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Probability Notes With Summary

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Published in: Mathematics | Science
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Probability Notes with Summary.

Upendra K / Bangalore

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Qualification: M.Tech (National Institute of Technology - 2010)

Teaches: Algebra, Business Mathematics, Mathematics, Statistics, Biology, Chemistry, Physics, KVPY Exam, SSC Exams, BITSAT, CET, IIT JEE Mains, Electrical, Electronics, Instrumentation, AIEED, AITMT, CAT Exam, CMAT

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  1. ROBABILIT INTRODUCTION: 1.1 In our daily life, most of us go through some common phrases like ' most probably' , ' most likely' , ' almost uncertain' etc. We will find that all of them involve an element of uncertainty after reading these phrases carefully. The measure of uncertainty is known as the theory of probability, Probabilities are ratios that is usually written as a proper fraction. It is also expressed as a decimal or a percentage. It is a measure of how likely it is for an event E to happen. It is denoted by P(E). number of favourable outcomes for event E n(E) total number of possible outcomes Where n(E) is the number of favourable outcomes in event E and n(S) is the total number of possible outcomes in the sample space S. Now a days, there is great mathematical interest and big practical importance of theory of probability. 1.2 ASIC TERMS & CONCEPTS 1. EXPERIMENT} A process that results in some well defined outcome is known as an experiment. E.g.- (a) when a coin is tossed , we get either a head or a tail i.e. its outcome is a head or a tail which is well defined. (b) when a die is thrown the possible outcomes are 1, 2, 3, 4, 5 and 6 which are
  2. 3. Also well defined. 2. Random experiment, when all the outcomes of the experiment are known in advance but any specific outcome of the experiment is not known in advance is called random experiment. e.g.- tossing of a coin and throwing a die is random experiment because we know in advance that only two possible outcomes head and tail in case of a coin and there are only six possible outcomes 1 , 2 , 3 , 4, 5 and 6 in case of a die. Event: an outcome of a random experiment is called an event. e.g.- if a card is drawn from a well shuffled pack of 52 playing cards , any one of them can be outcome. Therefore, there are 52 events of random experiment of drawing a card from a pack of 52 playing cards. 4. Equally likely outcomes, it is very clear in advance that the coin on toss will give outcome only either a head or a tail. It means there are equal chances for the coin to display with its head or a tail as for each 50%. So, we can say that the outcomes head and tail are equally likely . easurement of robabili 1.3 The probability of an event designates the possibility of happening. If in a random experiment, total number of events (outcomes) are 'a' out of which 'b' events are favourable to a particular event E. then P(E) = Probability of happening event E a number of outcomes favourable to E total number of all possible outcomes xample•, A die is rolled once and odd number is required on the upper face of it. then
  3. In this experiment, total number of outcomes = 6 (1,2,3,4,5 and 6) Number offavourable outcomes = 3 ( 1 or 3 or 5) . probability of getting an odd number on upper face number of favourable outcomes 3 total number of possible outcomes 6 1. Experimental probability; 1 2 A probability is called an experimental probability when it is based on an actual experiment. e.g.- if a coin is tossed 100 times and the outcomes of this experiment are 52 gand that of a tail is 48 heads and 48 tails , the probability of a head IS 100 100 since these probabilities are based on the actual experiment of tossing a coin 100 times. So, they are experimental probabilities. theoretical probability 2. A probability is called theoretical probability if a repetition of an experiment can be avoided for calculating the exact probability. 1. 2. 3. 4. Tip for Students An adequate recording of the outcome is required for the finding the experimental probability. The experimental probability can be applied to every event associated With an experiment which can be repeated a large number of times. In theoretical probability, we make certain assumptions and one of these assumptions is that the outcomes are equally likely. number of favourable outcomes probability of an event = number of all possible outcomes
  4. am lel A bag contains a black ball, a red ball and a green ball, all the balls are identical in shape n size. Robert takes out a ball from the bag at random. Find the probability that the ball drawn is: a.) Green ball b.) Red ball c.) Black ball olution When Robert takes out a ball at random, the outcomes of the experiment are equally likely. Total number of possible outcomes = 3 a.) Number of favourable outcome (getting a green ball) The probability of drawing a green ball number of favourable outcomes total number of outcomes 1 b.) P (drawing a red ball) = 3 1 c.) P (drawing a black ball) = 3 REMEMBER: 1 3 a.) The sum of probabilities of all the elementary events of an experiment is always one. In above example, P(red ball) + P(black ball) + P(green ball) = 3 3 3 b.) (i) A deck of playing cards consists of 52 cards which are divided into 4 Suits of 13 cards each. (ii) The four suits in a deck of cards are : spades , hearts , diamonds And clubs. Thus a deck consists of 13 cards of spades , 13 cards of
  5. 1.4 Hearts, 13 cards of diamonds and 13 cards of clubs. (iii) the 13 cards of each suit are : an ace , a king , a queen , a jack , and 2. (iv) kings, queens and jacks are called face cards. A deck of 52 cards contains 4 kings, 4 queens and 4 jacks. . total number of face cards in a deck of 52 cards are = 4 + 4+ 4 = 12. (v) clubs and spades are of black colour whereas hearts and diamonds are of red colour. ROPERTIES OF PROBABILITY 1. If the probability of an event = 0 i.e.- P (E) = 0 then Eis called an impossible event. e.g. The probability of throwing a die and getting 7 is Q, which is 0. 6 2. If the probability of an event = 1 i.e.- P (E) = 1 then Eis called a certain event. e.g. The probability of throwing a die and getting a number smaller than 7 is é , which 6 is 1. 3. 0 P (E) 1. Any probability lie between 0 and 1 inclusive. Probability of no Event can be less than 0 and more than 1. It means the closer to zero, then the more unlikely the event is to happen ; the closer to 1, then the more likely the event is to happen. -1
  6. P (E) = 1 - P(notE) = 1 -P (E) E and É (not E) are called complementary events. É contains all the outcomes of the sample space that are not in E . 1.5 OSSIBILITY DIAGRAMES AND TREE DIAGRAMS When an experiment involving two events is complex, we use the Possibility Diagram or the Tree Diagram to list the possible outcomes of the sample space. POSSIBILITY DIAGRAMES: We use a possibility diagram to represent the sample space when a random experiment involves two stages. XAMPLE2 A bag contains 5 red balls and 8 black balls. A ball is taken out at random. Find the probability that it is a) Red b) black c) green olutio there are total 13 balls in the bag so there are 13 possible outcomes. a) P(a red ball) = ± 13 8 b) black ball) = — 13 = O it is an impossible event since there is no any c)P(a green ball) = 13 green ball in the bag. The probability of an impossible event is zero.
  7. In a single throw of a die, find the probability of getting a number a) Greater than 3 olution b) less than or equal to 3 c) not greater than 3 Total possible outcomes are 6 (1,2,3,4,5 and 6) in a single throw of a die. a) Numbers greater than 3 are 4 , 5 and 6 out of 1,2,3,4,5 and 6. . total number of favourable outcomes = 3 (4, 5 and 6) number of favourable outcomes P (greater than 3) = number of all possible outcomes 3 6 1 2 b) numbers less than or equal to 3 are 1 , 2 and 3 out of possible outcomes 1, 2, 3, 4, 5, and 6. . total number of favourable outcomes — -3 (1,2 and 3) number of favourable outcomes P (less than or equal to 3) = number of all possible outcomes 3 1 6 2 c) numbers not greater than 3 are 1, 2 and 3 out of possible outcomes 1, 2, 3, 4, 5 and 6. . total number of favourable outcomes = 3 ( 1 , 2 and 3) number of favourable outcomes P (not greater than 3) = number of all possible outcomes 3 6 1 2
  8. Tip for Students: In a single throw of a die, to get a number less than or equal to 3 and To get a number not greater than 3 mean the same. . P (less than or equal to 3) = P (not greater than 3) In a tennis match between pinto and zain, the probability of winning of Pinto is 0.54. find the probability of: a) Winning of zain b) Not winning of pinto a) P(winninng of pinto) + P(not winninng of pinto) 0.54 + P(not winninng of pinto) = 1 P(not winninng of pinto) = 1 - 0.54 = 0.46 b) P(winninng of zain)= P(not winninng of pinto) = 0.46 Tips for Students: 1. tossing a coin once: The total possible outcomes are: head (H) and tail(T) Number of possible outcomes = 21 = 2 2. tossing a coin two times or tossing of two coins simultaneously: Number of possible outcomes = 22 = 4 possible outcomes are : i)head on first coin and head on 2nd coin i.e. HH
  9. ii)head on first coin and tail on 2nd coin i.e.HT iii)tail on first coin and head on 2nd coin i.e. TH iv)tail on first coin and tail on 2nd coin i.e. TT tossing of 2 coins simultaneously or tossing one coin twice gives same outcomes. Similarly, tossing of 3 coins simultaneously or tossing one coin three times gives same outcomes. 3. Tossing a coin 3 times or tossing 3 coins simultaneously : total number of possible outcomes = 23 = 8 possible outcomes are AND TTT. In general, if a coin tossed n times or n coins are tossed simultaneously total number of possible outcomes = 2n 4.throwing or rolling a die once : total possible outcomes are : 1, 2 , 3 , 4, 5 and 6 number of possible outcomes = 61 = 6 5. throwing 2 dice simultaneously or 1 die 2 times number of possible outcomes = 62 = 6x6 = 36 in general , if a die is rolled n times or n dice are rolled simultaneously, total number of possible outcomes = 6n A pair of fair dice is tossed. Let X denote the sum shown on both dice. Find the probability that b) x > 9. c) x < 13. d) x = 10. e) x = 13.
  10. OLUTION Since one die can have 6 possible outcomes, two dice can have 6X6 = 36 possible outcomes. A possibility diagram is drawn to show all the possible sums. 1st DIE nd DIE a) c) 13) 36 = 36 36 d) = 10) 36 e) = 13) 36 36 12 TREE DIAGRAMES When a random experiment consists of the two or more stages, we use a probability tree diagram to represent the process.
  11. Three unbiased coins are tossed. Find the probability of getting a) 3 tails. OLUTION FIRST COIN b) 2 head and 1 tail. SECOND COIN THIRD COIN OUTCOM HI-IH HHT HTH HTT THH THT TTH TTT
  12. There are 8 equal likely outcomes. a) b) P(3tails) P (TTT) 1 8 E = {HHT, HTH, THH} 3 P (2 head and 1 tail) 8 1.6 DDITION OF PROBABILITIES For two mutually exclusive events A and B, the probability of A or B occurring is Mutually exclusive means that the events cannot happen at the same time. A single die is thrown. Find the probability of a 3 or 5 showing up. OLUTION If you throw a die, the outcome will be 1, 2, 3, 4, 5 or 6, but only one of these can occur, e.g. the score cannot be 3 and 5 at the same time. The outcomes 1, 2, 3, 4, 5 and 6 are therefore mutually exclusive.
  13. 1 1 6 6 2 6 1 3 A bag contains 7 red balls, 5 blue balls, 4 green balls and 2 yellow balls. A ball is chosen at random from the bag. Find the probability that the chosen ball is a) c) Blue not Red b) Red, Green or Yellow OLUTION Total number of balls in the bag = 18 Total number of possible outcomes = 18 a) 5 P (A Blue Ball) 18 b) P (a Red, green or Yellow Ball) = P( a Red Ball) + P ( a Green Ball) + P ( a Yellow Ball ) c) 7 4 2 18 18 18 13 18 7 P (Red) 18
  14. a) A loses c) Neither A nor B wins. P ( Not Red) 1 - PC Red) 7 1- 18 11 18 MPLE9 1 The probabilities of three teams, A, B and C, winning a football match are - respectively. Find the probability that 1- 1 1 - and - 5 2 b) Either B or C wins a) b) c) P (A loses ) 3 4 P ( B or C wins ) P (A or B wins ) - P( A wins ) 1 4 P( B wins ) + P( C wins ) 1 1 5 2 7 10 P (A wins ) + B wins )
  15. P ( Neither A Nor B wins ) XAMPLEIO 1 1 5 4 9 20 1 1- 11 20 - P (A or B wins ) 9 20 A bag contains 6 red marbles, 4 green marbles and 2 black marbles. A marble is chosen at random from the bag. Find the probability that it is a) red c) Red or black Solution b) black d) neither red nor green Total number of marbles in the bag Total number of possible outcomes - 6 a) P(a red marble) 12 2 b) P(a black marble) 12 1 2 1 6 c) P(either a red or black marble) = P(a red marble) + P(a black marble) = 12 - 12 1 1 2 4 6 6 2 3
  16. d) P(a green marble) 12 3 P(either a red or green marble) = P(a red marble) + P(a green marble) 2 3 5 6 PC neither a red nor green marble) - 1 — P(either a red or green marble) 5 6 6 XAMPLEII A card is selected at random from 30 cards numbering from 1 to 30. Find the Probability that the number on the card a) is odd c)is even and divisible by 5 olution Sample Space = b) is divisible by 4 d) is even or divisible by 5 {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30} a) Total number of odds on the card from 1 to 30 = 15 Favourable outcomes 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, } 15 1 P(an odd number) = 30 2 b) Favourable outcomes ={4, 8, 12, 16, 20, 24, 28} 7 P(a number divisible by 4) = 30 c) Favourable outcomes 20, 30} 3 P(an even number and divisible by 5) = 30 d) Favourable outcomes = 1 10
  17. {2, 4, 5, 6, 8, 10, 12, 14, 15, 16, 18, 20, 22, 24, 25, 26, 28, 30} 18 P(an even number or divisible by 5) = 30 3 5 There are 15 women, 12 men, 8 girls and 10 boys in a group. A person is chosen at random from the group. Another person is chosen at random from the remaining people in the group. Find the probability that a) The first person chosen is a male b) The first person is an adult and the second is a child c) An adult is chosen on both times d) Neither one of the person chosen is an adult olution Total number of the people = 15 + 12 + 8 + 10 = 45 a) P(lst person chosen is a male) 12 10 (favourable outcomes = 12 men and 10 boys) 45 45 22 45 b) P(lst person is an adult and 2nd is a child) 27 18 (favourable outcomes = 27 women & men and 18 girls & boys) 45 44 27 110 c) P(an adult on both times) 27 26 45 44 39 110 d) P(Neither person is an adult)
  18. = P( both are children) 18 17 45 44 17 110 XAMPLE13 Bag A contains 5 black balls and 3 white balls. Bag B contains 4 black balls and 6 white balls. If a ball is taken at random from each bag , find the probability that a) Both balls are white b) both balls are of the same colour c)two balls are of different colours. olution Bag A 5 8 8 Bag B 4 10 6 10 4 10 6 10
  19. a) P(both balls are white)= P(WW) 3 6 -x— 8 10 9 40 b) P(both balls are of the same colour)= P(BB or WW) = P(BB) + P(WW) c) P(two balls are of different colour) 5 4 8 10 4 40 19 40 9 40 = 1 - P(both balls of the same colour) 40 21 40 Sid , Kim, Hardy and Anna are playing a game of Monopoly. The probabilities of each of them winning the game are shown in table below. Sid 0.38 a) Calculate the value of x. Kim 0.27 Hard x Anna 0.22 b) Find the probability that either one of the girls will win the game. Roony and Kate will challenge each other at a game of chess. The probability That Roony will win the game is 0.46. c) Find the probability that Sid will win the Monopoly game and Kate will win the
  20. Game of chess. OLUTION a) x = 1-0.38-0.27 -0.22 = 1 - 0.87 (Total probabilities is always equal to 1) = 0.13 b) P(either one of the girls will win) = P(Kim wins or Anna wins) = P(Kim wins) + P(Anna wins) = 0.27 + 0.22 = 0.49 c) P(Kate wins chess) = 1 - 0.46 (Total probabilities is always equal to 1) = 0.54 P(Sid wins Monopoly and Kate wins chess) = P(Sid wins) x P(Kate wins) = 0.38 x 0.54 = 0.205 a) 3 biased coins are tossed and the number of tails appearing in each outcome is recorded in the table below: Number of Tails Probability 2 9 1 1 9 2 5 9 3 x
  21. i) ii) iii) Calculate the value of x. Find the probability that at least one tail appears. Find the probability that all heads or tails appear. b) John, Smith and Brian played a game of Chance. John is thrice as likely to win as Smith and Smith is twice as likely to win as Brian. Find the probability that John will win ii) Brian will not win either Smith or Brian will win i) ii) OLUTION 2 a) i)x=l 9 8 9 1 9 1 9 5 e (Total probabilities is always equal to 1) 9 ii) iii) P(at least 1 tail) = 1 - P(no tails) 9 7 9 P(all heads or all tails) = P(no tails) + P(all tails) 2 1 1 3 ( P(all tails) = P(3 tails)=
  22. i) ii) iii) b) Let the probability that Brian will win the game be y. Probability that Smith will win = 2y Probability that John will win = 6y Total probabilities = 1 y + 2y + 6y = 1 1 P(John will win) = 6y 1 9 2 3 P(Brian will win) 1 9 P(either Smith or Brian will win) = P(Smith will win) + P(Brian will win) 2 1 1 3 XAMPLE16 A particular biased die was thrown and the probability of each number appearing is recorded in the table below. Number Probabili 1 0.25 2 0.16 3 0.12 4 0.32 5 0.08 6 z a) Calculate the value of z b) If the particular die is thrown once , find the probability that the number on the die is
  23. i) an odd number ii) a number other than 4 c) If the particular die is thrown twice , find the probability of getting a sum of 9. a) z = 1-0.25-0.16- = 1 - 0.93 = 0.07 0.12 - 0.32 - 0.08 b) i) P(an odd number) = or 5) = 0.25 + 0.12 + 0.08 = 0.45 ii) P(a no. other than 4) = 1- 0.32 = 0.68 c) P(sum of 9) = 0.07 + 0.07 xo. 12 + 0.32 xo. 08+0.08 xo. 32 = 0.0084 + 0.0084 + 0.0256 + 0.0256 = 0.068 A biased spinner is used in a board game. When the spinner is spun once, The probability of scoring a 1 is 0.18 The probability of scoring a 2 is 0.26
  24. The probability of scoring a 3 is 0.14 The probability of scoring a 4 is 0.32 a) Calculate the probability of scoring a 5 with one spin. b) The spinner is used in a board game called 'Land Mines'. In the game, a player moves his counter forward by the score on the spinner. if the player's counter lands on a bomb, he is out of the game. 92 93 94 95 96 BOMB 97 BOMB 98 BOMB 99 Finish 100 i) Jakson's counter is now on number 93. Find the probability that he will not be out of the game after one spin. ii) Smith's counter is now on number 92. Find the probability that he will finish the game with two spins. a) 1-0.18 - 0.26 - 0.14-0.32 = 1 - 0.90 (Total probabilities is always equal to 1) = 0.10 b) i) P(Jakson will not out of game) ii) = 0.18 + 0.26 = 0.44 P(Smith will finish in two spins)
  25. 1.7 = and 5) = 0.14 x 0.10 = 0.014 ULTIPLICATION OF PROBABILITIES If A and B are independent events, the probability of both events A and B occurring is and B) = (B) Events A and B are independent events if the occurrence of one of them does not influence the occurrence of the other. M 18 Rosy has 12 tops in the colours: 3 red, 4 pink, 3 blue and 2 green. She has 7 jeans in the colours: 4 black and 3 blue. She selects a top and a jeans at random. Find the probability that she selects a red tops and black jeans. OLUTION The events are independent as the colours of tops selected has no effect on the colour of jeans selected. 4 P (a pink top) 12 4 P (a black jeans) 7 1 3 P (a pink top and a black jeans) P (A pink top) X P (A black jeans) 1 4 -x- 3 7 4 21
  26. XAMPLE19 A bag contains 3 red marbles and 2 green marbles. A marble is taken at random from the bag and its colour is noted. The first marble is replaced in the bag before the second marble is taken. Find the probability that a) Two red marbles are taken c) Both marbles are of the same colour OLUTION b) The second marble is red d) The two marbles are of different colour. A tree diagram is constructed to list the outcomes. FIRST MARBLE SECOND MARBLE OUTCOME a) 3 5 2 5 P (Two red marbles) 3 3 -x- 5 5 3 5 2 5 3 5 2 5 To calculate the probability of an outcome, Multiply the probabilities along the branches leading to the outcome.
  27. 25 b) P (Second marble is red) P(RR or GR) c) P (Both marbles are of the same colour) d) 13 25 P ( Both marbles are of the different colour) 1 - P (Both marbles are of the same colour) 13 25 12 25
  28. ALTERNATIVE METHOD: P (Both marbles are of the different colour) P (RG Or GR) 3 3 -x—+-x 5 5 12 25 A bag contains 5 red balls, 4 blue balls and 1 green ball. Two balls are drawn at random, one after the other, and are not replaced. Find the probability that a) The first ball drawn is red, c) Both balls are of the same colour, OLUTION Draw a tree diagram to illustrate. b) Both balls are blue, d) The two balls are of different colours.
  29. FIRST BALI, 10 10 10 XAMPLE21 SECOND BALL
  30. The probability that Boris oversleeps is 0.4. If he oversleeps, the probability that he is late is 0.7. If does not oversleep, the probability that he is late is 0.2. a) Complete the probability tree diagram. b) Find the probability that he is late. 0.4 a) 0.4 (0.6) 0.7 OVERSLEEPS OVERSLEEPS Late (0.3) (0.2) DOES NOT OVER SLEEP (0.8) Late Not Late Late Not Late
  31. b) P (late) P (Oversleeps, late) + P (Does not oversleep, late) (0.4 x 0.7) + (0.6 x 0.2) 0.28 + 0.12 0.40 There are 7 orange and 5 purple counters in a bag. A counter is drawn at random, replaced before another counter is drawn. Find the probability that a) Both counters orange b) At least one counter is purple c) Both counters are of the same colour d) Both counters are of different colour. Éirst counter 7 12 5 12 o Second counter O 7 12 5 12 7 O 12 5 12
  32. P(both counters are orange) = P(OO) 7 7 12 12 49 144 a) P(at least one counter is purple) = 1 - P(none of the counters is purple) = 1 - P(OO) 49 144 95 144 b) P(both counters are of the same colour) = P(OO or PP) = P(OO) + P(PP) 49 5 x 144 12 12 25 49 144 74 144 144 37 72 c) P(both counters are of different colours) = 1 - P(both counters are of the same colour) 72 35 72
  33. XAMPLE23 Three fair coins are tossed. Find the probability of getting a) 3 Tails c) at least 1 Head olution b) 2 Tails and 1 Head A tree diagram is constructed to list the outcomes. a) FIRST COI Tails) = P (TTT) SECOND colN THIRD COIN
  34. 1 1 1 2 2 2 1 8 b) Tails and 1 Head) = P(TTH , THT , HTT) = P(TTH) + P(THT) + P(HTT) 1 1 8 3 8 c) P(at least 1 Head) = 1 - P(no Heads) = 1 - P(TTT) 8 7 8 1 Jack is playing a game of darts. The probability that he hits the target is - . 4 a) If he throws 4 darts , find the probability that he will hit the target i) Four times ii) At least twice. b) Find the probability that his dart will hit the target after exactly n times. Find the value of 'n'. OLUTION a) i) P(hit4 times) 1 1 1 1
  35. 1 256 ii)P(hit at least twice) = 1 - P(will not hit) - P(will hit once) 2) P(hit, miss, miss, miss), P(miss, miss, hit, miss), P(miss, miss, miss, hit), P(miss,hit,miss,miss) 108 256 256 189 256 67 256 July either drives or takes the Cab to work each morning. The probability that 3 she drives is - 8 when she drives , the probability that she is not late is — 10 4 when she takes the Cab, the probability that she is late is — 10 a) find the probability that on a particular morning, i) July drives and is late She is not late ii) b) If July goes to work from Monday to Friday, find the probability that She will take the Cab on each of these days. olution
  36. MPLE2 The diagram shows three identical smaller circles inside a larger circle. 0 is the centre of the large and one of the small circles. A point is chosen at random inside the large circle. Find the probability that it lies inside the shaded region. olution Let the radius of the smaller circle be x cm.
  37. Radius of the larger circle - - 3x cm Area of smaller circle Area of a circle 9 IT x 2 cm2 Where r Radius Area of shaded region IT x2 cm2 Area Of shaded region 91Tx2 -3XIT x 2 cm2 6 IT x 2 cm2 P ( Inside shaded region) Area of shadded region Area of larger circle 2 6m 2 9m 2 3 Tip for Students: We can also write the formula for probability as: Area where E occurs Total area
  38. xam le26 A card is drawn at random from a well shuffled deck of playing cards. Find the probability that the card drawn is (i) A card of spades or an ace (ii) a red king (iii) neither a king nor a queen (iv) either a king or a queen. Total number of outcomes 52 [since there are 52 cards in pack of playing cards] (i) (ii) (iii) Favourable outcomes = 13 + 4 — I 16 [Since, there are 13 spadescardsand 4 aces. But 1 aceappears in 13 spade cards] 16 4 Required probability 52 13 Favourable outcomes 2 [since there are 2 red kings in playing cards] 2 1 . required probability 52 26 Number of king cards and queen cards = 4 + 4 8 [since,there are 4 kingsand4queensinthe packofcards] Favourable outcomes 52 - 8 44 44 11 . required probability 52 13 By equation (iii) favourable outcomes 8 2 . required probability 52 13 xam le27 8 From a pack of 52 playing cards; Jacks, queens, kings and aces or red colour are removed. From the remaining, a card is drawn at random. Find the probability that the card drawn is (i) a black queen (ii) a red card (iii) a black jack (iv) a picture card. [jacks, queens and kings are picture cards]
  39. Total number of outcome 52 Cards removed 2 + 2 + 2 + 2 8 [sincethere are 2 jacks„2 queens,2kings and2aces of red colour] (i) (ii) (iii) (iv) . remaining number of cards 52=8 = 44 total number of outcomes 44 Favourable outcomes — 2 [ thereare 2 blackqueens] 2 1 . required probability 22 44 Favourable outcomes number red cards left 26 — 8 18 9 . probability for a red card 44 22 Favourable outcomes number of black jacks = 2 [ out of total 4 jacks, two have been removed] 2 1 :.required probability 22 44 Number of picture cards left 2 + 2 + 2 backs,queens, kings are picture cards] 6 3 . required probability 22 44 18 xam le28 A box contains 19 balls bearing numbers l, 2, 3 19. A ball is drawn at random from the box. Find the probability that the number on the ball is (i) a prime number (ii) divisible by 3 or 5 (iii) neither divisible by 5 nor by 10 (iv) an even number Solution Total number of outcomes = 19 (i) (ii) Prime numbers from I to 19 are 2, 3, 5, 7, Il, 13, 19 P (a prime number) 19 Numbers divisible by 3 are 3, 6, 9, 1,2 15, 18 6 numbers divisible by 5 are 5, 10, 15 = 3 numbers divisible by 3 and 5 is 15 — I . numbers divisible by 3 or 5 6 + 3 — I 8 8
  40. (i) (ii) (iii) xam le30 111 35 —7 81 P (numbers divisible by 3 or 5) 19 Numbers divisible by 5 are 5, 10, 15 = 3 numbers divisible by 10 is 10 — I :.numbers divisible by 5 and 10 is 10 — I :.numbers divisible by 5 or 10 is — 3 + I . numbers which are neither divisible by 5 nor by 10 P (number neither divisible by 5 nor by 10) 2-: 19 3 16 Find the probability that a number selected at random from the numbers 35 is a (i) prime number (ii) multiple of 7 (iii) multiple of 3 or 5. olution Prime numbers are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31 11 [ total number of outcomes If P (Prime number) Multiples of 7 are: 7, 14, 21, 28, 35=5 5 (a multiple of 7) Multiples of '3' are 3, 6, 1 9, Multiples of '5' are 5, 10, 15 Multiples of '3' and '5' are 15, Multiples of 3 or 5 — Il + 7 16 P (multiples of 3 or 5) 35 351 33 35 30 2=16 11 7 2 (a).Cards marked with numbers 3, 4, 5 .50 are placed in a box and mixed thoroughly. One card is drawn at random from the box. Find the probability that number on the drawn card is (i) divisible by 7 (ii) a number which is a perfect square. Total number of outcomes 50 2 48
  41. (i) (ii) xam le31 Cards in the box having number divisible by 7 are 7, 14, 21, 28, 35 42, 49 favourable outcomes in this case 7 required probability 48 Cards in the box having perfect squares are 4, 9, 16, 25, 36, 49 which are 6 in number. :.favourable outcomes 6 6 required probability (b).Cards marked with number 13, 14, 15, 60 are placed in a box and mixed thoroughly. One card is drawn at random from the box. Find the probability that the number on the drawn card is, (i) divisible by 5 (ii) a number which is a perfect square. Total number of outcomes 60 — 12 48 (i) (ii) Cards in the box having numbers divisible by 5 are 15, 20, 25, 30,35, 40 45, 50, 55, 60 which are 10 in number. 10 5 . required probability Cards in the box having numbers, which are perfect squares are . 36, 49 which are 4 in numbers 4 1 required probability 12 16, 25, x m 1 32 A bag contains 5 red balls, 4 green balls, and 7 white balls. A ball is drawn at random from the box. Find the probability that the ball drawn is (a) white (b) neither red nor white. Total number of balls in the bag — 5 + 4 + 7 — 16 total number of possible outcomes = 16 a ball is drawn at random . all equally likely outcomes will be taken] (a) Favourable outcomes for a white ball = 7
  42. 7 . required proability for a white ball 16 Favourable outcomes for neither red nor white ball number of green balls = 4 4 required probability 16 1 33 x 1 4 From a well shuffled pack of 52 cards, black aces and black queens are removed. From the remaining cards, a card is drawn at random. Find the probability of drawing a King or a Queen. Sjtnirjn Total number of cards 52 Number of black aces 2 Number of black queens 2 . Cards left 52-2-2=48 Total number of equally likely cases 48 Number o Kings and Queens left in the 48 cards 6 favourable cases — 6 6 required probability 48 am e 4 1 6 A die is thrown once. Find the probability of getting a number less than or equal to 4. olution Possible outcomes are : l, 2, 3, 4, 5, 6, which are 6 in number. Favourable outcomes are : l, 2, 3, 4 which are 4 in number. 2 4 Required probability 6 3 xam le35 (a) Once card is drawn from a well shuffled deck of 52 cards. What is the probability of drawing an ace? (b) One card is drawn from a well shuffled deck of 52 cards. What is the probability of drawing a ace? (c) One card is drawn froma well shuffled deck of 52 cards. Find the probability of getting:
  43. (i) a king of red suit or colour (ii) a face card (iii) a red face card (iv) a queen of black suit (v) a jack of hearts (vi) a spade (vii) an ace of hearts (viii) the queen of diamonds. (Note: King, queen and jack are called face cards) olution (a) Total number of possible outcomes = 52 [yanyoneoutof 52cardscanbedrawnl Favourable outcomes are : 4 [ number of aces in the deck 4] 4 required probability 52 1 13 (b) Total number of possible outcomes 52 Favourable outcomes are — 52 — 4 48 [vnonacecards—totalcards. numberofacecards] 12 Required probability 52 13 (c) Total number of possible outcomes 52 (i) (ii) (iii) (iv) Favourable outcomes 2 i •Mhere are 2 kings of red suit] Favourable outcomes 12 [ there are 3 + 3 6 red face cards] 6 3 . required probability 52 13 Favourable outcomes — 6 [•:there are 3 + 3 6 3 Required probability 52 26 Favourable outcomes — 2 2 Required probability 52 6 red face cards] ['.•there are 2 queens of black suit] 1 26
  44. (v) (vi) (vii) [•:there is onlyljack of hearts] Favourable outcomes I 1 Required probability 52 Favourable outcomes = 13 [Ythere are 13 spade cards] 13 1 Required probability 52 4 [•:there is only I ace of hearts] Favourable outcomes = I 1 Required probability 52 A bag contains 5 red balls, 8 white balls, 4 green balls and 7 black balls, if one ball is drawn at random, find the probability that it is (i) black (ii) red (iii) not green. Total number of outcomes 5 + 8 + 4 + 7 24 (i) (ii) (iii) Example37: Favourable outcomes Required probability Favourable outcomes Required probability Favourable outcomes Required probability [•:there are 7 black balls] 7 7 24 [Vthere are 5 red balls] 5 5 5 + 8 + 7 20 [•.•total no. reen balls 20 5 24 6 4 A bag contains 5 red balls and some blue balls. (i) If the probability of drawing a blue ball from the bag is thrice that of a red ball, find the number of blue balls in the bag. (ii) If the probability of drawing a blue ball from the bag is four times that of a red ball. Find the number of blue balls in the bag. Solution: let the number of blue balls be x. then, total number of balls in the bag 5 red balls + x blue balls
  45. (5 + x ) balls As per given condition, x Or , x 5 15 Hence, the number of blue balls (ii) as per given condition , x 5 Or, x 20 Hence, the number of blue balls Example38: 15 20. A bag contains 6 red balls and some blue balls. If the probability of drawing a blue ball from the bag is twice that of a red ball, find the number of blue balls in the bag. Solution: let the number of blue balls be x. then, total number of balls in the bag As per given condition, 6 red balls + x blue balls (6 + x ) balls x Or, x 6 2.— 12 Hence, the number of blue balls =12. Example39: A die is thrown once. Find the probability of getting (i) a prime number (ii) a number between 2 and 6. (iii) a number greater than 4 (iv) an odd number. Solution: (i) Possible outcomes are: l, 2, 3, 4, 5, 6 which are six in number. (prime numbers from I to 6 are 2, 3 , 5).
  46. Favourable outcomes are: 2, 3 or 5 which are three in number. Number of favourable outcomes 3 Required probability Total number of possible outcome 6 1 2 (ii) Possible outcomes are : l, 2, 3, 4, 5, 6 which are six in number. (iii) Favourable outcomes are 3,4 or 5 which are three in number. 3 1 Required probability Possible outcomes are l, 2, 3, 4, 5, 6 which are six in number. Favourable outcomes are 5 or 6 which are two in number. 2 1 Reqd. probability 6 3 : If A is the event of getting a number greater than 4, then B will be the event of getting a number not greater than 4 i.e., getting a number less than or equal to 4. 2 1 3' 3 2 1 so, P(A) + P(B) 3 3 thus , P(A) 1 - P(B) Example40: 1. From a well-shuffled pack of cards, a card is drawn at random. Find the probability of getting a black queen. Solution: Since, there are 2 cards of black queen. 2 the probability of drawing a black queen 52 Example41 : 1 26 A die is thrown once. Find the probability of getting a number divisible by 2. Solution: Let A be an event "getting a number divisible by 2". Favourable cases for a number divisible by 2 are 2, 4 and 6. Required probability P(A) 3 6 1 2
  47. Example42: Die is thrown once. Find the probability of getting an even prime number. Solution: Let A be an "getting an even prime number". Favourable case for an even prime number is 2, . Required probability P(A) Example43 : 1 6 A box contains cards bearing numbers from 6 to 70. If one card is drawn at random from the box, find the probability that it bears (i) a one digit number, (ii) a number divisible by 5. Solution: Total number of cards in the bag 65 (i) Since one digit numbers are 6, 7, 8,9 there are four cards having one digit numbr Favourable cases P (card having one digit number) Totalnumber ofcases 4 65 (ii) Since the numbers divisible by 5 from 6 to 70 are 10, 15, 20, 25, 30, 35 40, 45, 50, 55, 60, 65, 70 There are 13 cards having number divisible by 5 Favourable cases P (card having a number divisible by 5) Total number of cases Example44: What is the probability of having 53 Mondays in a leap year? Solution: 13 65 1 5 We know that every leap year has 52 complete weeks and 2 days. The remaining 2 days can be any one of the following cases MT, TW, WT, TF, FS, SS, SM Favourable cases are SM and MT 2 required probability 7
  48. Example45: From a bag containing 5 red, 8 black and 7 blue balls,a ball is selected at random. Find the probability that (i) it is not a red ball,(ii) it is not a blue ball. Solution: Total number of balls in the bag 5 + 8 + 7 = 20. 15 3 (i) P (ball is not red) 20 4 13 (ii) P (ball is not blue) Example46: A box has cards numbered 14 to 99. Cards are mixed thoroughly and a card is drawn from the bag at random. Find the probability that the number on the card drawn from the box is (i) An odd number (ii) A perfect square number (iii) A number divisible by 7. Solution: Total number of cards = 86. i.e., total number of equally likely cases 86 (i) Favourable cases are 15, 17, l, 21, 23 99 i.e., 43 in number. 99 84 (n-l)2 43 1 P (an odd number) 86 2 (ii) Favourable cases are 16, 25, 36, 49, 64, 81 i.e., 8 in number. 8 4 P(a perfect square number) 86 43 (iii) Favourable cases are 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91,98 i.e 13 in number, 7 P(a number divisible by 7 86 Example47: A card is drawn from a pack of 100 cards numbered I to 100. Find the
  49. Probability of drawing a number that is a square. Solution: Total number of all possible outcomes = 100 Favourable outcomes are : I , 4 , 9 , 16 , 25 , 36 the number of favourable outcomes = 10 49 64 81 100 number of favourable outcomes Required probability 10 100 total number of outcomes 1 10 Example48: In a musical chairs game, a person has been advised to stop playing The music at any time within 45 seconds after its start. What is the Probability that the music will stop within the first 20 seconds? Solution: The Favourable results = 0 sec to 20 sec and Total results 0 sec to 45 sec We can say that, The favourable outcomes Total number of outcomes . Required probability Example49: 20 45 20 4 9 Sam and Jade are friends. What is the probability that both will have a.) Different birthdays b.) Same birthdays Solution:
  50. a.) The number of favourable outcomes 365 364 Total number of outcomes 365 364 :.probability(of different birthdays) 365 364 b.) probability(of same birthdays)= I 365 Example50: 1 365 Offices in Jakarta are open for five days in a week (Monday to Friday). Two employees of an office remain absent for one day in the same particular week. Find the probability that they remain absent on: a.) the same day b.) consecutive day c.) Different day Solution: Total number of possible outcomes = 5 >< 5 = 25 Let the five days of the week denoted as Monday by M, Tuesday by T, Wednesday by W, Thursday by Th and Friday by F. Then a.) Favourable outcomes are TT WW,Th Th and FF i.e. 5 in all. 5 1 Probability (absent on the same day) 25 5 b.) Favourable outcomes are : MT , T M , TW, WT ,W Th, Th W, Th F and F Th i.e. 8 in all. 8 Probability(absent on the consecutive day) 25 c.) Probability(absent on the different day) I — probability(absent on the same day) 1- 5 5 UMMARY AND KEY POINT
  51. 1. The measure of uncertainty is known as the theory of probability. Probabilities are ratios that is usually written as a proper fraction. It is also expressed as a decimal or a percentage. It is a measure of how likely it is for an event E to happen. It is Denoted by P(E). number of favourable outcomes for event E n(E) it(S) total number of possible outcomes Where n(E) is the number of favourable outcomes in event E and n(S) is the total number of possible outcomes in the sample space S. 2. Experiment: A process that results in some well defined outcome is known as an experiment. E.g.- when a coin is tossed , we get either a head or a tail i.e. its outcome is a head or a tail which is well defined. 3. Random experiment: when all the outcomes of the experiment are known in advance but any specific outcome of the experiment is not known in advance is called random experiment. e.g.- throwing a die is random experiment because we know in advance that there are only six possible outcomes 1, 2, 3 , 4, 5 and 6 in case of a die. 4. Event: an outcome of a random experiment is called an event. e.g.- if a card is drawn from a well shuffled pack of 52 playing cards , any one of them can be outcome. so, there are 52 events of random experiment of drawing a card from a pack of 52 playing cards. 6. Equally likely outcomes: it is very clear in advance that the coin on toss will give outcome only either a head or a tail. It means there are equal chances for the coin to display with its head or a tail as for each 50%.
  52. So, we can say that the outcomes head and tail are equally likely. 6. If in a random experiment, total number of events (outcomes) are 'a' out of which 'b' events are favourable to a particular event E. then P(E) = Probability of happening event E a b number of outcomes favourable to E total number of all possible outcomes 7. Experimental probability: A probability is called an experimental probability when it is based on an actual experiment. e.g.- if a coin is tossed 100 times and the outcomes of this experiment are 52 52 and that of a tail is . 48 heads and 48 tails , the probability of a head IS — 100 100 8. Theoretical probability: A probability is called theoretical probability if a repetition of an experiment can be avoided for calculating the exact probability. 9. An adequate recording of the outcome is required for the finding the experimental probability. 10. The experimental probability can be applied to every event associated with an experiment which can be repeated a large number of times. 11. In theoretical probability, we make certain assumptions and one of these assumptions is that the outcomes are equally likely. number of favourable outcomes 12. probability of an event = number of all possible outcomes 13. The sum of probabilities of all the elementary events of an experiment is always one.
  53. 14.(i) A deck of playing cards consists of 52 cards which are divided into 4 Suits of 13 cards each. (ii) The four suits in a deck of cards are: spades, hearts, diamonds And clubs. Thus a deck consists of 13 cards of spades, 13 cards of Hearts , 13 cards of diamonds and 13 cards of clubs. (iii) the 13 cards of each suit are : an ace , a king , a queen , a jack , and 2. (iv) kings , queens and jacks are called face cards. A deck of 52 cards Contains 4 kings, 4 queens and 4 jacks. . total number of face cards in a deck of 52 cards are = 4 + 4+ 4 = 12. (v) clubs and spades are of black colour whereas hearts and diamonds are of red colour. 15. Properties of Probability: a.) If the probability of an event = impossible event. O i.e.- = O then Eis called an e.g. The probability of throwing a die and getting 7 is g, which is O. 6 b.) If the probability of an event = 1 i.e.- P (E) = 1 then Eis called a certain event. e.g. The probability of throwing a die and getting a number smaller than 7 is é = 1 6 c.) O P (E) 1. Any probability lie between O and 1 inclusive. Probability of no Event can be less than O and more than 1. It means the closer to zero, then the more unlikely the event is to happen ; the closer to 1, then the more likely the event is to happen.
  54. 1 P (E) = 1 - P(notE) = 1 -P (E) E and É (not E) are called complementary events. É contains all the outcomes of the sample space that are not in E . 16. Possibility Diagrams and Tree Diagrams: When an experiment involving two events is complex, we use the Possibility Diagram or the Tree Diagram to list the possible outcomes of the sample space. a.) POSSIBILITY DIAGRAMES: We use a possibility diagram to represent the sample space when a random experiment involves two stages. b.) TREE DIAGRAMES: When a random experiment consists of the two or more stages, we use a probability tree diagram to represent the process. 17. ADDITION OF PROBABILITIES: For two mutually exclusive events A and B, the probability ofA or B occurring is P(AorB) = Mutually exclusive events means that the events cannot happen at the same time. 18. MULTIPLICATION OF PROBABILITIES: IfA and B are independent events, the probability of both events A and B occurring is P(AandB)
  55. Events A and B are independent events if the occurrence of one of them does not influence the occurrence of the other. 19. Important Results: a) Tossing a coin once: The possible outcomes are: head (H) and tail(T) Number of possible outcomes = 21 = 2 b.) Tossing a coin two times or tossing of two coins simultaneously: total number of possible outcomes = 22 = 4 possible outcomes are : i)head on first coin and head on 2nd coin i.e. HH ii)head on first coin and tail on 2nd coin i.e. HT iii)tail on first coin and head on 2nd coin i.e. TH iv)tail on first coin and tail on 2nd coin i.e. TT the total possible outcomes are : HH , HT , TH and T T. Tossing of 2 coins simultaneously or tossing one coin twice gives same outcomes. Similarly , tossing of 3 coins simultaneously or tossing one coin three times gives same outcomes. c.) Tossing a coin 3 times or tossing 3 coins simultaneously: total number of possible outcomes = 23 = 8 The possible outcomes are AND TTT. d.)Throwing or rolling a die once: possible outcomes are : 1, 2 , 3 , 4, 5 and 6
  56. Total number of possible outcomes = 61 = 6 e.) Throwing 2 dice simultaneously or 1 die 2 times number of possible outcomes = 62 = 6x6 = 36 f.) In general , if a coin tossed n times or n coins are tossed simultaneously the total number of possible outcomes = 2n g.) In general , if a die is rolled n times or n dice are rolled simultaneously, the total number of outcomes = 6n

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