Welcome dear students! Today we are going to learn about Probability from Class 10 Maths.
The theory of probabilities and the theory of errors now constitute a formidable body of great mathematical interest and of great practical importance, as noted by R.S. Woodward. Let us begin with section 14.1, Probability, A Theoretical Approach.
Let us consider the following situation. Suppose a coin is tossed at random. When we speak of a coin, we assume it to be fair, that is, it is symmetrical so that there is no reason for it to come down more often on one side than the other. We call this property of the coin as being unbiased. By the phrase random toss, we mean that the coin is allowed to fall freely without any bias or interference. We know, in advance, that the coin can only land in one of two possible ways, either head up or tail up. We dismiss the possibility of its landing on its edge, which may be possible, for example, if it falls on sand. We can reasonably assume that each outcome, head or tail, is as likely to occur as the other. We refer to this by saying that the outcomes head and tail, are equally likely.
For another example of equally likely outcomes, suppose we throw a die once. For us, a die will always mean a fair die. What are the possible outcomes? They are 1, 2, 3, 4, 5, 6. Each number has the same possibility of showing up. So the equally likely outcomes of throwing a die are 1, 2, 3, 4, 5 and 6.
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Are the outcomes of every experiment equally likely? Let us see. Suppose that a bag contains 4 red balls and 1 blue ball, and you draw a ball without looking into the bag. What are the outcomes? Are the outcomes, a red ball and a blue ball equally likely? Since there are 4 red balls and only one blue ball, you would agree that you are more likely to get a red ball than a blue ball. So, the outcomes, a red ball or a blue ball, are not equally likely. However, the outcome of drawing a ball of any colour from the bag is equally likely. So, all experiments do not necessarily have equally likely outcomes. However, in this chapter, from now on, we will assume that all the experiments have equally likely outcomes.
In Class 9, we defined the experimental or empirical probability P(E) of an event E as P(E) = (Number of trials in which the event happened) / (Total number of trials). The empirical interpretation of probability can be applied to every event associated with an experiment which can be repeated a large number of times. The requirement of repeating an experiment has some limitations, as it may be very expensive or unfeasible in many situations. Of course, it worked well in coin tossing or die throwing experiments. But how about repeating the experiment of launching a satellite in order to compute the empirical probability of its failure during launching, or the repetition of the phenomenon of an earthquake to compute the empirical probability of a multi storeyed building getting destroyed in an earthquake?
In experiments where we are prepared to make certain assumptions, the repetition of an experiment can be avoided, as the assumptions help in directly calculating the exact theoretical probability. The assumption of equally likely outcomes is one such assumption that leads us to the following definition of probability of an event.
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The theoretical probability, also called classical probability, of an event E, written as P(E), is defined as P(E) = (Number of outcomes favourable to E) / (Number of all possible outcomes of the experiment), where we assume that the outcomes of the experiment are equally likely. We will briefly refer to theoretical probability as probability. This definition of probability was given by Pierre Simon Laplace in 1795.
Probability theory had its origin in the 16th century when an Italian physician and mathematician J. Cardan wrote the first book on the subject, The Book on Games of Chance. Since its inception, the study of probability has attracted the attention of great mathematicians. James Bernoulli, A. de Moivre, and Pierre Simon Laplace are among those who made significant contributions to this field. Laplace's Theorie Analytique des Probabilites, published in 1812, is considered to be the greatest contribution by a single person to the theory of probability. In recent years, probability has been used extensively in many areas such as biology, economics, genetics, physics, sociology and so on.
Let us find the probability for some of the events associated with experiments where the equally likely assumption holds.
Example 1: Find the probability of getting a head when a coin is tossed once. Also find the probability of getting a tail. Solution: In the experiment of tossing a coin once, the number of possible outcomes is two, Head and Tail. Let E be the event getting a head. The number of outcomes favourable to E, that is, of getting a head, is 1. Therefore, P(E) = P(head) = 1/2. Similarly, if F is the event getting a tail, then P(F) = P(tail) = 1/2.
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Example 2: A bag contains a red ball, a blue ball and a yellow ball, all the balls being of the same size. Kritika takes out a ball from the bag without looking into it. What is the probability that she takes out the yellow ball, the red ball, and the blue ball? Solution: Kritika takes out a ball from the bag without looking into it. So, it is equally likely that she takes out any one of them. Let Y be the event the ball taken out is yellow, B be the event the ball taken out is blue, and R be the event the ball taken out is red. Now, the number of possible outcomes is 3. The number of outcomes favourable to the event Y is 1. So, P(Y) = 1/3. Similarly, P(R) = 1/3 and P(B) = 1/3.
Remarks: An event having only one outcome of the experiment is called an elementary event. In Example 1, both the events E and F are elementary events. Similarly, in Example 2, all the three events, Y, B and R are elementary events. In Example 1, we note that P(E) + P(F) = 1. In Example 2, we note that P(Y) + P(R) + P(B) = 1. Observe that the sum of the probabilities of all the elementary events of an experiment is 1. This is true in general also.
Example 3: Suppose we throw a die once. What is the probability of getting a number greater than 4? What is the probability of getting a number less than or equal to 4? Solution: Let E be the event getting a number greater than 4. The number of possible outcomes is six: 1, 2, 3, 4, 5 and 6, and the outcomes favourable to E are 5 and 6. Therefore, the number of outcomes favourable to E is 2. So, P(E) = 2/6 = 1/3. Let F be the event getting a number less than or equal to 4. Number of possible outcomes is 6. Outcomes favourable to the event F are 1, 2, 3, 4. So, the number of outcomes favourable to F is 4. Therefore, P(F) = 4/6 = 2/3.
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Are the events E and F in the example above elementary events? No, they are not because the event E has 2 outcomes and the event F has 4 outcomes.
From Example 1, we note that P(E) + P(F) = 1, where E is the event getting a head and F is the event getting a tail. From Example 3, we also get P(E) + P(F) = 1, where E is the event getting a number greater than 4 and F is the event getting a number less than or equal to 4. Note that getting a number not greater than 4 is same as getting a number less than or equal to 4, and vice versa. In both cases, F is the same as not E. We denote the event not E by E bar. So, P(E) + P(not E) = 1, that is, P(E) + P(E bar) = 1, which gives us P(E bar) = 1 – P(E). In general, it is true that for an event E, P(E bar) = 1 – P(E). The event E bar, representing not E, is called the complement of the event E. We also say that E and E bar are complementary events.
Before proceeding further, let us try to find the answers to the following questions. What is the probability of getting a number 8 in a single throw of a die? What is the probability of getting a number less than 7 in a single throw of a die? Let us answer the first one. We know that there are only six possible outcomes in a single throw of a die. These outcomes are 1, 2, 3, 4, 5 and 6. Since no face of the die is marked 8, so there is no outcome favourable to 8, that is, the number of such outcomes is zero. In other words, getting 8 in a single throw of a die, is impossible. So, P(getting 8) = 0/6 = 0. That is, the probability of an event which is impossible to occur is 0. Such an event is called an impossible event.
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Let us answer the second one. Since every face of a die is marked with a number less than 7, it is sure that we will always get a number less than 7 when it is thrown once. So, the number of favourable outcomes is the same as the number of all possible outcomes, which is 6. Therefore, P(E) = P(getting a number less than 7) = 6/6 = 1. So, the probability of an event which is sure or certain to occur is 1. Such an event is called a sure event or a certain event.
Note: From the definition of the probability P(E), we see that the numerator, which is the number of outcomes favourable to the event E, is always less than or equal to the denominator, which is the number of all possible outcomes. Therefore, 0 ≤ P(E) ≤ 1.
Now, let us take an example related to playing cards. Have you seen a deck of playing cards? It consists of 52 cards which are divided into 4 suits of 13 cards each, namely spades, hearts, diamonds and clubs. Clubs and spades are of black colour, while hearts and diamonds are of red colour. The cards in each suit are ace, king, queen, jack, 10, 9, 8, 7, 6, 5, 4, 3 and 2. Kings, queens and jacks are called face cards.
Example 4: One card is drawn from a well shuffled deck of 52 cards. Calculate the probability that the card will be an ace, and not be an ace. Solution: Well shuffling ensures equally likely outcomes. There are 4 aces in a deck. Let E be the event the card is an ace. The number of outcomes favourable to E is 4. The number of possible outcomes is 52. Therefore, P(E) = 4/52 = 1/13. Let F be the event card drawn is not an ace. The number of outcomes favourable to the event F is 52 – 4 = 48. The number of possible outcomes is 52. Therefore, P(F) = 48/52 = 12/13. Remark: Note that F is nothing but E bar. Therefore, we can also calculate P(F) as follows: P(F) = P(E bar) = 1 – P(E) = 1 – 1/13 = 12/13.
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Example 5: Two players, Sangeeta and Reshma, play a tennis match. It is known that the probability of Sangeeta winning the match is 0.62. What is the probability of Reshma winning the match? Solution: Let S and R denote the events that Sangeeta wins the match and Reshma wins the match, respectively. The probability of Sangeeta winning is P(S) = 0.62. The probability of Reshma winning is P(R) = 1 – P(S), because the events R and S are complementary. So, P(R) = 1 – 0.62 = 0.38.
Example 6: Savita and Hamida are friends. What is the probability that both will have different birthdays, and the same birthday, ignoring a leap year? Solution: Out of the two friends, one girl, say, Savita's birthday can be any day of the year. Now, Hamida's birthday can also be any day of 365 days in the year. We assume that these 365 outcomes are equally likely. If Hamida's birthday is different from Savita's, the number of favourable outcomes for her birthday is 365 – 1 = 364. So, the probability that Hamida's birthday is different from Savita's birthday is 364/365. The probability that Savita and Hamida have the same birthday is 1 – P(both have different birthdays) = 1 – 364/365 = 1/365.
Example 7: There are 40 students in Class 10 of a school of whom 25 are girls and 15 are boys. The class teacher has to select one student as a class representative. She writes the name of each student on a separate card, the cards being identical. Then she puts cards in a bag and stirs them thoroughly. She then draws one card from the bag. What is the probability that the name written on the card is the name of a girl, and a boy? Solution: There are 40 students, and only one name card has to be chosen. The number of all possible outcomes is 40. The number of outcomes favourable for a card with the name of a girl is 25. Therefore, the probability of drawing a card with the name of a girl is 25/40 = 5/8. The number of outcomes favourable for a card with the name of a boy is 15. Therefore, the probability of drawing a card with the name of a boy is 15/40 = 3/8. Note: We can also determine the probability of drawing a boy by taking 1 – P(not Boy) = 1 – P(Girl) = 1 – 5/8 = 3/8.
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Example 8: A box contains 3 blue, 2 white, and 4 red marbles. If a marble is drawn at random from the box, what is the probability that it will be white, blue, and red? Solution: Saying that a marble is drawn at random is a short way of saying that all the marbles are equally likely to be drawn. Therefore, the number of possible outcomes is 3 + 2 + 4 = 9. Let W denote the event the marble is white, B denote the event the marble is blue and R denote the event the marble is red. The number of outcomes favourable to the event W is 2. So, P(W) = 2/9. Similarly, P(B) = 3/9 = 1/3, and P(R) = 4/9. Note that P(W) + P(B) + P(R) = 1.
Example 9: Harpreet tosses two different coins simultaneously. What is the probability that she gets at least one head? Solution: We write H for head and T for tail. When two coins are tossed simultaneously, the possible outcomes are (H, H), (H, T), (T, H), (T, T), which are all equally likely. The outcomes favourable to the event E, at least one head, are (H, H), (H, T) and (T, H). So, the number of outcomes favourable to E is 3. Therefore, P(E) = 3/4. Note: You can also find P(E) as follows: P(E) = 1 – P(E bar), where E bar is no head. P(no head) is 1/4. So, 1 – 1/4 = 3/4.
Did you observe that in all the examples discussed so far, the number of possible outcomes in each experiment was finite? If not, check it now. There are many experiments in which the outcome is any number between two given numbers, or in which the outcome is every point within a circle or rectangle. Can you now count the number of all possible outcomes? As you know, this is not possible since there are infinitely many numbers between two given numbers, or there are infinitely many points within a circle. So, the definition of theoretical probability which you have learnt so far cannot be applied in the present form. What is the way out?
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To answer this, let us consider the following example. Example 10: In a musical chair game, the person playing the music has been advised to stop playing the music at any time within 2 minutes after she starts playing. What is the probability that the music will stop within the first half minute after starting? Solution: Here the possible outcomes are all the numbers between 0 and 2. Imagine a number line from 0 to 2. Let E be the event that the music is stopped within the first half minute. The outcomes favourable to E are points on the number line from 0 to 1/2. The distance from 0 to 2 is 2, while the distance from 0 to 1/2 is 1/2. Since all the outcomes are equally likely, we can argue that, of the total distance of 2, the distance favourable to the event E is 1/2. So, P(E) = (Distance favourable to the event E) / (Total distance in which outcomes can lie) = (1/2) / 2 = 1/4.
Can we now extend the idea of Example 10 for finding the probability as the ratio of the favourable area to the total area? Example 11: A missing helicopter is reported to have crashed somewhere in the rectangular region shown in Figure 14.2. What is the probability that it crashed inside the lake shown in the figure? Solution: Imagine a rectangle with length 9 km and width 4.5 km. Inside it, there is a smaller rectangular lake with length 3 km and width 2.5 km. The helicopter is equally likely to crash anywhere in the region. Area of the entire region where the helicopter can crash is 4.5 × 9 km² = 40.5 km². Area of the lake is 2.5 × 3 km² = 7.5 km². Therefore, the probability that the helicopter crashed in the lake is 7.5 / 40.5 = 75 / 405 = 5 / 27.
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Example 12: A carton consists of 100 shirts of which 88 are good, 8 have minor defects and 4 have major defects. Jimmy, a trader, will only accept the shirts which are good, but Sujatha, another trader, will only reject the shirts which have major defects. One shirt is drawn at random from the carton. What is the probability that it is acceptable to Jimmy, and it is acceptable to Sujatha? Solution: One shirt is drawn at random from the carton of 100 shirts. Therefore, there are 100 equally likely outcomes. The number of outcomes favourable to Jimmy is 88. Therefore, the probability that the shirt is acceptable to Jimmy is 88/100 = 0.88. The number of outcomes favourable to Sujatha is 88 + 8 = 96, because she only rejects shirts with major defects. So, the probability that the shirt is acceptable to Sujatha is 96/100 = 0.96.
Example 13: Two dice, one blue and one grey, are thrown at the same time. Write down all the possible outcomes. What is the probability that the sum of the two numbers appearing on the top of the dice is 8, 13, and less than or equal to 12? Solution: When the blue die shows 1, the grey die could show any one of the numbers 1, 2, 3, 4, 5, 6. The same is true when the blue die shows 2, 3, 4, 5 or 6. The possible outcomes are listed as ordered pairs where the first number is the blue die and the second is the grey die. They are: (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6). Note that the pair (1,4) is different from (4,1). So, the number of possible outcomes is 6 × 6 = 36. The outcomes favourable to the event the sum of the two numbers is 8 are (2,6), (3,5), (4,4), (5,3), (6,2). The number of outcomes favourable to this event is 5. Hence, the probability is 5/36. As you can see, there is no outcome favourable to the event the sum of two numbers is 13. So, the probability is 0/36 = 0. All the outcomes are favourable to the event sum of two numbers less than or equal to 12. So, the probability is 36/36 = 1.
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Now, let us move on to Exercise 14.1. I will solve every question step by step.
Question 1: Complete the following statements. (i) Probability of an event E + Probability of the event not E = 1. (ii) The probability of an event that cannot happen is 0. Such an event is called an impossible event. (iii) The probability of an event that is certain to happen is 1. Such an event is called a sure event or a certain event. (iv) The sum of the probabilities of all the elementary events of an experiment is 1. (v) The probability of an event is greater than or equal to 0 and less than or equal to 1.
Question 2: Which of the following experiments have equally likely outcomes? Explain. (i) A driver attempts to start a car. The car starts or does not start. This does not have equally likely outcomes because the car's starting depends on its mechanical condition, fuel, and other factors, making starting more or less likely than not starting. (ii) A player attempts to shoot a basketball. She or he shoots or misses the shot. This does not have equally likely outcomes because it depends on the player's skill, distance, and fatigue. (iii) A trial is made to answer a true false question. The answer is right or wrong. This has equally likely outcomes if the answer is guessed randomly, as there are only two choices and no reason to prefer one over the other. (iv) A baby is born. It is a boy or a girl. This is generally considered to have equally likely outcomes, as biological factors make the probability of a boy or girl approximately equal.
Question 3: Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game? Tossing a coin is considered fair because a coin is symmetrical and unbiased. When tossed randomly, the two possible outcomes, head or tail, are equally likely to occur, giving each team an equal chance of 1/2 to win the toss.
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Question 4: Which of the following cannot be the probability of an event? Options are 2/3, –1.5, 15%, 0.7. The probability of an event must be between 0 and 1, inclusive. –1.5 is less than 0, so it cannot be a probability. The correct option is B.
Question 5: If P(E) = 0.05, what is the probability of not E? Using the complement rule, P(not E) = 1 – P(E) = 1 – 0.05 = 0.95.
Question 6: A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out an orange flavoured candy, and a lemon flavoured candy? (i) Since there are no orange flavoured candies, the event is impossible. The probability is 0. (ii) Since all candies are lemon flavoured, the event is certain. The probability is 1.
Question 7: It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday? The events are complementary. So, the probability that they have the same birthday equals 1 – 0.992 = 0.008.
Question 8: A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is red, and not red? Total balls equal 3 + 5 = 8. (i) Probability of drawing a red ball is 3/8. (ii) Probability of drawing a ball that is not red is 5/8.
Question 9: A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be red, white, and not green? Total marbles equal 5 + 8 + 4 = 17. (i) Probability of red is 5/17. (ii) Probability of white is 8/17. (iii) Probability of not green equals 1 – probability of green. Probability of green is 4/17. So, 1 – 4/17 = 13/17.
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Question 10: A piggy bank contains hundred 50 paise coins, fifty 1 rupee coins, twenty 2 rupee coins and ten 5 rupee coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin will be a 50 paise coin, and will not be a 5 rupee coin? Total coins equal 100 + 50 + 20 + 10 = 180. (i) Probability of a 50 paise coin is 100/180 = 5/9. (ii) Probability of not a 5 rupee coin equals 1 – probability of a 5 rupee coin. Probability of a 5 rupee coin is 10/180 = 1/18. So, 1 – 1/18 = 17/18.
Question 11: Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish. What is the probability that the fish taken out is a male fish? Total fish equal 5 + 8 = 13. Probability of a male fish is 5/13.
Question 12: A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, and these are equally likely outcomes. What is the probability that it will point at 8, an odd number, a number greater than 2, and a number less than 9? Total outcomes equal 8. (i) Probability of 8 is 1/8. (ii) Odd numbers are 1, 3, 5, 7. There are 4 of them. Probability is 4/8 = 1/2. (iii) Numbers greater than 2 are 3, 4, 5, 6, 7, 8. There are 6. Probability is 6/8 = 3/4. (iv) Numbers less than 9 are all 8 numbers. Probability is 8/8 = 1.
Question 13: A die is thrown once. Find the probability of getting a prime number, a number lying between 2 and 6, and an odd number. Total outcomes equal 6: 1, 2, 3, 4, 5, 6. (i) Prime numbers are 2, 3, 5. There are 3. Probability is 3/6 = 1/2. (ii) Numbers lying between 2 and 6 are 3, 4, 5. There are 3. Probability is 3/6 = 1/2. (iii) Odd numbers are 1, 3, 5. There are 3. Probability is 3/6 = 1/2.
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Question 14: One card is drawn from a well shuffled deck of 52 cards. Find the probability of getting a king of red colour, a face card, a red face card, the jack of hearts, a spade, and the queen of diamonds. Total cards equal 52. (i) Red kings are king of hearts and king of diamonds. There are 2. Probability is 2/52 = 1/26. (ii) Face cards are kings, queens, and jacks. There are 3 per suit times 4 suits, so 12 face cards. Probability is 12/52 = 3/13. (iii) Red face cards are hearts and diamonds face cards. There are 6. Probability is 6/52 = 3/26. (iv) Jack of hearts is exactly 1 card. Probability is 1/52. (v) Spades are 13 cards. Probability is 13/52 = 1/4. (vi) Queen of diamonds is exactly 1 card. Probability is 1/52.
Question 15: Five cards, the ten, jack, queen, king and ace of diamonds, are well shuffled with their face downwards. One card is then picked up at random. What is the probability that the card is the queen? If the queen is drawn and put aside, what is the probability that the second card picked up is an ace, and a queen? Total cards initially equal 5. (i) Probability of the queen is 1/5. (ii) After removing the queen, 4 cards remain. (a) There is 1 ace left. Probability is 1/4. (b) There are 0 queens left. Probability is 0/4 = 0.
Question 16: 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one. Total pens equal 12 + 132 = 144. Good pens equal 132. Probability is 132/144 = 11/12.
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Question 17: A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective? Suppose the bulb drawn in the first part is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective? (i) Total bulbs equal 20. Defective equal 4. Probability is 4/20 = 1/5. (ii) One non defective bulb is removed. Remaining bulbs equal 19. Remaining non defective bulbs equal 16 – 1 = 15. Probability is 15/19.
Question 18: A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears a two digit number, a perfect square number, and a number divisible by 5. Total discs equal 90. (i) Two digit numbers are from 10 to 90. Count is 90 – 9 = 81. Probability is 81/90 = 9/10. (ii) Perfect squares between 1 and 90 are 1, 4, 9, 16, 25, 36, 49, 64, 81. There are 9. Probability is 9/90 = 1/10. (iii) Numbers divisible by 5 are 5, 10, 15, ..., 90. Count is 90 / 5 = 18. Probability is 18/90 = 1/5.
Question 19: A child has a die whose six faces show the letters as given below: A, B, C, D, E, A. The die is thrown once. What is the probability of getting A, and D? Total faces equal 6. (i) A appears on 2 faces. Probability is 2/6 = 1/3. (ii) D appears on 1 face. Probability is 1/6.
Question 20: Suppose you drop a die at random on the rectangular region shown in Figure 14.6. What is the probability that it will land inside the circle with diameter 1 metre? The figure shows a rectangle measuring 3 metres by 2 metres. Inside it is a circle with diameter 1 metre, so radius is 0.5 metres. Area of rectangle equals 3 × 2 = 6 m². Area of circle equals π × (0.5)² = 0.25π m². Probability equals Area of circle / Area of rectangle = 0.25π / 6 = π / 24.
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Question 21: A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that she will buy it, and she will not buy it? Total pens equal 144. Good pens equal 144 – 20 = 124. (i) Probability she will buy it (good pen) is 124/144 = 31/36. (ii) Probability she will not buy it (defective pen) is 20/144 = 5/36.
Question 22: Refer to Example 13. Complete the following table for the sum on 2 dice: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. The probabilities given are 1/36 for 2, 5/36 for 8, 1/36 for 12. We need to fill the rest. Total outcomes equal 36. Sum 2: (1,1) -> 1 outcome -> 1/36 Sum 3: (1,2), (2,1) -> 2 outcomes -> 2/36 Sum 4: (1,3), (2,2), (3,1) -> 3 outcomes -> 3/36 Sum 5: (1,4), (2,3), (3,2), (4,1) -> 4 outcomes -> 4/36 Sum 6: (1,5), (2,4), (3,3), (4,2), (5,1) -> 5 outcomes -> 5/36 Sum 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) -> 6 outcomes -> 6/36 Sum 8: (2,6), (3,5), (4,4), (5,3), (6,2) -> 5 outcomes -> 5/36 Sum 9: (3,6), (4,5), (5,4), (6,3) -> 4 outcomes -> 4/36 Sum 10: (4,6), (5,5), (6,4) -> 3 outcomes -> 3/36 Sum 11: (5,6), (6,5) -> 2 outcomes -> 2/36 Sum 12: (6,6) -> 1 outcome -> 1/36 (ii) A student argues that there are 11 possible outcomes 2 through 12, so each has probability 1/11. Do you agree? No, I do not agree. The outcomes are not equally likely. As shown in the table above, some sums like 7 have 6 favourable outcomes, while sums like 2 or 12 have only 1. Therefore, their probabilities differ.
Question 23: A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result, that is, three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game. Total outcomes for 3 tosses are 2³ = 8: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. Winning outcomes are HHH and TTT. That is 2 outcomes. Probability of winning is 2/8 = 1/4. Probability of losing is 1 – probability of winning = 1 – 1/4 = 3/4.
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Question 24: A die is thrown twice. What is the probability that 5 will not come up either time, and 5 will come up at least once? Total outcomes equal 36. (i) Outcomes where 5 does not come up: Each die has 5 options (1,2,3,4,6). So 5 × 5 = 25 outcomes. Probability is 25/36. (ii) Probability that 5 will come up at least once equals 1 – probability that 5 does not come up either time. So, 1 – 25/36 = 11/36.
Question 25: Which of the following arguments are correct and which are not correct? Give reasons. (i) If two coins are tossed simultaneously there are three possible outcomes, two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is 1/3. This argument is not correct. The outcomes are not equally likely. The actual equally likely outcomes are HH, HT, TH, TT. The event one of each includes HT and TH, so it has probability 2/4 = 1/2, while two heads and two tails each have probability 1/4. (ii) If a die is thrown, there are two possible outcomes, an odd number or an even number. Therefore, the probability of getting an odd number is 1/2. This argument is correct. The odd numbers are 1, 3, 5 and the even numbers are 2, 4, 6. Each group has 3 outcomes out of 6, so the probability is 3/6 = 1/2. The outcomes are equally likely.
Now, let us read a note to the reader. The experimental or empirical probability of an event is based on what has actually happened while the theoretical probability of the event attempts to predict what will happen on the basis of certain assumptions. As the number of trials in an experiment, go on increasing we may expect the experimental and theoretical probabilities to be nearly the same.
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Let us conclude with the summary of this chapter. 1. The theoretical probability of an event E, written as P(E), is defined as P(E) = (Number of outcomes favourable to E) / (Number of all possible outcomes of the experiment), where we assume that the outcomes of the experiment are equally likely. 2. The probability of a sure event or certain event is 1. 3. The probability of an impossible event is 0. 4. The probability of an event E is a number P(E) such that 0 ≤ P(E) ≤ 1. 5. An event having only one outcome is called an elementary event. The sum of the probabilities of all the elementary events of an experiment is 1. 6. For any event E, P(E) + P(E bar) = 1, where E bar stands for not E. E and E bar are called complementary events.
Thank you for listening! Keep revising and practicing. Goodbye! [CHAPTER_COMPLETE]