = (E ∪ F)´ (by Demorgan's Law) 
  Now P(E ∪ F) = P(E) + P(F) –  P(E ∩ F) or P(E ∪ F) = 0.05 + 0.10 –  0.02 = 0.13 
  Therefore P(E´ ∩ F´) = P(E ∪ F)´ = 1 –  P(E ∪ F) = 1 –  0.13 = 0.87 
    
(b) P (atleast one of them will not qualify) = 1 –  P(both of them will qualify) = 1 –  0.02 = 0.98
   
(c) The event only one of them will qualify the examination is same as the event either
  (Anil will qualify, and Ashima will not qualify) or (Anil will not qualify and Ashima will qualify) 
  i.e., E ∩ F´ or E´ ∩ F, where E ∩ F´ and E´ ∩ F are mutually exclusive. 
  Therefore,  P(only one of them will qualify) = P(E ∩ F´ or E´ ∩ F) 
= P(E ∩ F´) + P(E´ ∩ F) 
= P (E) – P(E ∩ F) + P(F) – P (E ∩ F) 
= 0.05 – 0.02 + 0.10 – 0.02 = 0.11 
    

Example-13 :-  A committee of two persons is selected from two men and two women. What is the probability that the committee will have
(a) no man?
(b) one man?
(c) two men?

Solution :-
  The total number of persons = 2 + 2 = 4. Out of these four person, two can be selected in ⁴C₂ ways. 

(a) No men in the committee of two means there will be two women in the committee. 
  Out of two women, two can be selected in ²C₂ = 1 way.
  Therefore, P(no man) = ²C₂/⁴C₂ = 1/(2 x 3) = 1/6

(b) One man in the committee means that there is one woman. 
  One man out of 2 can be selected in ²C₁ ways and one woman out of 2 can be selected in ²C₁ ways.
  Together they can be selected in ²C₁ × ²C₁ ways. 
  Therefore, P(one man) = (²C₁ × ²C₁)/⁴C₂ = (2 x 2)/(2 x 3) = 2/3

(c) Two men can be selected in ²C₂ way.
  Therefore, P(two man) = ²C₂/⁴C₂ = 1/(2 x 3) = 1/6
   
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