Exercise 14.1 — Probability

Classical probability and complementary events — coins, dice, cards, marbles, defective items, and one non-examinable geometric-probability question

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13 marksNCERT Cl-10 Maths, Ex 14.1, Q1
Complete: (i) P(E) + P(not E) = ___. (ii) The probability of an event that cannot happen is called ___. (iii) The probability of an event that is certain to happen is called ___. (iv) The sum of the probabilities of all elementary events of an experiment is ___. (v) The probability of an event is greater than or equal to ___ and less than or equal to ___.
Hint. These five blanks are the chapter's foundational vocabulary — get them exactly right and everything else follows.
(i) P(E) + P(not E) = 1, since E and 'not E' between them cover every possible outcome exactly once.

(ii) An event that cannot happen has probability 0 — this is called an impossible event.

(iii) An event certain to happen has probability 1 — this is called a sure (or certain) event.

(iv) The probabilities of all elementary events add up to 1, because together they account for the entire sample space.

(v) The probability of any event satisfies 0 ≤ P(E) ≤ 1.

✦ Answer: (i) 1 (ii) impossible event (iii) sure/certain event (iv) 1 (v) 0 and 1
Where students slip. Writing '100%' for parts (i) or (iv) instead of the number 1. Probability in this chapter is expressed as a number between 0 and 1, not as a percentage, unless the question itself uses percentages.
Another way. All five blanks are consequences of one idea: the sample space accounts for everything that could happen, so probabilities over it must total exactly 1, and no single piece can exceed that total or fall below zero.
23 marksNCERT Cl-10 Maths, Ex 14.1, Q2
Which of the following experiments have equally likely outcomes? Explain. (i) A driver attempts to start a car — it starts or does not. (ii) A player shoots a basketball — scores or misses. (iii) A true-false question is answered — right or wrong. (iv) A baby is born — boy or girl.
Hint. Having exactly two outcomes does not automatically make them equally likely — ask whether anything favours one outcome over the other.
(i) Not equally likely. Whether a car starts depends on its mechanical condition, fuel, battery charge, and so on — one outcome is typically far more probable than the other, and that likelihood isn't fixed at 1/2 by the nature of the experiment.

(ii) Not equally likely. A player's chance of scoring depends on skill, distance, and practice — nothing makes 'score' and 'miss' inherently equal in probability.

(iii) Equally likely — but only under the assumption that the answer is a random guess with no knowledge of the subject. With that assumption (which the question intends), right and wrong are symmetric, each with probability 1/2.

(iv) Equally likely (as a standard modelling assumption). Biologically the split isn't exactly 50-50, but for this chapter's purposes, a baby being a boy or a girl is treated as equally likely.

✦ Answer: (i) No (ii) No (iii) Yes (iv) Yes
Where students slip. Answering all four 'yes' simply because each has exactly two outcomes. The number of outcomes tells you nothing about whether they're equally likely — that requires a genuine symmetry in the situation, which (i) and (ii) lack.
Another way. Ask a diagnostic question for each: 'Is there any reason, in principle, that one outcome should occur more often than the other?' If yes (as in a car's mechanical reliability, or a player's aim), the outcomes are not equally likely.
32 marksNCERT Cl-10 Maths, Ex 14.1, Q3
Why is tossing a coin considered a fair way of deciding which team gets the ball at the start of a football game?
Hint. 'Fair' here means neither team is favoured — connect that directly to the coin's two possible outcomes.
A coin toss has exactly two possible outcomes, heads or tails, and — for a fair (unbiased) coin — both are equally likely, each with probability 1/2.

Since neither team has any way to influence which face lands up, and both outcomes are equally probable, assigning one team to heads and the other to tails gives both teams an identical, unbiased chance of winning the toss. That symmetry is exactly what makes it 'fair'.

✦ Answer: Because a fair coin gives exactly two equally likely outcomes (P = 1/2 each), neither team has an advantage — the method is unbiased.
Where students slip. Answering only 'because it has two outcomes' without mentioning that those two outcomes are *equally likely*. A coin weighted to favour one side would still have two outcomes, but the toss would no longer be fair.
Another way. Contrast with an unfair method: rolling a die and giving the ball to whichever team guesses closer to the result would not be fair in the same way, since a die has six outcomes and the assignment of 'closer' could be structured to favour one side.
42 marksNCERT Cl-10 Maths, Ex 14.1, Q4
Which of the following cannot be the probability of an event? (A) 2/3 (B) −1.5 (C) 15% (D) 0.7
Hint. Every valid probability must fall between 0 and 1 inclusive — check each option against that range.
Step 1 — Convert each option to a plain number between 0 and 1 where needed, and check the range 0 to 1. (A) 2/3 ≈ 0.667 — within range ✓ (B) −1.5 — negative, outside the range ✗ (C) 15% = 0.15 — within range ✓ (D) 0.7 — within range ✓

Step 2 — Only option (B) falls outside 0 ≤ P(E) ≤ 1, since a probability can never be negative.

✦ Answer: (B) −1.5 cannot be a probability.
Where students slip. Flagging (A) as suspicious because 2/3 isn't a 'clean' decimal. Any fraction between 0 and 1 is a perfectly valid probability — the issue is only ever about the number being outside [0, 1], as with a negative value or a value over 1.
Another way. Recall the definition: probability is a count of favourable outcomes divided by a count of total outcomes, both non-negative, with favourable never exceeding total. A negative value like −1.5 simply cannot arise from that construction.
51 markNCERT Cl-10 Maths, Ex 14.1, Q5
If P(E) = 0.05, what is the probability of 'not E'?
Hint. Straight application of the complementary-events formula.
Step 1 — Apply P(E) + P(not E) = 1. P(not E) = 1 − P(E) = 1 − 0.05

✦ Answer: 0.95
Where students slip. Computing 1/0.05 or some other operation instead of the simple subtraction 1 − P(E). The complementary rule is always a subtraction, never a division.
Another way. Think of it as a near-certain event: if E happens only 5% of the time, 'not E' must happen the other 95% of the time — 0.95 follows immediately from that framing.
62 marksNCERT Cl-10 Maths, Ex 14.1, Q6
A bag contains lemon-flavoured candies only. Malini takes out one candy without looking. What is the probability that she takes out (i) an orange-flavoured candy? (ii) a lemon-flavoured candy?
Hint. There are no orange candies in the bag at all — think about what that means for the count of favourable outcomes.
(i) Since the bag contains lemon candies only, there are zero orange candies among the total. The number of favourable outcomes for 'orange-flavoured' is 0. P(orange) = 0/total = 0 — an impossible event.

(ii) Every candy in the bag is lemon-flavoured, so however many candies there are, all of them are favourable outcomes for 'lemon-flavoured'. P(lemon) = total/total = 1 — a certain event.

✦ Answer: (i) 0 (ii) 1
Where students slip. Trying to assign some small positive probability to the orange outcome 'just in case'. If the bag genuinely contains lemon candies only, the orange outcome literally cannot occur — its probability is exactly 0, not merely small.
Another way. This question doesn't need the total number of candies at all — whatever that number is, 0 out of it is 0, and all of it out of itself is 1. The answer is determined purely by the words 'lemon-flavoured candies only'.
72 marksNCERT Cl-10 Maths, Ex 14.1, Q7
In a group of 3 students, the probability that 2 students do not have the same birthday is 0.992. What is the probability that 2 students have the same birthday?
Hint. 'Same birthday' and 'not the same birthday' are complements of each other.
Step 1 — Recognise that the two events are complements: either the two students share a birthday, or they don't — there's no third possibility.

Step 2 — Apply P(E) + P(not E) = 1, since the two events between them cover every possibility for the pair. P(same birthday) = 1 − P(not same birthday) = 1 − 0.992

✦ Answer: 0.008
Where students slip. Overcomplicating this by trying to compute the probability from scratch using the number of days in a year. The question already gives you one probability directly — recognising the complementary relationship is all that's needed.
Another way. This is the classic birthday-problem setup: the 'not sharing a birthday' probability is usually easier to compute directly (via 365/365 × 364/365 for the second student), while 'sharing' is easier via the complement — exactly why the question hands you the harder one and asks for the easier one via subtraction.
82 marksNCERT Cl-10 Maths, Ex 14.1, Q8
A bag contains 3 red balls and 5 black balls. A ball is drawn at random. What is the probability that it is (i) red? (ii) not red?
Hint. Find the total first, then either count black balls directly or use the complement.
Step 1 — Find the total number of balls. 3 + 5 = 8

(i) P(red) = favourable/total = 3/8

✦ P(red) = 3/8

(ii) 'Not red' means black, and there are 5 black balls out of 8. P(not red) = 5/8

✦ P(not red) = 5/8

Check. P(red) + P(not red) = 3/8 + 5/8 = 8/8 = 1, which is what we expect since between them the two events cover every ball in the bag.
Where students slip. Forgetting to add both colours to find the total, and instead dividing by just one of the two counts (say, dividing by 5 instead of 8).
Another way. Get part (ii) via the complement rule instead of counting black balls directly: P(not red) = 1 − P(red) = 1 − 3/8 = 5/8 — useful practice for when the 'not' count isn't given as cleanly.
93 marksNCERT Cl-10 Maths, Ex 14.1, Q9
A box contains 5 red, 8 white and 4 green marbles. One marble is taken out at random. What is the probability that it will be (i) red? (ii) white? (iii) not green?
Hint. Total the three colours first; for part (iii), decide whether to count directly or use the complement.
Step 1 — Find the total. 5 + 8 + 4 = 17

(i) P(red) = 5/17

(ii) P(white) = 8/17

(iii) 'Not green' means red or white, which together number 5 + 8 = 13. P(not green) = 13/17

✦ Answer: (i) 5/17 (ii) 8/17 (iii) 13/17, since red and white together make up every marble that isn't green.
Where students slip. Computing part (iii) as 1 − P(red) or 1 − P(white) alone, forgetting that 'not green' includes *both* other colours, not just one.
Another way. Cross-check part (iii) via the complement of green: P(green) = 4/17, so P(not green) = 1 − 4/17 = 13/17 — the same answer, and a good habit whenever 'not [category]' appears with more than two categories present.
103 marksNCERT Cl-10 Maths, Ex 14.1, Q10
A piggy bank contains a hundred 50p coins, fifty ₹1 coins, twenty ₹2 coins and ten ₹5 coins. If it's equally likely that any one coin falls out, what is the probability that the coin (i) will be a 50p coin? (ii) will not be a ₹5 coin?
Hint. Add up all four coin counts for the total; for part (ii), the complement is quicker than counting three categories directly.
Step 1 — Find the total number of coins. 100 + 50 + 20 + 10 = 180

(i) P(50p coin) = 100/180 = 5/9

✦ P(50p) = 5/9

(ii) First find P(₹5 coin) = 10/180 = 1/18. P(not ₹5) = 1 − 1/18 = 17/18, since every coin that isn't a ₹5 coin falls into this event.

✦ P(not ₹5) = 17/18
Where students slip. Trying to add up 50p + ₹1 + ₹2 coins (100+50+20 = 170) directly for part (ii) instead of simplifying with the complement — both routes give the same 170/180 = 17/18, but the complement is faster and less error-prone with four categories in play.
Another way. Direct count for part (ii): coins that are not ₹5 total 100+50+20 = 170, so P = 170/180 = 17/18 — matches the complement method exactly, which is itself a good consistency check.
112 marksNCERT Cl-10 Maths, Ex 14.1, Q11 (Fig. 14.4)
Gopi buys a fish for his aquarium. The shopkeeper takes one fish at random from a tank containing 5 male and 8 female fish. What is the probability the fish taken out is male?
Hint. Simple ratio of one category over the total.
Step 1 — Find the total number of fish. 5 + 8 = 13

Step 2 — Apply the classical probability formula. P(male) = 5/13

✦ Answer: 5/13
Where students slip. Dividing 5 by 8 (male count over female count) instead of 5 by the total of 13. The denominator in classical probability is always the *total* number of outcomes, not a count of the complementary category.
Another way. As a check, P(female) = 8/13, and 5/13 + 8/13 = 13/13 = 1, confirming every fish has been accounted for exactly once.
124 marksNCERT Cl-10 Maths, Ex 14.1, Q12 (Fig. 14.5)
A spinning arrow rests on one of the numbers 1-8, all equally likely. What is the probability it points at (i) 8? (ii) an odd number? (iii) a number greater than 2? (iv) a number less than 9?
Hint. Total outcomes are 8 throughout. List the favourable set explicitly for each part before counting.
Total outcomes: 8 (the numbers 1 through 8, each equally likely).

(i) Only one outcome, 8 itself, is favourable. P(8) = 1/8

(ii) Odd numbers in the set: {1, 3, 5, 7} — 4 outcomes. P(odd) = 4/8 = 1/2

(iii) Numbers greater than 2: {3, 4, 5, 6, 7, 8} — 6 outcomes. P(>2) = 6/8 = 3/4

(iv) Numbers less than 9: {1, 2, 3, 4, 5, 6, 7, 8} — all 8 outcomes, since every number on the spinner is less than 9. P(<9) = 8/8 = 1

✦ Answer: (i) 1/8 (ii) 1/2 (iii) 3/4 (iv) 1 (a certain event)
Where students slip. In part (iii), including 2 itself in the 'greater than 2' count. 'Greater than' is a strict inequality — 2 is excluded, so the set starts at 3.
Another way. Part (iv) is worth recognising instantly as a certain event without any counting: since every possible outcome (1 through 8) is less than 9, the probability must be exactly 1.
132 marksNCERT Cl-10 Maths, Ex 14.1, Q13
A 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.
Hint. List the favourable outcomes for each part explicitly — 'between 2 and 6' is easy to get wrong on the boundaries.
Total outcomes: 6 (the faces 1 through 6).

(i) Prime numbers among 1-6: {2, 3, 5} — 3 is prime, 1 is not (by definition primes must be greater than 1). P(prime) = 3/6 = 1/2

(ii) 'Between 2 and 6' means strictly between, excluding the endpoints: {3, 4, 5} — 3 outcomes. P(between 2 and 6) = 3/6 = 1/2

(iii) Odd numbers: {1, 3, 5} — 3 outcomes. P(odd) = 3/6 = 1/2

✦ Answer: (i) 1/2 (ii) 1/2 (iii) 1/2, since each set happens to contain exactly half of the six faces.
Where students slip. Including 1 as a prime number, or including 2 and/or 6 in the 'between 2 and 6' set. Neither 1 is prime, nor do the endpoints count for 'between'.
Another way. It's a pleasant coincidence that all three parts give 1/2 here — worth noticing, but each set (primes, the middle three numbers, odds) arrives at that fraction independently, not because of any deeper connection between them.
143 marksNCERT Cl-10 Maths, Ex 14.1, Q14
One card is drawn from a well-shuffled deck of 52. 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.
Hint. Know the deck structure cold: 4 suits of 13 each, 2 red suits and 2 black, and 3 face cards (J, Q, K) per suit.
Total outcomes: 52.

(i) King of red colour. There are 2 red suits (hearts, diamonds), each with one king: 2 favourable. P = 2/52 = 1/26

(ii) Face card. Each of the 4 suits has 3 face cards (J, Q, K): 4 × 3 = 12 favourable. P = 12/52 = 3/13

(iii) Red face card. 2 red suits × 3 face cards each = 6 favourable. P = 6/52 = 3/26

(iv) The jack of hearts. Exactly one such card exists in the deck. P = 1/52

(v) A spade. One full suit: 13 favourable. P = 13/52 = 1/4

(vi) The queen of diamonds. Exactly one such card. P = 1/52

✦ Answer: (i) 1/26 (ii) 3/13 (iii) 3/26 (iv) 1/52 (v) 1/4 (vi) 1/52, since each fraction is simply the favourable count over the deck's 52 cards.
Where students slip. Miscounting face cards as including the ace, giving 4 per suit instead of 3. In standard usage, face cards are only jack, queen and king — the ace is a separate category.
Another way. Part (iii) can be seen as half of part (ii): exactly half the face cards (6 of 12) are red, since 2 of the 4 suits are red — consistent with 6/52 = 3/26 being exactly half of 12/52 = 3/13.
154 marksNCERT Cl-10 Maths, Ex 14.1, Q15
Five cards — ten, jack, queen, king, ace of diamonds — are shuffled face down. (i) What is the probability the card is the queen? (ii) If the queen is drawn and set aside, what is the probability the second card is (a) an ace? (b) a queen?
Hint. After the queen is removed, both the total and the composition of the remaining pile change for part (ii).
(i) Total outcomes: 5 cards, exactly one of which is the queen. P(queen) = 1/5

(ii) After the queen is drawn and set aside, 4 cards remain: ten, jack, king, ace.

(a) Exactly one of these 4 remaining cards is the ace. P(ace | queen removed) = 1/4

(b) The queen has already been set aside — it is no longer among the remaining 4 cards. P(queen | queen removed) = 0/4 = 0, since the queen has already been set aside and can no longer be drawn

✦ Answer: (i) 1/5 (ii)(a) 1/4 (ii)(b) 0
Where students slip. Using 5 as the total for part (ii) instead of 4. Once the queen is set aside, it is genuinely gone from the pile — the second draw happens from the remaining 4 cards only.
Another way. Part (ii)(b) is worth pausing on conceptually: it's not a small probability, it's exactly zero, because the queen has been physically removed from the pile — there is no queen left to possibly draw.
162 marksNCERT Cl-10 Maths, Ex 14.1, Q16
12 defective pens are accidentally mixed with 132 good ones. One pen is taken out at random. Find the probability that it is a good one.
Hint. Total the two categories first.
Step 1 — Find the total number of pens. 12 + 132 = 144

Step 2 — Apply the classical probability formula. P(good) = 132/144

Step 3 — Simplify. Both numbers divide by 12. 132/144 = 11/12, since both numbers share a factor of 12.

✦ Answer: 11/12
Where students slip. Leaving the answer as 132/144 unsimplified when a cleaner equivalent fraction is expected, or simplifying incorrectly by dividing by the wrong common factor.
Another way. Via the complement: P(defective) = 12/144 = 1/12, so P(good) = 1 − 1/12 = 11/12 — same answer, and arguably quicker since 12/144 reduces to a recognisable 1/12 immediately.
173 marksNCERT Cl-10 Maths, Ex 14.1, Q17
(i) A lot of 20 bulbs contains 4 defective ones. One bulb is drawn at random. Find the probability it is defective. (ii) Suppose that bulb is not defective and is not replaced. Now one bulb is drawn from the rest. Find the probability this bulb is not defective.
Hint. Part (ii) draws from a changed pool — one bulb (a good one) has already been removed.
(i) Total: 20 bulbs, 4 defective. P(defective) = 4/20 = 1/5

(ii) The bulb removed in part (i) was not defective (given), and it was not replaced. So the remaining pool has: total = 20 − 1 = 19 bulbs defective = still 4 (none were removed) good = 16 − 1 = 15

P(not defective, i.e. good, on the second draw) = 15/19

✦ Answer: (i) 1/5 (ii) 15/19
Where students slip. Keeping the total at 20 for part (ii), forgetting that one bulb has physically left the pool. Whenever a question says 'not replaced', both the total and the relevant category count must be updated for the next draw.
Another way. Track it as a running tally: start 20 total (4 defective, 16 good) → remove 1 good → 19 total (4 defective, 15 good) → P(good now) = 15/19. Writing out the before/after tally explicitly avoids losing track of which count changed.
183 marksNCERT Cl-10 Maths, Ex 14.1, Q18
A box contains 90 discs numbered 1 to 90. One disc is drawn at random. Find the probability it bears (i) a two-digit number, (ii) a perfect square number, (iii) a number divisible by 5.
Hint. Count each favourable set carefully — two-digit numbers run from 10 to 90, and perfect squares need to be listed out.
Total outcomes: 90.

(i) Two-digit numbers. These run from 10 to 90 inclusive. count = 90 − 10 + 1 = 81 P = 81/90 = 9/10

(ii) Perfect squares from 1 to 90. List them: 1, 4, 9, 16, 25, 36, 49, 64, 81 — that's 9² = 81 as the last one, since 10² = 100 exceeds 90. count = 9 P = 9/90 = 1/10

(iii) Numbers divisible by 5. These are 5, 10, 15, ..., 90. count = 90/5 = 18 P = 18/90 = 1/5

✦ Answer: (i) 9/10 (ii) 1/10 (iii) 1/5
Where students slip. Counting two-digit numbers as 90 − 10 = 80, forgetting the '+1' needed when both endpoints of a range are included. The correct count of integers from a to b inclusive is (b − a + 1).
Another way. For part (ii), instead of listing squares, note that the largest n with n² ≤ 90 is n = 9 (since 9² = 81 and 10² = 100), so there are exactly 9 perfect squares from 1 to 90 — a quick way to get the count without writing out the list.
192 marksNCERT Cl-10 Maths, Ex 14.1, Q19
A die's six faces show the letters A, B, C, D, E, A (note: A appears on two faces). The die is thrown once. Find the probability of getting (i) A, (ii) D.
Hint. This die is not the usual 1-6 one — one letter is repeated, which changes its probability.
Total outcomes: 6 faces.

(i) The letter A appears on 2 of the 6 faces (this is the detail that makes the die non-standard). P(A) = 2/6 = 1/3

(ii) The letter D appears on exactly 1 face. P(D) = 1/6

✦ Answer: (i) 1/3 (ii) 1/6 — different values precisely because A occupies two faces while D occupies only one.
Where students slip. Treating every letter as equally represented and answering 1/6 for both parts. The whole point of this question is that A is *not* equally represented — it occupies two faces, not one.
Another way. As a check, add up probabilities for all distinct letters: P(A)=2/6, P(B)=1/6, P(C)=1/6, P(D)=1/6, P(E)=1/6. Sum = 2/6+1/6+1/6+1/6+1/6 = 6/6 = 1 ✓ — confirming all six faces are accounted for.
203 marksNCERT Cl-10 Maths, Ex 14.1, Q20 (Fig. 14.6) — marked by NCERT as 'not from the examination point of view'
A die is dropped at random on a rectangular region 3 m by 2 m. What is the probability it lands inside a circle of diameter 1 m marked within the rectangle? (This question is starred in the textbook as background only, not examinable.)
Hint. This is geometric probability — the ratio of two areas, not a ratio of counts.
Step 1 — Find the area of the rectangle (the total 'sample space' here). area = 3 × 2 = 6 m²

Step 2 — Find the area of the circle (the 'favourable' region). diameter = 1 m, so radius = 0.5 m area = πr² = π(0.5)² = 0.25π m²

Step 3 — The probability is the ratio of favourable area to total area. P = 0.25π / 6 = π/24

Step 4 — Evaluate numerically. ≈ 3.1416/24 ≈ 0.1309

✦ Answer: π/24 ≈ 0.131
Where students slip. Treating this like the rest of the exercise and looking for a count of discrete outcomes. Here, the die can land anywhere in a continuous region, so probability is measured by area, not by counting points.
Another way. This kind of problem — probability as a ratio of geometric measures rather than a ratio of counts — is called geometric probability. It is genuinely useful (it's how Buffon's needle problem and many simulation techniques work), which is presumably why NCERT kept the question even while marking it non-examinable.
212 marksNCERT Cl-10 Maths, Ex 14.1, Q21
A lot of 144 ball pens has 20 defective ones, the rest good. Nuri buys the pen if good, refuses it if defective. The shopkeeper hands her one at random. Find (i) P(she buys it), (ii) P(she does not buy it).
Hint. 'Buys it' means the pen is good — translate the word problem into the right category first.
Step 1 — Find the number of good pens. 144 − 20 = 124

(i) 'She will buy it' happens exactly when the pen is good. P(buys) = 124/144 = 31/36

(ii) 'She will not buy it' happens exactly when the pen is defective. P(not buy) = 20/144 = 5/36

Check. 31/36 + 5/36 = 36/36 = 1, since buying and not buying are complementary events that account for every pen in the lot.

✦ Answer: (i) 31/36 (ii) 5/36
Where students slip. Missing the translation step — reading 'buys it' as a separate random event unrelated to defectiveness, rather than recognising it is simply another name for 'the pen is good'.
Another way. Once (i) is found, (ii) follows immediately by the complement: P(not buy) = 1 − 31/36 = 5/36, without needing to separately reduce 20/144.

225 marksNCERT Cl-10 Maths, Ex 14.1, Q22 (refers to Example 13 — two dice thrown together)

Two dice are thrown together. (i) Complete the probability table for each possible sum (2 through 12). (ii) A student argues there are 11 possible sums (2 through 12), so each has probability 1/11. Do you agree? Justify your answer.

Hint. The two dice are distinguishable for counting purposes — (1,2) and (2,1) are different outcomes even though they give the same sum.

(i) The complete table.

Step 1 — There are 6 × 6 = 36 equally likely outcomes when two dice are thrown, since each die independently shows one of 6 faces.

Step 2 — Count how many of the 36 outcomes give each sum.

Sum	Outcomes giving that sum	Count	Probability
2	(1,1)	1	1/36
3	(1,2),(2,1)	2	2/36
4	(1,3),(2,2),(3,1)	3	3/36
5	(1,4),(2,3),(3,2),(4,1)	4	4/36
6	(1,5),(2,4),(3,3),(4,2),(5,1)	5	5/36
7	(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)	6	6/36
8	(2,6),(3,5),(4,4),(5,3),(6,2)	5	5/36
9	(3,6),(4,5),(5,4),(6,3)	4	4/36
10	(4,6),(5,5),(6,4)	3	3/36
11	(5,6),(6,5)	2	2/36
12	(6,6)	1	1/36

Check. 1+2+3+4+5+6+5+4+3+2+1 = 36, and 36/36 = 1 ✓

(ii) Do you agree with the student?

No. The student's error is treating the 11 sums as if they were the 11 equally likely elementary outcomes — they are not. The genuinely equally likely outcomes are the 36 ordered pairs (die 1 result, die 2 result), each with probability 1/36. The sums group these 36 outcomes unevenly: sum 7 is built from 6 of the 36 pairs, while sum 2 is built from only 1. Since the sums don't each get an equal share of the 36 underlying outcomes, they cannot each have equal probability.

✦ Answer: Disagree. There are 36 equally likely elementary outcomes, not 11 — the sums 2 through 12 arise from very different numbers of those 36 outcomes (from 1 up to 6), so their probabilities are not equal.

Where students slip. Assuming that because there are 11 *possible values* for the sum, each value must be equally probable. Having 11 distinct outcomes for the sum tells you nothing about their relative likelihood — that depends entirely on how many of the 36 underlying dice-pairs produce each sum.

Another way. A physical check: if you actually rolled two dice many times, you would see 7 come up roughly six times as often as 2 or 12 — that empirical observation is exactly predicted by the counts 6 versus 1 out of 36, and would visibly contradict the student's claim of equal 1/11 probabilities.

233 marksNCERT Cl-10 Maths, Ex 14.1, Q23
A game tosses a coin 3 times. Hanif wins if all three tosses match (all heads or all tails), and loses otherwise. Find the probability that Hanif loses.
Hint. List the full sample space of 3 tosses — there are only 8 outcomes, so this is manageable directly.
Step 1 — List the sample space for 3 tosses. Each toss is H or T, so there are 2³ = 8 equally likely outcomes: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT

Step 2 — Identify the winning outcomes: all three the same. HHH and TTT — 2 outcomes. P(win) = 2/8 = 1/4

Step 3 — Hanif loses on every other outcome, so use the complement. P(lose) = 1 − 1/4 = 3/4

✦ Answer: 3/4
Where students slip. Undercounting the sample space as having only 4 outcomes (as if the order of heads and tails across the three tosses didn't matter). Each toss is a separate, ordered event, giving 2³ = 8 distinct outcomes, not fewer.
Another way. Count the losing outcomes directly instead: all 8 outcomes except HHH and TTT are losses, giving 6 losing outcomes directly, so P(lose) = 6/8 = 3/4 — the same answer without going through the complement formula explicitly.
243 marksNCERT Cl-10 Maths, Ex 14.1, Q24
A die is thrown twice. What is the probability that (i) 5 does not come up either time? (ii) 5 comes up at least once?
Hint. For part (i), treat the two throws as independent and multiply. For part (ii), the complement of 'at least once' is exactly 'not at all', which is part (i).
Total outcomes: 6 × 6 = 36, treating the two throws (or equivalently, two dice thrown together) the same way.

(i) 5 does not come up on either throw. For the first throw, 5 outcomes avoid a 5 (namely 1,2,3,4,6); same for the second throw. favourable = 5 × 5 = 25 P(no 5 at all) = 25/36

(ii) 5 comes up at least once. This is the complement of 'no 5 at all'. P(at least one 5) = 1 − 25/36 = 11/36

✦ Answer: (i) 25/36 (ii) 11/36, since 'at least once' and 'not at all' are complementary events that together cover every one of the 36 outcomes.
Where students slip. Trying to count 'at least once' by directly listing every way a 5 could appear (only on the first throw, only on the second, or on both) and risking double-counting or missing a case. The complement route sidesteps that entirely.
Another way. Direct count for part (ii), as a check: outcomes with at least one 5 are those where the first die is 5 (6 outcomes: (5,1) through (5,6)) plus those where the second die is 5 but the first isn't (5 outcomes: (1,5),(2,5),(3,5),(4,5),(6,5)), total 6+5 = 11, giving 11/36 — matches the complement method exactly.
253 marksNCERT Cl-10 Maths, Ex 14.1, Q25
Which of these arguments are correct, and which not? Give reasons. (i) Two coins tossed together give three outcomes — two heads, two tails, or one of each — so each has probability 1/3. (ii) A die is thrown; there are two outcomes, odd or even, so P(odd) = 1/2.
Hint. Check each argument against the actual equally-likely sample space, not against however many named outcomes are listed.
(i) Not correct. The three named outcomes — two heads, two tails, one of each — are not equally likely. The genuine sample space for two coins is {HH, HT, TH, TT}, four equally likely outcomes. 'One of each' actually covers two of these (HT and TH), while 'two heads' and 'two tails' each cover only one. So: P(two heads) = 1/4, P(two tails) = 1/4, P(one of each) = 2/4 = 1/2

These are not all 1/3 — the argument's error is treating three unequal groupings as if grouping them into three named categories automatically made them equally probable.

(ii) Correct — but check why, not just that it happens to work. The sample space for one die is {1,2,3,4,5,6}, six equally likely outcomes. Odd numbers {1,3,5} and even numbers {2,4,6} each contain exactly 3 of the 6 outcomes, so they genuinely are equally likely here. P(odd) = 3/6 = 1/2 ✓

Unlike part (i), grouping into two categories here happens to preserve equal likelihood, because odd and even numbers are represented by the same count (3 each) among the six faces.

✦ Answer: (i) Incorrect — the three groupings are not equally likely, since HT and TH both fall under 'one of each'. (ii) Correct — odd and even each cover exactly 3 of the 6 equally likely outcomes.
Where students slip. Applying the same verdict to both parts, either accepting both arguments' reasoning as valid or rejecting both. The method used in each argument (count the named outcomes, assume they're equally likely) is the same, but it happens to fail in (i) and succeed in (ii) — the reasoning must be checked against the true sample space every time, not assumed to work.
Another way. The general test: to say n named outcomes are 'equally likely', each one must correspond to the same number of equally-likely elementary outcomes. In (i), the three groups correspond to 1, 1, 2 of the 4 elementary outcomes — unequal. In (ii), the two groups correspond to 3, 3 of the 6 elementary outcomes — equal. That count is what actually decides the question, not the number of named outcomes.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 10 Mathematics textbook, Reprint 2026-27 (this chapter is unchanged by rationalisation — a single exercise, 14.1, with 25 questions; one question, on geometric probability, is marked by the textbook itself as not from the examination point of view). Questions are referenced from the NCERT textbook for identification.

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