Probability concepts by Sibanand Pattnaik  Part 2

Equally likely events/outcomes
In an experiment, two or more event/outcomes are said to be equally likely, if they have the same chances associated with them. i.e. no one of them has more chance of occurrence than others.
Let S be the sample space & E be the event, such that n(S) = n and n(E) = m.
If each out come is equally likely, then it follows that P(E) = m/n = Number of outcomes favourable to E/ Total possible outcomesProbability of the event ‘A or B’
Let us now find the probability of event ‘A or B’, i.e., P (A ∪B)
Let A = {HHT, HTH, THH} and B = {HTH, THH, HHH} be two events associated with ‘tossing of a coin thrice’
Clearly A ∪B = {HHT, HTH, THH, HHH}
Now P (A ∪B) = P(HHT) + P(HTH) + P(THH) + P(HHH)
If all the outcomes are equally likely, then
P(A B )= 1/8 + 1/8 +1/8 + 1/8 = 4/8 = ½
Also P(A) = P(HHT) + P(HTH) + P(THH) = 3/8
and P(B ) = P(HTH) + P(THH) + P(HHH) = 3/ 8
Therefore P(A) + P(B ) = 3/8 + 3/8 = 6/8
It is clear that P(A ∪ B ) ≠P(A) + P(B)
The points HTH and THH are common to both A and B . In the computation of P(A) + P(B) the probabilities of points HTH and THH, i.e., the elements of A ∩B are included twice. Thus to get the probability P(A ∪ B ) we have to subtract the probabilities of the sample points in A ∩B from P(A) + P(B )
Thus we observe that, P(A∪B) = P(A) + P(B) − P(A∩B)Probability of event ‘not A’
Consider the event A = {2, 4, 6, 8} associated with the experiment of drawing a card from a deck of ten cards numbered from 1 to 10. Clearly the sample space is S = {1, 2, 3, ...,10}
If all the outcomes 1, 2, ...,10 are considered to be equally likely, then the probability of each outcome is
1/ 10
Now P(A) = P(2) + P(4) + P(6) + P(8) =1/10 +1/10 +1/10+1/10 = 2/5
Also event ‘not A’ = A′ = {1, 3, 5, 7, 9, 10}
Now P(A′) = P(1) + P(3) + P(5) + P(7) + P(9) + P(10) = 3/5
Thus, P(A′) = 3/5 = 1 2/5 = 1 P(A)
So P( A′ ) = P(not A) = 1 – P(A)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 cardWhen 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) = 14/52 =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)
11/4 = 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)
11/2 =1/2A 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.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 = 7/9Two 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 examinationLet 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 = 0.11A 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?The total number of persons = 2 + 2 = 4. Out of these four person, two can be selected in 4 C2 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 2 C2 =1 way.
Therefore P( no man) = 2c2/4c2 = 1/6(b) One man in the committee means that there is one woman. One man out of 2 can be selected in 2
C1 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 = 2/3(c) Two men can be selected in 2C2 way.
Hence P (Two men) 2c2/4c2 = 1/6On 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}Thus 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.
Here F = {ABCD, DABC, ABDC, ADBC}Thus P(f) = n(f)/n(s) = 4/24 = 1/ 6
(iii) 2/24 = 1/12
(iv) 12/24 = 1/2
(v) 6/24 = 1/4
I have written all the cases, but you all can solve directly also.
let say 5 person a,b,c,d,e can be arranged in 5! ways ; no. of ways in which a is before b = 5!/2!; no. of ways in which a is before b & b is before c = 5!/3! ; no. of ways in which a is before b & b is before c & c is before d = 5!/4! ; no. of ways in which a is before b & b is before c & c is before d & d is before e = 5!/5! = 1;
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 = 52 C7
(i) Number of hands with 4 Kings = 4 C4 × 48C3 (other 3 cards must be chosen from the rest 48 cards)
Hence P (a hand will have 4 Kings) = 4 C4 × 48C3 / 52 C7 = 1/7735(ii) Number of hands with 3 Kings and 4 nonKing cards = 4C3 × 48C4
Therefore P (3 Kings) =4C3 × 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 = 46/7735In 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 5 P3 = 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 = 1/10In a given race the odds in favour of four horses A, B, C, D are 1:3, 1:4, 1:5, 1:6 respectively. Assuming that, a dead heat is impossible, find the chance that one of them wins the race.
Let P(A), P(B), P(C) and P(D) be the responsibilities of winning of the horses A, B, C and D respectively. Then
P(A) = 1/4, P(B) = 1/5, P(C) = 1/6, P(D) = 1/7.
Since the above events are mutually exclusive, the chance that one of them wins
= P(AUBUCUD) = P(A) + P(B) + P(C) + P(D)
= (1/4) + (1/5) + (1/6) + (1/7) .