The word probability is neither new nor unknown. In general we can say probability is the chance of occurrence or non-occurrence of some specific thing.
In the present world of uncertainity, probability acquired added significance.
While manufacturing consumer goods, the manufacturer wants to know in advance, the probability of marketing the product.
Politicians want to asses the probability of coming to power of a certain alliance.
Medicines are prepared after calculating the probability of its effectiveness . ]A new heroine is promoted in the cinema world after evaluating the probability of her acceptanceamong the viewers.
In this way, probability is the most useful tool to survive in the present era where uncertainity is the only certain thing.
In the field of economics, politics, culture, industry, education, health, insurance, game, media, everywhere, everyone require to know probability.
Here, I want to introduce, explain and develop this pragmatic portion of mathematics in my own style and certainly not in the text book style. But those who are learning probability for any academic course, they can also able to learn from here.
Whenever all possible outcomes of an experiment is known, then the ratio of number of favourable outcomes to that of all possible outcomes is known as the measure of probability.
But this classical approach of probabilitycannot help usin lot of cases. Even after knowing these limitations, we have to start from the classical approach. We can discuss about the complexities at a later stage.
Since we have to deal with numbers, first of all it is better to have a glance towards the counting principles. Once again I want to remind the readers that I will follow a lucid style such that one can understand and follow it easily. My purpose of writing is to make the user capable of solving problems.
First of all I will ask you to remember that
If an operation can be done in m ways and a succeeding operation, depending upon the first operation can be done in n ways, then these two operations can be done in m x n ways.
And, if an operation can be done in m ways and a succeeding operation, independent of the first operation can be done in n ways, then these two operations can be done in m + n ways.
This logic can be extended to any number of operations.
In general you have to remember that whenever "and" is there between two operations there you have to multiply.
And wherever OR IS THEREBETWEEN TWO OPERATIONS, THERE YOU HAVE TO ADD.
PERMUTATION means arranging few things where we have to consider different order as different arrangement.
abc, acb, bac,..all are different arrangements.
COMBINATION means arrangement of things where there is no significance of order. abc, bac, cab,all are same arrangement.
Instead of discussing much, I will proceed further with the help of certain examples :
1.
Consider all possible permutations of the letters of the word ENDEANOEL
Match the statements/ Expressions in column I with the statements/ Expressions in column II
column I column II
(A) The number of permutations containing the word ENDEA is (p) 5 !
(B) The number of permutations in which the letter E occurs (q) 2 x 5 !
in the first and the last positions is
(C) The number of permutations in which none of the letters
D, L, N occurs in the last five positions is (r) 7 x 5 !
(D) The number of permutations in which the letters A, E, O
occur only in odd positions is (s) 21 x 5 !
[IIT 2008]
Solution :
(A) Consider ENDEA as a single (combined) letter, then total number of letters become 5.
Hence number of permutations are 5!
(B) Since E occurs in the first and last positions, the remaining 7 letters can be arranged in 7! ways ; but 2 N's cannot be arranged in 2 ways;
Hence number of permutations are 7!/2 ways
(C) divide the letters in two categories : (D,L,N,N) and (EEEAO)
In the last five positions the letters (EEEAO) can be arranged in 5!/3! ways
And in the first four positionsthe letters (D,L,N,N) can be arranged in 4!/2! ways
Hence number of permutations are 20 x 12 = 240 ways.
Now you will be surely able to write down the final answer.
Since it is a matching, you need not require to do the (D), but you can do it to satisfy yourself.
2.
The letters of the word COCHIN are permuted and all the permutations are arranged in an alphabetical order as in an English dictionary. The number of words that appear before the word COCHIN is
(A) 360 (B) 192 (C) 96 (D) 48
[IIT 2007]
Solution :
In alphabetical order keeping CC first, the number of permutations are 4! = 24.
keeping CH first, the number of permutations are 4! = 24.
keeping CI first, the number of permutations are 4! = 24.
keeping CN first, the number of permutations are 4! = 24.
keeping CO first, the first arrangement is COCHIN
Now you have to write down the final answer.
3. There are 11 points on a plane with 5 lying on one straight line and another 5 lying on a second straight line which is parallel to the first line. The remaining point is not collinear with any two of the previous 10 points. The number of triangles that can be formed with vertices chosen from these 11 points is
(A) 85 (B) 105 (C) 125 (D) 145 [ISI 2004]
Solution :
From the given conditions we can make a triangle in 3 ways :
(i) Taking any two points from the first straight line and the remaining one point from the remaining 6 points.
(ii) Taking any two points from the second straight line and the remaining one point from the remaining 6 points.
(iii) Taking the isolated point and any one from the first straight line and any one from the second straight line.
Now calculate and add the number of ways in three cases to get the final answer.
4. How many three digit numbers of distinct digits can be formed by using the digits 1,2,3,4,5,9 such that the sum of the digits is at least 12 ?
(A) 61 (B) 66 (C) 60 (D) 11 [ISI 2005]
Solution :
If the digit 9 is used, then we can choose any two digits from the remaining and the sum will be at least 12. 9 we can put in any of the three places and the other two digits we can put in 5 P2 ways. So the number of numbers where 9 is one of the digit is = 3 x (5x4) = 60.
Excluding 9, the sum can be 12, if we choose 3, 4 and 5 only. In this case number of numbers is = 3 P3 = 3x2 = 6.
Now you have to write down the final answer, yourself.
5. 20 persons are invited for a party. In how many different ways can they and the host be seated at circular table if two particular persons are to be seated on either side of the host?
(a) 20! (b) 2 . (18) ! (c) 18 ! (d) none of these [WBJEE 2007]
Solution : Those two particular person can be seated on either side of the host in 2 ways.
For circular permutations we have to fix those and then the remaining 18 persons can be permuted in 18! ways.
Now you are surely able to choose the correct answer.
6. How many words can be formed from the letters of the word DOGMATIC, if all the vowels remain together?
(a) 4140 (b) 4320 (c) 432 (d) 43 [ AMU 2003]
Solution : There are three vowels. Let us tie those together with a thread. It can be done by arranging among themselves in 3! ways.
Now, considering the tied letters as one, the total number of letters become 1 + 5 = 6.
These 6 letters can be arranged in 6! ways. We have to do the first and the second operations.
You are knowing that and implies multiplication.
Hence, choose the correct option.
7. At an election, a voter may vote for any number of candidates, not greater than the number to be elected. There are 10 candidates and 4 are to be elected. If a voter votes for at least one candidate, then the number of ways in which he can vote is [ AIEEE 2006 ]
(a) 5040 (b) 385 (c) 6210 (d) 1110
Solution :
He can vote in 10C1 + 10C2 + 10C3 + 10C4 ways [ why ? Justify yourself. ]
9. In how many ways 7 men and 7 women can sit on a round table such that no two women sit together?
(a)(7!)2 (b)7! 6! (c)(6!)2 (d) 7! [ UPSEE 2002 ]
Solution : In a round table we have to keep one person fixed. So first we can put 7 men in 6!ways.
Now in between 7 men there are 7 places only where we can put women maintaining the condition given. We can put those in 7! ways.
Hence the total number of ways = 7! 6!
And option (b) is the correct answer.
10. In a polygon of n sides has 275 diagonals, then n is equal to
(a)25 (b)35 (c)20 (d) 15 [ EAMCET 2007 ]
Solution : In a polygon of n sides, there are n vertices. By joining any two vertices we can get a straight line. So the no. of straight lines are nC2. Out of which n are sides. Hence the no. of diagonals are (nC2 - n)
By the problem (nC2 - n) = 275
[n (n 1)/2] n = 275
n (n 1)/2 = 275 + n
n (n 1) = 550 + 2n
n (n 3) = 550 = 25 22
Hence n = 25
And option (a) is the correct answer
Along with theoretical discussion, I always prefer to add numerical problems. Actually while going through a problem one can understand the subject easily. Here I will show you how to solve problems. As a model I have taken few randomly chosen sums from different highly reputed examinations in India.
1. In a box there are 2 red, 3 black and 4 white balls. Out of these 3 balls are drawn together. The probability of these being of same colour is
(a) 5/84 (b) 1/21 (c) 1/84 (d) None of these
[ AMU 2002 ]
Soln.
While doing these sums, first see carefully how balls are drawn. In this case the balls are drawn together.
Here there are two possibilities only. Either the balls are black or white; because there are only 2 red balls.
P(all balls black) = C(3,3) / C(9,3)
= 1 / [9.8.7 / 3.2.1]
= 1 / 84
P(all balls white) = C(4,3) / C(9,3)
= 4 / 84
Required probability = 1/84 + 4/84
= 5 / 84 [Take care, here or is there , so you have to add]
Hence option (a)is the correct answer.
2. Aishawarya studies either computer science or mathematics everyday. If she studies computer science on a day, then the probability that she studies mathematics the next day is 0.6. If she studies mathematics on a day, then the probability that she studies computer science the next day is 0.4. Given that Aishawarya studies computer science on Monday, what is the probability that she studies computer science on Wednesday?
(A)0.24 (B)0.36 (C)0.4(D) 0.6
[GATE (CS) 2008]
SOLUTION :
Given that Aishawarya studies computer science on Monday
On Tuesday, the probability that she studies computer science is 0.4
and in that case, the probability that she studies computer science on Wednesday is 0.4.
On Tuesday, the probability that she studies mathematics is 0.6 and in that case, the probability that she studies computer science on Wednesday is 0.4
Hence required probability is
= 0.4 x 0.4 + 0.6 x 0.4[why? Justify yourself]
= 0.16 + 0.24
= 0.4
Hence option (C) is the correct answer.
3. A six faced fair dice is thrown until 1 comes, then the probability that 1 comes in even number of trials is
(A) 5/11 (B) 5/6 (C) 6/11 (D) 1/6
[IIT 2005]
SOLUTION :
Here, the required probability is (first fail and second success) or (first fail and second fail and third fail and fourth success) or so on .
