Key Concepts & Formulae at a Glance
- Theoretical Probability:P(E) = Number of outcomes favourable to E ÷ Number of all possible outcomes, assuming all outcomes are equally likely.
- Range of Probability:0 ≤ P(E) ≤ 1. P(impossible event) = 0. P(surecertain event) = 1.
- Complementary Events:P(E) + P(not E) = 1, so P(not E) = 1 − P(E). "Not E" is written Ē.
- Elementary Events:An event with exactly one outcome. The probabilities of all elementary events of an experiment always sum to 1.
- Listing Outcomes Carefully:For compound experiments — e.g. two coins (4 outcomes: HH,HT,TH,TT) or two dice (36 ordered pairs) — don't collapse distinct orderings like (1,4) and (4,1) into one outcome.
- Geometric (AreaLength) Probability:For continuous outcomes: P(E) = favourable length or area ÷ total length or area.
Exercise 14.1
25 Questions
1Complete the following statements:
(i) Probability of an event E + Probability of the event 'not E' = ____.
(ii) The probability of an event that cannot happen is ____. Such an event is called ____.
(iii) The probability of an event that is certain to happen is ____. Such an event is called ____.
(iv) The sum of the probabilities of all the elementary events of an experiment is ____.
(v) The probability of an event is greater than or equal to ____ and less than or equal to ____.
(i) 1
(ii) 0; called an impossible event
(iii) 1; called a sure event (or certain event)
(iv) 1
(v) 0 and 1
(ii) 0; called an impossible event
(iii) 1; called a sure event (or certain event)
(iv) 1
(v) 0 and 1
2Which of the following experiments have equally likely outcomes? Explain.
(i) A driver attempts to start a car. The car starts or does not start.
(ii) A player attempts to shoot a basketball. Shehe shoots or misses the shot.
(iii) A trial is made to answer a true-false question. The answer is right or wrong.
(iv) A baby is born. It is a boy or a girl.
(i) Car starts / doesn't start
Not equally likely — whether the car starts depends on its mechanical condition, not on an even, unbiased chance.
(ii) Basketball shot goes in / misses
Not equally likely — the outcome depends heavily on the player's skill, not a fair 50-50 chance.
(iii) True-false answer right / wrong
Equally likely — with no other information, a guess has an even chance of being right or wrong.
(iv) Baby is a boy / girl
Equally likely (by the standard assumption used in this chapter).
3Why 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?
Because an unbiased coin toss has exactly two possible outcomes — head and tail — that are equally likely. Assigning one outcome to each team gives both teams exactly the same chance of winning the toss, so neither team has an advantage.
4Which of the following cannot be the probability of an event?
(A) 23 (B) −1.5 (C) 15% (D) 0.7
Every valid probability must satisfy 0 ≤ P(E) ≤ 1. (A) 23, (C) 15% = 0.15, and (D) 0.7 all lie in this range. (B) −1.5 is negative, which is outside the allowed range.
(B) −1.5 cannot be the probability of an event.
5If P(E) = 0.05, what is the probability of 'not E'?
P(not E) = 1 − P(E) = 1 − 0.05
P(not E) = 0.95.
6A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out
(i) an orange flavoured candy?
(ii) a lemon flavoured candy?
(i) There are no orange candies in the bag at all, so getting one is impossible.
(ii) Every candy in the bag is lemon flavoured, so getting one is certain.
(ii) Every candy in the bag is lemon flavoured, so getting one is certain.
(i) P(orange) = 0. (ii) P(lemon) = 1.
7It 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?
"Same birthday" and "not the same birthday" are complementary events.
P(same birthday) = 1 − P(not same birthday) = 1 − 0.992
P(same birthday) = 1 − P(not same birthday) = 1 − 0.992
P(same birthday) = 0.008.
8A 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 (i) red? (ii) not red?
Total balls = 3 + 5 = 8.
(i) P(red) = 38.
(ii) P(not red) = 1 − 38 = 58 (matches directly, since the 5 black balls are exactly the "not red" outcomes).
(i) P(red) = 38.
(ii) P(not red) = 1 − 38 = 58 (matches directly, since the 5 black balls are exactly the "not red" outcomes).
(i) P(red) = 38. (ii) P(not red) = 58.
9A 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 (i) red? (ii) white? (iii) not green?
Total marbles = 5 + 8 + 4 = 17.
(i) P(red) = 517.
(ii) P(white) = 817.
(iii) P(not green) = 1 − 417 = 1317.
(i) P(red) = 517.
(ii) P(white) = 817.
(iii) P(not green) = 1 − 417 = 1317.
(i) 517. (ii) 817. (iii) 1317.
10A piggy bank contains hundred 50p coins, fifty ₹1 coins, twenty ₹2 coins and ten ₹5 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 (i) will be a 50p coin? (ii) will not be a ₹5 coin?
Total coins = 100 + 50 + 20 + 10 = 180.
(i) P(50p coin) = 100180 = 59.
(ii) P(not ₹5 coin) = 1 − 10180 = 170180 = 1718.
(i) P(50p coin) = 100180 = 59.
(ii) P(not ₹5 coin) = 1 − 10180 = 170180 = 1718.
(i) P(50p) = 59. (ii) P(not ₹5) = 1718.
11Gopi 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 = 5 + 8 = 13. P(male fish) = 513.
P(male fish) = 513.
12A 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
(i) 8? (ii) an odd number? (iii) a number greater than 2? (iv) a number less than 9?
Total possible outcomes = 8.
(i) Only one outcome (8) is favourable: P(8) = 18.
(ii) Odd numbers: 1,3,5,7 — 4 outcomes: P(odd) = 48 = 12.
(iii) Numbers greater than 2: 3,4,5,6,7,8 — 6 outcomes: P(>2) = 68 = 34.
(iv) Every number from 1 to 8 is less than 9, so this is a certain event: P(<9) = 88 = 1.
(i) Only one outcome (8) is favourable: P(8) = 18.
(ii) Odd numbers: 1,3,5,7 — 4 outcomes: P(odd) = 48 = 12.
(iii) Numbers greater than 2: 3,4,5,6,7,8 — 6 outcomes: P(>2) = 68 = 34.
(iv) Every number from 1 to 8 is less than 9, so this is a certain event: P(<9) = 88 = 1.
(i) 18. (ii) 12. (iii) 34. (iv) 1.
13A die is thrown once. Find the probability of getting (i) a prime number; (ii) a number lying between 2 and 6; (iii) an odd number.
Total outcomes = 6.
(i) Prime numbers on a die: 2, 3, 5 — 3 outcomes: P(prime) = 36 = 12.
(ii) Numbers strictly between 2 and 6: 3, 4, 5 — 3 outcomes: P = 36 = 12.
(iii) Odd numbers: 1, 3, 5 — 3 outcomes: P(odd) = 36 = 12.
(i) Prime numbers on a die: 2, 3, 5 — 3 outcomes: P(prime) = 36 = 12.
(ii) Numbers strictly between 2 and 6: 3, 4, 5 — 3 outcomes: P = 36 = 12.
(iii) Odd numbers: 1, 3, 5 — 3 outcomes: P(odd) = 36 = 12.
(i) 12. (ii) 12. (iii) 12.
14One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting (i) a king of red colour (ii) a face card (iii) a red face card (iv) the jack of hearts (v) a spade (vi) the queen of diamonds
Total cards = 52.
(i) Red kings: king of hearts, king of diamonds — 2 outcomes: P = 252 = 126.
(ii) Face cards (jack, queen, king in all 4 suits): 4×3 = 12 outcomes: P = 1252 = 313.
(iii) Red face cards (jack, queen, king of hearts and diamonds): 2×3 = 6 outcomes: P = 652 = 326.
(iv) Only 1 jack of hearts: P = 152.
(v) 13 spades: P = 1352 = 14.
(vi) Only 1 queen of diamonds: P = 152.
(i) Red kings: king of hearts, king of diamonds — 2 outcomes: P = 252 = 126.
(ii) Face cards (jack, queen, king in all 4 suits): 4×3 = 12 outcomes: P = 1252 = 313.
(iii) Red face cards (jack, queen, king of hearts and diamonds): 2×3 = 6 outcomes: P = 652 = 326.
(iv) Only 1 jack of hearts: P = 152.
(v) 13 spades: P = 1352 = 14.
(vi) Only 1 queen of diamonds: P = 152.
(i) 126. (ii) 313. (iii) 326. (iv) 152. (v) 14. (vi) 152.
15Five 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.
(i) What is the probability that the card is the queen?
(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?
(i) 5 cards total, 1 queen: P(queen) = 15.
(ii) After removing the queen, 4 cards remain: ten, jack, king, ace.
(a) 1 ace among the 4 remaining: P(ace) = 14.
(b) No queen remains among the 4: P(queen) = 04 = 0.
(ii) After removing the queen, 4 cards remain: ten, jack, king, ace.
(a) 1 ace among the 4 remaining: P(ace) = 14.
(b) No queen remains among the 4: P(queen) = 04 = 0.
(i) 15. (ii)(a) 14. (ii)(b) 0.
1612 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 = 12 + 132 = 144. P(good pen) = 132144 = 1112.
P(good pen) = 1112.
17(i) 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?
(ii) Suppose the bulb drawn in (i) 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) 4 defective out of 20: P(defective) = 420 = 15.
(ii) Since the bulb drawn in (i) was good and not replaced, 19 bulbs remain: 15 good and 4 still defective. P(not defective) = 1519.
(ii) Since the bulb drawn in (i) was good and not replaced, 19 bulbs remain: 15 good and 4 still defective. P(not defective) = 1519.
(i) 15. (ii) 1519.
18A 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 (i) a two-digit number (ii) a perfect square number (iii) a number divisible by 5.
Total discs = 90.
(i) Two-digit numbers from 10 to 90: 90−10+1 = 81 numbers: P = 8190 = 910.
(ii) Perfect squares from 1 to 90: 1,4,9,16,25,36,49,64,81 — 9 numbers (10²=100 is too big): P = 990 = 110.
(iii) Multiples of 5 from 1 to 90: 5,10,...,90 — 18 numbers: P = 1890 = 15.
(i) Two-digit numbers from 10 to 90: 90−10+1 = 81 numbers: P = 8190 = 910.
(ii) Perfect squares from 1 to 90: 1,4,9,16,25,36,49,64,81 — 9 numbers (10²=100 is too big): P = 990 = 110.
(iii) Multiples of 5 from 1 to 90: 5,10,...,90 — 18 numbers: P = 1890 = 15.
(i) 910. (ii) 110. (iii) 15.
19A 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 (i) A? (ii) D?
Total faces = 6, with the letter A appearing on 2 of them (all others appear once).
(i) P(A) = 26 = 13.
(ii) P(D) = 16.
(i) P(A) = 26 = 13.
(ii) P(D) = 16.
(i) 13. (ii) 16.
20Suppose you drop a die at random on the rectangular region shown in the figure. What is the probability that it will land inside the circle with diameter 1 m? (The rectangular region is 3 m by 2 m.)
A circle of diameter 1 m sits inside a 3 m × 2 m rectangular region.
This is a geometric probability question — the die is equally likely to land anywhere in the rectangle, so probability = favourable area ÷ total area.
Area of rectangle = 3 × 2 = 6 m².
Radius of circle = 0.5 m, so area of circle = πr² = π×0.5² = 0.25π m².
P(lands inside circle) = 0.25π6 = π24. Using π = 227: (227)/24 = 22168 = 1184.
Area of rectangle = 3 × 2 = 6 m².
Radius of circle = 0.5 m, so area of circle = πr² = π×0.5² = 0.25π m².
P(lands inside circle) = 0.25π6 = π24. Using π = 227: (227)/24 = 22168 = 1184.
P(lands inside circle) = 1184 (≈ 0.131).
21A 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 (i) she will buy it? (ii) she will not buy it?
Good pens = 144 − 20 = 124.
(i) P(buy, i.e. good) = 124144 = 3136.
(ii) P(not buy, i.e. defective) = 20144 = 536.
(i) P(buy, i.e. good) = 124144 = 3136.
(ii) P(not buy, i.e. defective) = 20144 = 536.
(i) 3136. (ii) 536.
22Refer to Example 13 (two dice, one blue and one grey, thrown together). (i) Complete the following table:
Event 'Sum on 2 dice': 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 — Probability: 136, _, _, _, _, _, 536, _, _, _, 136
(ii) A student argues that 'there are 11 possible outcomes 2, 3, 4, ..., 12. Therefore, each of them has a probability 111.' Do you agree with this argument? Justify your answer.
| Sum | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| Ways | 1 | 2 | 3 | 4 | 5 | 6 | 5 | 4 | 3 | 2 | 1 |
| Probability | 136 | 236 | 336 | 436 | 536 | 636 | 536 | 436 | 336 | 236 | 136 |
(ii) No, this argument is not correct. The 36 individual (blue die, grey die) outcomes are equally likely, but the 11 possible sums are not — a sum of 7 can be made in 6 different ways (e.g. 1+6, 2+5, ..., 6+1), while a sum of 2 can only be made in 1 way (1+1). So P(7) = 636 is much larger than P(2) = 136, not equal.
The table above gives all 11 probabilities; treating all 11 sums as equally likely (each 111) is incorrect.
23A 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 i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.
Total outcomes of tossing a coin 3 times = 2³ = 8: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.
Winning outcomes (all same): HHH, TTT — 2 outcomes. P(win) = 28 = 14.
P(lose) = 1 − P(win) = 1 − 14
Winning outcomes (all same): HHH, TTT — 2 outcomes. P(win) = 28 = 14.
P(lose) = 1 − P(win) = 1 − 14
P(Hanif loses) = 34.
24A die is thrown twice. What is the probability that (i) 5 will not come up either time? (ii) 5 will come up at least once? [Hint: Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]
Total outcomes = 6 × 6 = 36.
(i) Outcomes where neither throw shows 5: each throw has 5 non-5 options, so 5×5 = 25 outcomes: P = 2536.
(ii) "At least one 5" is the complement of "no 5 at all": P = 1 − 2536 = 1136.
(i) Outcomes where neither throw shows 5: each throw has 5 non-5 options, so 5×5 = 25 outcomes: P = 2536.
(ii) "At least one 5" is the complement of "no 5 at all": P = 1 − 2536 = 1136.
(i) 2536. (ii) 1136.
25Which of the following arguments are correct and which are not correct? Give reasons for your answer.
(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 13.
(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 12.
(i)
Not correct. The actual equally likely outcomes are HH, HT, TH, TT (4 outcomes, each with probability 14). "One of each" is really two of these outcomes (HT and TH) combined, so it has probability 24 = 12, not the same probability as "two heads" (14) or "two tails" (14). Treating all three descriptions as equally likely outcomes is the error.
(ii)
Correct — though the reasoning needs to be stated carefully. It's not simply because there happen to be "2 outcomes" (that logic alone would be flawed, as in part (i)). It's correct here because a die's 6 faces split evenly into 3 odd numbers and 3 even numbers, so each of the two events genuinely does have probability 12.
Frequently Asked Questions
Experimental (empirical) probability is based on what actually happened in real trials — the fraction of trials in which an event occurred. Theoretical probability instead assumes all outcomes are equally likely and calculates the fraction of favourable outcomes out of all possible ones, without needing to run the experiment at all. As the number of real trials grows very large, experimental probability tends to approach the theoretical value.
The formula silently assumes each individual outcome has the same chance of happening. If that's not true — say, a biased coin, or a bag with far more red balls than blue ones — then simply counting outcomes without weighting them by how likely each one is would give a wrong answer. That's why the setup always has to justify equal likelihood first, often through phrases like "drawn at random" or "well-shuffled".
There are 36 equally likely ordered outcomes, like (1,6) and (6,1), when two dice are thrown. But different sums are reachable by different numbers of these combinations — only one combination gives a sum of 2 (namely 1+1), while six combinations give a sum of 7. So even though every individual (die1, die2) pair is equally likely, the sums built from them aren't, because some sums have more ways of occurring than others.
Two events are complementary when exactly one of them must happen and they can't both happen — like an event E and "not E". Since their probabilities always add up to 1, once you know P(E), you immediately get P(not E) = 1 − P(E) without any further counting. This is especially useful when "not E" is much easier to count than E itself, such as "at least one head" being easier via 1 − P(no heads at all).
Yes — a printable Formula Card covering all 7 units, including Probability, is available for ₹99.
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