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  2. 2. PROBABILITY • The word probability is related with chance of happening or non happening of an event. • Ex- “the probability that it will rain today” - “probability of getting a head or tail by tossing a coin. The credit for origin of probability goes to the European Gamblers of 17th century. They used to gamble on games of chances such as throwing a dice, tossing up a coin etc.
  3. 3. EVEVNTS 1. EXHAUSTIVE EVENTS - A : the event of getting 1 B : the event of getting 2 F : the event of getting 6 The six Events "A", "B", "C", "D", "E", "F" together are called exhaustive events. [One of these events will occur whenever the experiment is conducted.] 2. EQUALLY- LIKELY EVENTS - In the experiment of tossing a coin: A : the event of getting a "HEAD" and B : the event of getting a "TAIL" Events "A" and "B" are said to be equally likely events [Both the events have the same chance of occurrence].
  4. 4. 5. DEPENDENT EVENT -A box contains 3 white marbles and 4 black marbles. What is the probability of drawing 2 black marbles and 1 white marble in succession without replacement? On the first draw the probability of drawing a black marble is p1= 4/7 On the second draw the probability of drawing a black marble is p2= 3/6= 1/2 On the third draw the probability of drawing a white marble is p3= 3/5 Therefore, the probability of drawing 2 black marbles and 1 white marble is p1*p2*p3= 4/7*1/2*3/5= 6/35
  5. 5. 3. MUTUALLY EXCLUSIVE EVENTS - What is the probability of drawing either a king, a queen, or a jack from a deck of playing cards? -The individual probabilities are King- 4/52 Queen- 4/52 Jack- 4/52 Therefore, the probability of success is P= 4/52+4/52+4/52 = 12/52 = 3/13 4. INDEPENDENT EVENTS -A card is chosen at random from a deck of 52 cards. It is then replaced and a second card is chosen. What is the probability of choosing a jack and an eight? Probabilities:   P(jack) = 4/52 P(8) = 4/52 P(jack and 8) = P(jack) * P(8)    = 4/52* 4/52  = 16/2704  = 1/169
  6. 6. TECHNIQUES OF COUNTING • Counting has its importance in probability because the chance of occurrence of an event has to be assessed by counting the results favorable to the happenings of the event from amongst all possible results. • Some of the techniques are:- Factorial - Permutations
  7. 7. FACTORIAL • A factorial is derived by multiplication. Given any whole number (not a fraction), the factorial can be found by multiplying all the whole numbers together from the given down to one. Here are two examples: • If three is the given whole number, then three factorial is 3 x 2 x 1 or 6. • If six is the given whole number, then six factorial is 6 x 5 x 4 x 3 x 2 x 1 or 720. A factorial is designated by an exclamation point (! ), so a three factorial would be designated 3!.
  8. 8. PERMUTATIONS • Permutation refers to the different arrangement of objects in a set where all elements are different. • These arrangements are to be done without repetition of any individual object. Ex- if we had strips of red, yellow and green color, we can obtain different flags by putting them in different ways.
  9. 9. Q. How many different ways can a set of four state flags be arranged? _ 4 _   _ 3 _   _ 2 _   _ 1 _ The first slot has four choices, the next slot has three choices, the third has two choices and there is only one choice of a flag for the last slot. We multiply these together to get 4*3*2*1 = 24 possible different arrangements of the four flags. This permutation is done without replacement .
  10. 10. • It is represented by the following formulae: nPr= n!/n!-r! n= total elements r= elements selected. Q. Taking all the letters used in JINDAL, find the number of different words if: (a) Every word begins with J; and (b) None of the words begin with J.
  11. 11. (a) There are 6 alphabets and first word should always be ‘J’. Remaining 5 words can be written as5p5= 5!=5*4*3*2*1= 120 words. (b) All the 6 alphabets are different, words can be formed using all the six alphabets. 6p6= 6!= 6*5*4*3*2*1= 720 words. - It includes all the words starting from ‘J’. Therefore, words not starting with ‘J’ are =720-120= 600 words.
  12. 12. • Number of permutations of such items in which some items are common. The formula used is:n!/p!q!r! Q. How many permutations can be made out of letters of the word ‘AGGRAWAL’? There are eight letters in this word and G is repeated twice and A is repeated thrice.
  13. 13. The formula will be:n!/p!q! Where: n=8; p=2; q=3 8!/2!/3!= 8*7*6*5*4*3*2*1/(2*1) (3*2*1) 3360 ways (Ans).
  14. 14. THANK YOU..!