Exercise 14.1 answers: Probability

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(i) Probability of an event E + Probability of the event ‘not E’

1. An event E either happens or does not happen.
2. 'Not E' is exactly the case when E does not happen.
3. So E and 'not E' together cover every possible outcome, with none left over and none counted twice.
4. That means P(E) + P(not E) = 1.

Answer1

(ii) The probability of an event that cannot happen is

1. If an event can never happen, no outcome favours it.
2. So the number of favourable outcomes is 0, and P(event) = 0.
3. Such an event is called an impossible event.

Answer0; impossible event

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

1. If an event is certain to happen, every outcome favours it.
2. So favourable outcomes = total outcomes, and P(event) = 1.
3. Such an event is called a sure event (also called a certain event).

Answer1; sure event

(iv) sum of the probabilities of all the elementary events of an experiment

1. An elementary event has exactly one outcome.
2. Every outcome of the experiment belongs to exactly one elementary event, and none is left out or repeated.
3. So the probabilities of all the elementary events add up to 1.

Answer1

(v) The probability of an event is greater than or equal to

1. The favourable outcomes for any event are some of the total outcomes, never more.
2. So the favourable count can be as low as 0 (no outcome favours it) and as high as the total (every outcome favours it).
3. That makes 0 ≤ P(event) ≤ 1.

Answergreater than or equal to 0 and less than or equal to 1

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(i) The car starts or does not start

1. Whether a car starts depends on its condition — fuel, battery, engine and so on.
2. Nothing says starting and not starting are equally likely; it usually starts more often than not.

AnswerNot equally likely.

(ii) She/he shoots or misses the shot

1. Whether the player scores depends on her/his skill and the difficulty of the shot.
2. There is no reason the two results should be equally likely.

AnswerNot equally likely.

(iii) The answer is right or wrong

1. If the student is only guessing, with no knowledge of the answer, nothing favours right over wrong.
2. Read this way, the two results are equally likely.

AnswerEqually likely (when the answer is a blind guess).

(iv) It is a boy or a girl

1. Before birth, nothing distinguishes a boy from a girl for this purpose.
2. So, taken as an idealisation, the two are equally likely.

AnswerEqually likely.

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1. A coin has two faces, head and tail, that are alike in shape, size and weight.
2. Nothing about the coin or a fair toss makes one face more likely to land up than the other.
3. So the two results, head and tail, are equally likely — each has probability 1/2.
4. Because neither team is favoured over the other, tossing a coin is considered a fair way to decide.

In shortBecause the coin's two faces are alike, a toss gives head and tail equal chance, so it favours neither team.

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1. A probability is always between 0 and 1, both included, because favourable outcomes are only part of the total.
2. 2/3 is between 0 and 1 — allowed.
3. −1.5 is negative, so it is less than 0 — not allowed.
4. 15% = 0.15, which is between 0 and 1 — allowed.
5. 0.7 is between 0 and 1 — allowed.

Answer(B) −1.5, because a probability can never be negative.

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1. E and 'not E' together cover every outcome, so P(E) + P(not E) = 1.
2. P(not E) = 1 − P(E) = 1 − 0.05.
3. P(not E) = 0.95, which is 19/20.

AnswerP(not E) = 0.95 = 19/20

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(i) an orange flavoured candy?

1. Every candy in the bag is lemon flavoured; there is no orange candy at all.
2. So no outcome favours 'orange candy'.
3. P(orange candy) = 0.

Answer0

(ii) a lemon flavoured candy?

1. Every single candy in the bag is lemon flavoured.
2. So every outcome favours 'lemon candy'.
3. P(lemon candy) = 1.

Answer1

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1. 'Same birthday' and 'not the same birthday' are complementary events — together they cover every case.
2. So P(same birthday) = 1 − P(not the same birthday).
3. P(same birthday) = 1 − 0.992 = 0.008.

Answer0.008

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(i) red?

1. Total balls in the bag = 3 red + 5 black = 8.
2. Since the balls are drawn at random, all 8 are equally likely.
3. Favourable outcomes for 'red' = 3.
4. P(red) = 3/8.

Answer3/8

(ii) not red?

1. 'Not red' means the ball is black; there are 5 black balls.
2. P(not red) = 5/8.
3. Check: 3/8 + 5/8 = 1, as it should for complementary events.

Answer5/8

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(i) red?

1. Total marbles = 5 + 8 + 4 = 17.
2. Favourable for red = 5.
3. P(red) = 5/17.

Answer5/17

(ii) white?

1. Favourable for white = 8, out of 17.
2. P(white) = 8/17.

Answer8/17

(iii) not green?

1. P(green) = 4/17.
2. 'Not green' is the complement of 'green'.
3. P(not green) = 1 − 4/17 = 13/17.

Answer13/17

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(i) will be a 50 p coin?

1. Total coins = 100 + 50 + 20 + 10 = 180.
2. All coins are equally likely to fall out.
3. Favourable for 50 p = 100.
4. P(50 p coin) = 100/180 = 5/9.

Answer5/9

(ii) will not be a ₹ 5 coin?

1. P(₹5 coin) = 10/180 = 1/18.
2. 'Not a ₹5 coin' is the complement.
3. P(not ₹5 coin) = 1 − 1/18 = 17/18.

Answer17/18

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1. Total fish in the tank = 5 male + 8 female = 13.
2. The fish is taken out at random, so all 13 are equally likely.
3. Favourable outcomes for 'male fish' = 5.
4. P(male fish) = 5/13.

Answer5/13

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(i) 8?

1. There are 8 equally likely numbers: 1 to 8.
2. Only one of them is 8.
3. P(8) = 1/8.

Answer1/8

(ii) an odd number?

1. Odd numbers from 1 to 8: 1, 3, 5, 7 — that is 4 numbers.
2. P(odd) = 4/8 = 1/2.

Answer1/2

(iii) a number greater than 2?

1. Numbers greater than 2, from 1 to 8: 3, 4, 5, 6, 7, 8 — that is 6 numbers.
2. P(greater than 2) = 6/8 = 3/4.

Answer3/4

(iv) a number less than 9?

1. Every number from 1 to 8 is less than 9.
2. So all 8 outcomes are favourable.
3. P(less than 9) = 8/8 = 1 (a sure event).

Answer1

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(i) a prime number;

1. A die has 6 equally likely outcomes: 1, 2, 3, 4, 5, 6.
2. Prime numbers among these: 2, 3, 5 — that is 3 numbers.
3. P(prime) = 3/6 = 1/2.

Answer1/2

(ii) a number lying between 2 and 6;

1. 'Between 2 and 6' means strictly between, so 3, 4, 5 — that is 3 numbers.
2. P(between 2 and 6) = 3/6 = 1/2.

Answer1/2

(iii) an odd number.

1. Odd numbers on a die: 1, 3, 5 — that is 3 numbers.
2. P(odd) = 3/6 = 1/2.

Answer1/2

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(i) a king of red colour

1. A deck has 52 cards, all equally likely since it is well shuffled.
2. There are 4 kings, 2 of them red (hearts, diamonds).
3. P(red king) = 2/52 = 1/26.

Answer1/26

(ii) a face card

1. Face cards are king, queen and jack — 3 per suit, 4 suits, so 12 face cards.
2. P(face card) = 12/52 = 3/13.

Answer3/13

(iii) a red face card

1. Half the face cards are red: 6 red face cards.
2. P(red face card) = 6/52 = 3/26.

Answer3/26

(iv) the jack of hearts

1. There is exactly one jack of hearts in the deck.
2. P(jack of hearts) = 1/52.

Answer1/52

(v) a spade

1. Each suit has 13 cards, and spades is one suit.
2. P(spade) = 13/52 = 1/4.

Answer1/4

(vi) the queen of diamonds

1. There is exactly one queen of diamonds in the deck.
2. P(queen of diamonds) = 1/52.

Answer1/52

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(i) What is the probability that the card is the queen?

1. There are 5 cards: ten, jack, queen, king, ace — all equally likely.
2. Only 1 of them is the queen.
3. P(queen) = 1/5.

Answer1/5

(ii)(a) an ace?

1. The queen is drawn and put aside, so 4 cards remain: ten, jack, king, ace.
2. One card is now drawn from these 4.
3. P(ace) = 1/4.

Answer1/4

(ii)(b) a queen?

1. The remaining 4 cards are ten, jack, king, ace — there is no queen left.
2. So no outcome favours 'queen'.
3. P(queen) = 0/4 = 0.

Answer0

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1. Total pens = 12 defective + 132 good = 144.
2. All 144 pens are equally likely to be picked, since defective and good ones look alike.
3. Favourable outcomes for 'good pen' = 132.
4. P(good pen) = 132/144 = 11/12.

Answer11/12

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(i) What is the probability that this bulb is defective?

1. There are 20 bulbs in all, 4 of them defective.
2. P(defective) = 4/20 = 1/5.

Answer1/5

(ii) the bulb drawn in (i) is not defective and is not replaced

1. The bulb drawn was good and is not put back, so 19 bulbs remain.
2. Of these 19, the defective ones are still 4, so the good ones are 19 − 4 = 15.
3. P(not defective) = 15/19.

Answer15/19

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(i) a two-digit number

1. There are 90 discs, numbered 1 to 90, all equally likely.
2. Two-digit numbers run from 10 to 90, that is 90 − 10 + 1 = 81 numbers.
3. P(two-digit) = 81/90 = 9/10.

Answer9/10

(ii) a perfect square number

1. Perfect squares from 1 to 90: 1, 4, 9, 16, 25, 36, 49, 64, 81 — that is 9 numbers.
2. P(perfect square) = 9/90 = 1/10.

Answer1/10

(iii) a number divisible by 5

1. Multiples of 5 from 1 to 90: 5, 10, 15, ..., 90 — that is 90/5 = 18 numbers.
2. P(divisible by 5) = 18/90 = 1/5.

Answer1/5

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(i) A?

1. The die still has 6 faces, so the denominator is 6, even though only 5 letters are used.
2. The letter A is written on 2 of the 6 faces.
3. P(A) = 2/6 = 1/3.

Answer1/3

(ii) D?

1. The letter D is written on only 1 face.
2. P(D) = 1/6.

Answer1/6

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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 its radius is 1/2 m.
3. Area of the circle = π × (1/2)² = π/4 m².
4. Since the die can land anywhere on the rectangle with equal chance, the probability is a share of area: (circle area) ÷ (rectangle area).
5. P(inside circle) = (π/4) ÷ 6 = π/24.

Answerπ/24

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(i) She will buy it?

1. Total pens = 144, of which 20 are defective, so good pens = 144 − 20 = 124.
2. Nuri buys the pen exactly when it is good.
3. P(buys it) = 124/144 = 31/36.

Answer31/36

(ii) She will not buy it?

1. Nuri does not buy the pen exactly when it is defective.
2. P(does not buy) = 20/144 = 5/36.
3. Check: 31/36 + 5/36 = 1, as it should for complementary events.

Answer5/36

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(i) Complete the following table

1. Two dice give 6 × 6 = 36 equally likely ordered pairs.
2. For each sum, count the pairs that give it: sum 3 comes from (1,2),(2,1) — 2 pairs; sum 4 from 3 pairs; sum 5 from 4 pairs; sum 6 from 5 pairs; sum 7 from 6 pairs; sum 9 from 4 pairs; sum 10 from 3 pairs; sum 11 from 2 pairs.
3. Divide each count by 36 and simplify.

4. Sum	Probability
   2	1/36
   3	1/18
   4	1/12
   5	1/9
   6	5/36
   7	1/6
   8	5/36
   9	1/9
   10	1/12
   11	1/18
   12	1/36

AnswerTable above: 1/36, 1/18, 1/12, 1/9, 5/36, 1/6, 5/36, 1/9, 1/12, 1/18, 1/36 for sums 2 to 12.

(ii) Do you agree with this argument? Justify your answer.

1. The 11 sums are not equally likely — a sum of 7 comes from 6 pairs, while a sum of 2 comes from only 1 pair.
2. Since the outcomes named are not equally likely, 'favourable over total' cannot be applied to these 11 descriptions directly.
3. So each sum does not have probability 1/11; the table above shows the correct, unequal probabilities.

AnswerNo, the argument is not correct — the 11 sums are not equally likely outcomes.

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1. Tossing a coin 3 times gives 2 × 2 × 2 = 8 equally likely outcomes: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.
2. Hanif wins only on HHH or TTT — that is 2 outcomes.
3. P(Hanif wins) = 2/8 = 1/4.
4. 'Lose' is the complement of 'win'.
5. P(Hanif loses) = 1 − 1/4 = 3/4.

Answer3/4

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(i) 5 will not come up either time?

1. Throwing a die twice gives 6 × 6 = 36 equally likely ordered pairs.
2. On each throw, 5 of the 6 faces are not a 5, so pairs with no 5 at all number 5 × 5 = 25.
3. P(no 5 either time) = 25/36.

Answer25/36

(ii) 5 will come up at least once?

1. 'At least once' is the complement of 'not come up either time'.
2. P(at least one 5) = 1 − 25/36 = 11/36.

Answer11/36

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(i) two heads, two tails or one of each

1. Tossing two coins actually gives 4 equally likely outcomes if the coins are told apart: HH, HT, TH, TT.
2. 'Two heads' is 1 of these 4, 'two tails' is 1 of 4, but 'one of each' is 2 of 4 (HT and TH).
3. So the three descriptions are not equally likely: P(HH) = 1/4, P(TT) = 1/4, P(one of each) = 1/2, not 1/3 each.

AnswerNot correct — the three outcomes named are not equally likely.

(ii) an odd number or an even number

1. A die has 6 equally likely faces: 1, 2, 3, 4, 5, 6.
2. Odd numbers are 1, 3, 5 (3 of them) and even numbers are 2, 4, 6 (3 of them) — equally many.
3. So P(odd) = 3/6 = 1/2, which matches the given answer.

AnswerCorrect — odd and even numbers are equally likely (3 faces each) on a die.

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