Probability
Permutation
The arrangement of a number of things taking some or all of them at a time is called permutation. If there are ‘n’ number of things and we have to select ‘r’ things at a time then the total number of permutation is denoted by n =
For example if there are 3 candidates A ,B and C for the post of president and vice president of a college , since we have to select only 2 candidates , it can be done in 3! Ways. i.e. (A, B) (B, C) (A, C) (B, A) (C, B) and (C, A). Here order of arrangement matters.
Restricted Permutation:
Sometimes we have to find out the number of permutation keeping few specific objects at specific places. In this case, we find out the number of permutation of filling remaining vacant places by the remaining objects.
If r objects are taken out of n dissimilar objects
(i) A specific object is taken each time: if there are n objects . Suppose that is taken each time. If takes first place then the remaining (n-1) objects can be arranged in n-1 ways. Since can take any place so number of permutation is r n-1 .
(ii) Specific object never taken: then r objects are taken out of (n-1) objects, so number of permutation is n-1 .
{Note: n = n-1 + r n-1 }
Permutation of things when some are identical:
If we have n things in which p are exactly of one kind , q of second kind , r of third kind and the rest are different then the number of permutation of n things taken all at a time n =
183 IBPS PO EXAM 2013 : Quantitative Aptitude Example: In how many ways can the letters of the word LEADER be arranged?
Solution: The word LEADER contains total 6 letters namely 1L, 2E, 1A, 1D, 1R Therefore, the number of ways to arrange the letters of the word LEADER
= = 360.
Repetition of things:
The number of permutation formed by taking r things at a time out of n things in any object arrangement such that each object can be taken any number of time is .
Circular permutation:
If we fix one of the objects around the circumference of a circle then number of permutation of n different thing taken all at a time is (n-1)! Ways. It will be same as by putting (n-1) objects at (n-1) places.
But if we do not consider the direction i.e. clockwise and anticlockwise then the number of permutation is .
Combination
From a given group of object each of the number of groups which are formed by taking some objects or all objects at a time without caring about the sequence of the objects is called combination. The number of combination formed by taking r objects at a time out of n object is denoted by n where C expresses combination.
n =
For example if we have 3 objects A , B and C , 2 objects are taken out at a time then 3 combination are formed AB , BC and CA.
Note:
If r= 0 , then n = = 1
184 IBPS PO EXAM 2013 : Quantitative Aptitude
If r= 1 , then n = = n
If r= n , then n = = 1
n = n
Example: Find the value of .
Solution: We have, = = = 100 {Because 1! =1}
Restricted combination:
The combination of r object out of n objects on which p specific objects:
1. Are always included is n-p . We have to keep aside p specific objects and to select remaining (n-p).
2. Are never included is n-p . Since p specific object are never included we have to form the combination taking r obects out of (n-p) objects.
The number of ways to select some or all thing out of any number of given thing:
There are 2 ways to select anything i.e. either it will be selected or not. Therefore number of ways to select n things is 2 2 ……… n times = . In these empty selection is also include.
For non- empty selection is -1.
Note: n + n +………. + n = -1.
185 IBPS PO EXAM 2013 : Quantitative Aptitude Difference between permutation and combination:
Suppose there are 5 objects out of which 2 have to be chosen.
Permutation Combination
Number of required way
=
= = 5 4 = 20
=
= = = 10
So it is clear that in permutation order matters while in combination order does not matter.
Probability
The mathematical measure of the uncertainty is called probability. For example, consider the following questions:
(a) Will it rain today?
(b) Which of the three candidates will win?
(c) On throwing a dice, the number obtained will be even or odd?
(d) On tossing a coin, head will occur or tail will occur?
The answer to all these question is not sure i.e. there is uncertainty .We study the uncertainty of the result of such question in the theory of probability , which may not have one result but more than one result are possible .
Random experiment:
The experiments in which the outcomes cannot be predicted before hand is called random experiments. When these kind of experiment are repeated under identical condition, they do not produce the same outcome every time and there may be many possible outcome which depends upon chance and cannot be predicted. For example, on tossing a coin either the head will come up or the tail will come up, we cannot predict it. This is an example of random event.
Sample Space:
The set of all possible outcomes of experiments is called the sample space and it is denoted by S. And the subset of a sample space is called an event. That is, every subset A of the sample
186 IBPS PO EXAM 2013 : Quantitative Aptitude
space S is an event of that random experiment. For example, in an experiment of tossing a coin, if h is obtained then it is a random event, since here S = {H, T} and {H} S
Now, the probability of any event A can be defined as the ratio between the number of favourable outcomes to the event A and the number of total equiprobable outcomes, that is
P(A) =
Here it should be noted that the probability of a certain or sure event is 1 and that of impossible event is 0.
Now, since the probability of an event to occur is =
So the probability of an event A not to occur is = 1 – Mutually Exclusive events:
Two events A and B are said to be mutually exclusive if they cannot occur together, that is simultaneously. For example , on throwing a dice , the events A = { 2,4,6 } and B = { 1,3,5 } are mutually exclusive events , i.e. A B = .
In term of probability if A and B are mutually exclusive events, then P (A B) = P (A) + P (B) and,
P (S) = P (A) + P (A’) = 1 where A’ is Complement of A.
Example: If a dice is thrown once then the probability of the number appearing on dice is more than 2?
(a) 1/3 (b) 1/2 (c) 2/3 (d) 4.3/4
Solution: As there are 6 faces on a dice,
So the total number of possible events are 1, 2 , 3 ……. 6 , that is = 6 Now the number more than are 3, 4 , 5 and 6
So total number of favourable events =
187 IBPS PO EXAM 2013 : Quantitative Aptitude Probability of an event =
Required probability = =
Example : An urn contains 3 green, 6 red, and 4 black balls. 3 balls are drawn. Find the probability that all 3 balls are of same colour?
(a) 3/44 (b) 25/286 (c) 15/286 (d) 5/286
Solution: Total number of balls in an urn is 13.
Number of ways 3 balls can be drawn out of 13 balls = = = 286
Numbers of ways 3 green balls are drawn = = = 1
Numbers of ways 6 red balls are drawn = = = 20
Numbers of ways 4 black balls are drawn = = = 4 Now,
The required probability = + + =
188 IBPS PO EXAM 2013 : Quantitative Aptitude