Chapter 7 exercise answers: The Mathematics of Maybe: Introduction to Probability

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(i) The next Monday will come after Sunday.

1. The days of the week always run in the same order: Sunday, Monday, Tuesday, and so on.
2. So the day after a Sunday is always a Monday. Nothing is left to chance, and the event sits at 1 on the scale.

AnswerCertain

(ii) It will snow in Mumbai in July.

1. Snow needs air colder than 0 °C. July is the monsoon month in Mumbai, and it stays hot and humid, around 25–30 °C.
2. So snow in Mumbai in July cannot happen. The event sits at 0 on the scale.

AnswerImpossible

(iii) An elephant will walk through your classroom today.

1. Nothing makes this completely impossible: elephants do sometimes wander into villages and schools near forests.
2. But on an ordinary school day it is extremely unusual, so it sits very close to 0, far below the middle of the scale.
3. Some people would call it impossible. The chapter keeps 'impossible' for events that cannot happen at all, which is why 'less likely' fits better. Whichever label you choose, give your reason.

AnswerLess likely

(iv) You will greet at least one friend at school tomorrow.

1. On most school days you meet and greet at least one friend.
2. It is not guaranteed: you might be unwell, or tomorrow might be a holiday. So it sits above the middle of the scale but short of 1.

AnswerMore likely

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(i) Calculate the probability that a randomly picked sweet … is green.

1. The sample has 10 + 8 + 7 + 5 = 30 sweets in all.
2. 8 of them are green.
3. Probability of green = 8/30, which simplifies to 4/15.

Answer4/15

(ii) estimate how many are likely to be yellow

1. The share of yellow sweets in the sample is 7/30.
2. Apply that same share to the full bag of 600 sweets.
3. 7/30 × 600 = 140.

Answer140

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(i) the probability that a randomly chosen student … prefers the Arts Club?

1. 14 + 11 + 9 + 6 = 40 students were surveyed in all.
2. 11 of them chose the Arts Club.
3. Probability = 11/40.

Answer11/40

(ii) estimate how many students … prefer the Sports Club.

1. The share choosing the Sports Club in the sample is 9/40.
2. Scale this share up to the whole school of 800 students.
3. 9/40 × 800 = 180.

Answer180

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Open NCERT p. 165One way to think about it

(i) How many times did you get heads?

1. Toss the coin 20 times and write H or T after each toss.
2. Count the H's. The number depends on your own tosses, so there is no single right answer. For example, suppose one run of 20 tosses gave 11 heads.

In shortYour own count of heads (in the example, 11)

(ii) How many times did you get tails?

1. Every toss is either heads or tails, so tails = 20 − heads.
2. In the example: 20 − 11 = 9.

In short20 − your heads count (in the example, 9)

(iii) Calculate the experimental probability of getting heads.

1. Experimental probability of heads = number of heads ÷ number of tosses = heads ÷ 20.
2. In the example: 11/20 = 0.55.

In shortYour heads ÷ 20 (in the example, 11/20 = 0.55)

(iv) what is the probability of getting tails?

1. A coin has no memory, so each toss starts afresh, whatever happened before.
2. For a fair coin, the probability of tails on the next toss is 1/2, whatever your 20 tosses showed.

In short1/2

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Open NCERT p. 165One way to think about it

1. Toss the cup 100 times. After each toss, record whether it landed on its bottom, upside down on its top, or on its side.
2. Count each kind of landing. The three counts must add up to 100.
3. Experimental probability of each landing = its count ÷ 100.
4. Your counts will be your own, so there is no single right answer. For example, suppose the cup landed on its bottom 22 times, on its top 15 times and on its side 63 times. Then P(bottom) = 22/100 = 0.22, P(top) = 15/100 = 0.15 and P(side) = 63/100 = 0.63.
5. Check: 0.22 + 0.15 + 0.63 = 1. A cup is not symmetrical, so the three landings need not be equally likely, and only an experiment can give their probabilities.

In shortP(each landing) = its count ÷ 100 (in the example, 0.22, 0.15 and 0.63, which add up to 1)

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1. A fair die has 6 equally likely faces: 1, 2, 3, 4, 5, 6.
2. The even faces are 2, 4 and 6 — 3 out of 6.
3. Probability = 3/6 = 1/2.

Answer1/2

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(i) What is the experimental probability of rolling a '3'?

1. You rolled 12 times and a '3' turned up 3 times.
2. Experimental probability = 3/12 = 1/4.

Answer1/4

(ii) What is the theoretical probability of rolling a '3'?

1. A fair die has 6 equally likely faces, and exactly one of them is '3'.
2. Theoretical probability = 1/6.

Answer1/6

(iii) Why might these probabilities be different?

1. 12 rolls can only ever report multiples of 1/12; 1/6 equals 2/12, so 12 rolls COULD have matched it exactly, but this particular run of three '3's (1/4) simply missed, the way any short run can.
2. As the number of rolls grows, the experimental probability tends to drift closer to the theoretical 1/6, though it need not land on it exactly.

AnswerNo fixed number — expect the experimental probability to settle nearer to 1/6 as the roll count grows to 60, 600 and 6000.

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1. A single roll of a 6-sided die can show 1, 2, 3, 4, 5 or 6.
2. That is every possible outcome, listed once each.
3. So the sample space has 6 outcomes.

Answer6

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Open NCERT p. 167Checked by computerReads two ways: both answers shown

(i) Rolling a die and tossing a coin together.

1. The die shows 1 to 6 and the coin shows H or T, so each outcome is a pair (number, coin).
2. 6 × 2 = 12 outcomes.

AnswerS = {(1, H), (1, T), (2, H), (2, T), (3, H), (3, T), (4, H), (4, T), (5, H), (5, T), (6, H), (6, T)}

(ii) Choosing a random integer between −5 and +5.

1. The book does not say whether −5 and +5 themselves may be chosen, so both readings are shown.
2. Read as strictly between −5 and +5 (the ends left out): the integers are −4 to 4, which is 9 outcomes.
3. Read with the ends included: the integers are −5 to 5, which is 11 outcomes.

AnswerRead as strictly between −5 and +5: S = {−4, −3, −2, −1, 0, 1, 2, 3, 4}, 9 outcomes. Read with −5 and +5 included: S = {−5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5}, 11 outcomes.

(iii) A box containing 5 green and 7 red balls.

1. Only the colour of the drawn ball matters, so the possible results are Green and Red.
2. These two outcomes are not equally likely (5 balls are green, 7 are red). If you need equally likely outcomes, name each ball instead: {G1, …, G5, R1, …, R7}, 12 outcomes.

AnswerS = {Green, Red}

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(i) List the sample space of all possible snack and drink combinations

1. There are 3 snacks and 2 drinks, so pair every snack with every drink: 3 × 2 = 6 combinations.

AnswerS = {(Samosa,Chai),(Samosa,Lassi),(Pakora,Chai),(Pakora,Lassi),(Bhaji,Chai),(Bhaji,Lassi)}

(ii) List the event 'Selecting Samosa as a snack.'

1. From the 6 combinations, keep only the ones where the snack is Samosa.
2. That leaves 2 combinations.

AnswerE = {(Samosa,Chai),(Samosa,Lassi)}

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(i) Draw a tree diagram showing all possible pairs of fruits.

1. From the start, draw two branches for basket A: Apple (1 of its 3 fruits) and Orange (2 of its 3 fruits).
2. From each of those, draw two branches for basket B: Banana and Mango.
3. Label the branches with their probabilities: Apple 1/3, Orange 2/3 on the first split; Banana 1/2, Mango 1/2 on every second split.

AnswerA tree with first branches Apple(1/3), Orange(2/3), and from each a second pair Banana(1/2), Mango(1/2) — four end-to-end paths.

(ii) List the sample space.

1. Basket A gives Apple or Orange; basket B gives Banana or Mango.
2. Pair each of A's two fruit-types with each of B's two fruit-types.

AnswerS = {(Apple,Banana),(Apple,Mango),(Orange,Banana),(Orange,Mango)}

(iii) What is the probability of picking one apple and one banana?

1. P(Apple) = 1/3, since 1 of the 3 fruits in basket A is an apple.
2. P(Banana) = 1/2, since 1 of the 2 fruits in basket B is a banana.
3. The two picks don't affect each other, so multiply along the path: 1/3 × 1/2 = 1/6.

Answer1/6

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(i) What are the possible outcomes of the pen colours?

1. The box has 3 + 4 + 2 = 9 pens. Write R = red, B = black, G = green.
2. Each pick gives R with probability 3/9, B with 4/9 and G with 2/9. The pen is put back, so your friend's pick has the same three chances.
3. Tree diagram: from a starting point draw 3 branches R (3/9), B (4/9), G (2/9) for your pick. From the end of each, draw the same 3 branches again for your friend's pick. That makes 3 × 3 = 9 paths.

Answer{(R,R), (R,B), (R,G), (B,R), (B,B), (B,G), (G,R), (G,B), (G,G)}, where the first letter is your pen and the second is your friend's

(ii) the probability that both you and your friend pick … same colour?

1. Same colour happens on 3 paths: (R,R), (B,B) and (G,G).
2. Multiply along each path: 3/9 × 3/9 = 9/81, 4/9 × 4/9 = 16/81, 2/9 × 2/9 = 4/81.
3. Add them: 9/81 + 16/81 + 4/81 = 29/81.
4. Counting paths (3 out of 9 = 1/3) would be wrong, because the 9 paths are not equally likely.

Answer29/81

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(i) The probability of an impossible event is _______.

1. An impossible event never happens, so no outcome favours it: 0 favourable outcomes.

Answer0

(ii) The set of all possible outcomes … is called the __________.

1. This complete list of everything that could happen is the term the chapter gives it.

Answersample space

(iii) The probability of an event that is certain to happen is _______.

1. An event certain to happen is satisfied by every outcome in the sample space, so its probability is 1.

Answer1

(iv) Tossing a fair coin has a probability of ______ for getting heads.

1. A fair coin has 2 equally likely faces, heads and tails.
2. P(heads) = 1/2.

Answer1/2

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1. '15' is already the frequency — a plain count of how many students said yes.
2. The blank wants a fraction or decimal, so it is asking for the relative frequency: how big a share of the 50 that 15 is.
3. Relative frequency = 15/50 = 3/10 (or 0.3).

Answerrelative frequency, 3/10 (0.3)

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(i) The car starts or does not start.

1. A car in working order starts on almost every try and fails only now and then, for example with a flat battery.
2. Nothing makes 'starts' and 'does not start' equally likely. Having two outcomes does not make each of them 1/2.

AnswerNo

(ii) Tossing a fair coin once.

1. A fair coin is symmetrical, so neither face is favoured: heads and tails each have probability 1/2.

AnswerYes

(iii) Rolling a fair 6-sided die.

1. A fair die is a symmetrical cube, so each of the 6 faces has probability 1/6.

AnswerYes

(iv) a bag that contains 3 red marbles and 7 blue marbles.

1. Each single marble is equally likely to be picked, but the outcomes here are the colours.
2. Red has 3 of the 10 marbles and blue has 7, so P(red) = 3/10 and P(blue) = 7/10. These are not equal.

AnswerNo

(v) A baby is born. It is a boy or a girl.

1. Boy and girl are close to even but not exactly: worldwide, about 105 boys are born for every 100 girls.
2. So the two outcomes are not exactly equally likely. (Some books treat them as roughly equal, about 1/2 each, to keep things simple.)

AnswerNo

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(i) What is the probability of getting at least one head?

1. List every outcome of tossing two coins: HH, HT, TH, TT — 4 equally likely outcomes.
2. 'At least one head' means one or more heads, so it includes HH, HT and TH — 3 outcomes.
3. P = 3/4.

AnswerS = {HH, HT, TH, TT}; P(at least one head) = 3/4

(ii) What is the probability of drawing a card with an even number?

1. The sample space is every card from 1 to 10.
2. The even-numbered cards are 2, 4, 6, 8, 10 — 5 out of 10.
3. P = 5/10 = 1/2.

AnswerS = {1,2,3,4,5,6,7,8,9,10}; P(even) = 1/2

(iii) What is the probability of getting a number greater than 4?

1. The sample space for one die roll is 1 to 6.
2. Numbers greater than 4 are 5 and 6 — 2 out of 6.
3. P = 2/6 = 1/3.

AnswerS = {1,2,3,4,5,6}; P(greater than 4) = 1/3

(iv) What is the probability that it is not red?

1. There are 3 + 2 + 1 = 6 balls in all; label them R1,R2,R3,B1,B2,G1 so every ball is equally likely.
2. 'Not red' keeps the 2 blue and 1 green balls — 3 out of 6.
3. P = 3/6 = 1/2.

AnswerS = {R1,R2,R3,B1,B2,G1}; P(not red) = 1/2

(v) What is the probability of getting exactly two heads?

1. Tossing three coins gives 8 equally likely outcomes, from HHH down to TTT.
2. Exactly two heads happens on HHT, HTH, THH — 3 out of 8.
3. P = 3/8.

AnswerS = {HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}; P(exactly two heads) = 3/8

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1. There are 3 equally likely candies in the bag.
2. Only 1 of them is strawberry.
3. P(strawberry) = 1/3.

Answer1/3

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1. Pair each of the 2 shirts with each of the 3 pants: 2 × 3 = 6 outfits.

2. Shirt	Pants
   Red	Jeans
   Red	Khakis
   Red	Shorts
   Blue	Jeans
   Blue	Khakis
   Blue	Shorts

Answer6 outfits: (Red,Jeans), (Red,Khakis), (Red,Shorts), (Blue,Jeans), (Blue,Khakis), (Blue,Shorts)

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(i) Less than 4000 km.

1. The record already covers all 1000 tyres, so each relative frequency here is itself the probability.
2. 20 of the 1000 tyres lasted less than 4000 km.
3. P = 20/1000 = 1/50.

Answer1/50

(ii) Between 4000 and 14000 km.

1. 'Between 4000 and 14000' covers the two middle columns: 4001-9000 and 9001-14000.
2. Add their counts: 210 + 325 = 535.
3. P = 535/1000 = 107/200.

Answer107/200

(iii) More than 14000 km.

1. 445 of the 1000 tyres lasted more than 14000 km.
2. P = 445/1000 = 89/200.

Answer89/200

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(i) What is the probability that it is a P, E or C?

1. PEACE has 5 letter-cards: P, E, A, C, E.
2. P, E or C matches 4 of the 5 cards — both E cards count, plus P and C.
3. P = 4/5.

Answer4/5

(ii) What is the probability that it is not an E?

1. Not-E keeps P, A and C — 3 of the 5 cards.
2. P = 3/5.

Answer3/5

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(i) 8?

1. There are 8 equally likely sectors.
2. Only one of them is 8.
3. P = 1/8.

Answer1/8

(ii) An odd number?

1. The odd numbers among 1-8 are 1, 3, 5, 7 — 4 sectors.
2. P = 4/8 = 1/2.

Answer1/2

(iii) A number greater than 2?

1. Numbers greater than 2 are 3, 4, 5, 6, 7, 8 — 6 sectors.
2. P = 6/8 = 3/4.

Answer3/4

(iv) A number less than 9?

1. Every one of the 8 numbers is less than 9.
2. P = 8/8 = 1.

Answer1

(v) A multiple of 3?

1. Multiples of 3 among 1-8 are 3 and 6 — 2 sectors.
2. P = 2/8 = 1/4.

Answer1/4

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(i) the probability of drawing a red ball … then a blue ball?

1. First draw: there are 9 balls, so P(red) = 4/9 and P(blue) = 5/9.
2. If a red ball is laid aside, 3 red and 5 blue are left, so the second-draw branches are red 3/8 and blue 5/8.
3. If a blue ball is laid aside, 4 red and 4 blue are left, so the branches are red 4/8 and blue 4/8.
4. The tree has 4 paths: red–red = 4/9 × 3/8 = 1/6, red–blue = 4/9 × 5/8 = 5/18, blue–red = 5/9 × 4/8 = 5/18, blue–blue = 5/9 × 4/8 = 5/18. They add up to 1.
5. Red then blue: 4/9 × 5/8 = 20/72 = 5/18.

Answer5/18

(ii) What is the probability of drawing 2 blue balls?

1. Follow the blue–blue path: 5/9 × 4/8 = 20/72 = 5/18.

Answer5/18

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Open NCERT p. 172Checked by computerAnswers can differ: one example

1. Two dice give 6 × 6 = 36 equally likely pairs, from (1,1) to (6,6). The sum is always at least 1 + 1 = 2 and at most 6 + 6 = 12.
2. Probability 0: 'the sum of the two dice is 13'. No pair gives 13, so 0 of the 36 pairs qualify: P = 0/36 = 0.
3. Probability 1: no single pair such as (3,4) can have probability 1, because each pair has probability 1/36. What can have probability 1 is an event that includes every pair.
4. 'The sum of the two dice lies between 2 and 12' (2 and 12 included) is true for all 36 pairs, so P = 36/36 = 1.
5. Many other answers are also correct. For example, 'both dice show 7' has probability 0, and 'each die shows a number from 1 to 6' has probability 1.

AnswerProbability 0: the sum of the two dice is 13. Probability 1: the sum of the two dice lies between 2 and 12 (2 and 12 included).

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(i) the probability that the sum is a prime number greater than 5?

1. Two dice give 6 × 6 = 36 equally likely ordered pairs: S = {(1,1), (1,2), …, (6,6)}.
2. The largest possible sum is 12, so the prime sums greater than 5 are 7 and 11.
3. Sum 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1), which is 6 pairs. Sum 11: (5,6), (6,5), which is 2 pairs.
4. P = 8/36 = 2/9.

AnswerS = the 36 pairs (1,1) to (6,6); P = 2/9

(ii) What is the probability that both are of different colours?

1. Write R = red, G = green, B = blue. The bag has 4 + 3 + 2 = 9 balls.
2. Sample space of colours (first ball, second ball): S = {(R,R), (R,G), (R,B), (G,R), (G,G), (G,B), (B,R), (B,G), (B,B)}. These 9 are not equally likely, so use a tree and multiply along the paths.
3. Same colour: R then R = 4/9 × 3/8 = 12/72; G then G = 3/9 × 2/8 = 6/72; B then B = 2/9 × 1/8 = 2/72. Total = 20/72 = 5/18.
4. Different colours = 1 − 5/18 = 13/18.

AnswerS = {(R,R), (R,G), (R,B), (G,R), (G,G), (G,B), (B,R), (B,G), (B,B)}; P = 13/18

(iii) the first coin shows heads and exactly two heads occur in total?

1. Three coins give 8 equally likely outcomes: S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.
2. First coin heads and exactly two heads in total: HHT and HTH. THH does not count, because its first coin is a tail.
3. P = 2/8 = 1/4.

AnswerS = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}; P = 1/4

(iv) What is the probability that the number is even?

1. Arranging 1, 2, 3, 4 with no repeats gives 4 × 3 × 2 × 1 = 24 equally likely numbers: 1234, 1243, …, 4321.
2. A number is even when its last digit is 2 or 4.
3. With last digit 2, the other three digits can be arranged in 3 × 2 × 1 = 6 ways. The same is true for last digit 4. So 12 numbers are even.
4. P = 12/24 = 1/2.

AnswerS = the 24 numbers 1234, 1243, …, 4321; P = 1/2

(v) the student guesses and gets exactly 2 answers correct?

1. Write C = correct and I = incorrect for each question: S = {CCC, CCI, CIC, ICC, CII, ICI, IIC, III}. These 8 are not equally likely.
2. On each question a guess is correct with probability 1/4 (1 of the 4 options) and incorrect with probability 3/4.
3. Exactly 2 correct happens on CCI, CIC and ICC. Each has probability 1/4 × 1/4 × 3/4 = 3/64.
4. P = 3/64 + 3/64 + 3/64 = 9/64.

AnswerS = {CCC, CCI, CIC, ICC, CII, ICI, IIC, III}; P = 9/64

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(i) the ball is returned, and a second ball is drawn and recorded.

1. Tree diagram: from a starting point draw 4 branches for the first ball: 1, 2, 3, 4.
2. The ball is put back, so from the end of each branch draw 4 branches again: 1, 2, 3, 4.
3. Each path gives one outcome (first number, second number), so there are 4 × 4 = 16 paths.

AnswerS = {(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4), (4,1), (4,2), (4,3), (4,4)}

(ii) Without replacing the first ball, the experimenter draws … a second ball.

1. Tree diagram: draw 4 branches for the first ball: 1, 2, 3, 4.
2. The first ball is not put back, so from each branch draw only 3 branches: the three numbers left. For example, after 1 the branches are 2, 3, 4.
3. That gives 4 × 3 = 12 paths, and no number appears twice in one outcome.

AnswerS = {(1,2), (1,3), (1,4), (2,1), (2,3), (2,4), (3,1), (3,2), (3,4), (4,1), (4,2), (4,3)}

(iii) What are the sizes of these two sample spaces?

1. With replacement: 4 × 4 = 16 outcomes.
2. Without replacement: 4 × 3 = 12 outcomes.

Answer16 and 12

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1. The coin gives H or T, and the card gives one of 1 to 6.
2. Pair each coin result with each card number: 2 × 6 = 12 outcomes.

AnswerS = {(H,1),(H,2),(H,3),(H,4),(H,5),(H,6),(T,1),(T,2),(T,3),(T,4),(T,5),(T,6)}

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1. The number of heads in 3 tosses can only be 0, 1, 2 or 3 — nothing else is possible.
2. List (i) {1,2,3} leaves out 0, which can happen (TTT) — not valid.
3. List (ii) {0,1,2} leaves out 3, which can happen (HHH) — not valid.
4. List (iii) {0,1,2,3,4} includes 4, but three coins can never give 4 heads — not valid.
5. List (iv) {0,1,2,3} includes every possible head-count exactly once, and nothing extra — this is the sample space.

Answer(iv) {0, 1, 2, 3}

Watch this explained “The rule nobody writes down”, 6:35 into Listing every outcome: the sample space · हिंदी में देखें

Open NCERT p. 173Checked by computer

1. The rectangle measures 3 m by 2 m, so its area is 3 × 2 = 6 m².
2. The circle has diameter 1 m, so radius 0.5 m, and area = π × (0.5)² = π/4 m².
3. Since the dye lands at random anywhere on the rectangle, the probability of landing in the circle is the ratio of the two areas: (π/4) ÷ 6 = π/24.

Answerπ/24