NDA Probability previous year questions with solutions

Answer: (d) 5/⁠324

Method. Count positive integer solutions of a + b + c + d = 7

Each die shows at least 1, and a total of 7 keeps each face at most 4, so every positive solution is a valid outcome.
Number of solutions = ^7−⁠1C₃ = ⁶C₃ = 20.
Required probability = 201296 = 5324.

Answer: (c) 37/⁠55

Method. Convert odds to probabilities; P(solved) = 1 − P(neither solves)

Odds 6 to 9 in favour of X: P(X) = 615 = 25.
Odds 6 to 5 against Y: P(Y) = 511.
P(neither solves) = 35 × 611 = 1855.
P(solved) = 1 − 1855 = 3755.

Answer: (a) 5/⁠84

Method. Fix the smallest number and choose the rest from the larger numbers

Total selections = ¹⁰C₅ = 252.
x₁ = 4: 4 is chosen and the other 4 come from {5, 6, …, 10}: ⁶C₄ = 15 ways.
P(x₁ = 4) = 15252 = 584.

Answer: (a) 5/⁠21

Method. Fix the middle number; choose two below it and two above it

x₃ = 6: two numbers from {1, …, 5} and two from {7, …, 10}.
Ways = ⁵C₂ × ⁴C₂ = 10 × 6 = 60.
P(x₃ = 6) = 60252 = 521.

Answer: (c) 1/⁠252

Method. Fix the three smallest; the two largest must exceed 8

x₁ = 2, x₂ = 3, x₃ = 8 forces x₄, x₅ to be 9 and 10: 1 way.
P = 1¹⁰C₅ = 1252.

Answer: (a) 58

Method. Count 4-digit numbers by their last digit

Total: 4-digit arrangements from 5 digits = 5 × 4 × 3 × 2 = 120, less those starting with 0, 4 × 3 × 2 = 24, giving 96.
Even numbers end in 0, 2 or 4.
Last digit 0: 4 × 3 × 2 = 24.
Last digit 2: first digit from {1, 3, 4} (3 ways), then 3 × 2 = 6 ways for the middle, giving 18. Last digit 4: likewise 18.
Even numbers = 24 + 18 + 18 = 60, so the probability is 6096 = 58.

Answer: (b) 38

Method. Digit-sum test: drop one digit from 0 + 1 + 2 + 3 + 4 = 10

The four digits used are the five digits minus one, with sum 10 − d. This is divisible by 3 only for d = 1 or d = 4.
Digits {0, 2, 3, 4}: 4! − 3! = 18 numbers (excluding a leading 0).
Digits {0, 1, 2, 3}: 18 numbers.
Favourable = 36 out of 96, so the probability is 3696 = 38.

Answer: (d) 516

Method. A number is divisible by 4 when its last two digits form a multiple of 4

Possible endings with distinct digits from 0–⁠4: 04, 20, 40, 12, 24, 32.
Endings containing 0 (04, 20, 40): the other 3 digits fill the first two places in 3 × 2 = 6 ways, giving 18.
Endings without 0 (12, 24, 32): the first digit is from the 2 remaining non-zero digits and the second from the 2 left, so 2 × 2 = 4 ways each, giving 12.
Favourable = 30 out of 96, so the probability is 3096 = 516.

Answer: (c) 14

Method. Divisible by 6 means even and divisible by 3

Divisible by 3 needs digit set {0, 2, 3, 4} or {0, 1, 2, 3} (previous item).
{0, 2, 3, 4}: ending 0 gives 3! = 6; ending 2 gives 2 × 2 = 4 (first digit 3 or 4); ending 4 gives 2 × 2 = 4. Total 14.
{0, 1, 2, 3}: ending 0 gives 6; ending 2 gives 2 × 2 = 4 (first digit 1 or 3). Total 10.
Favourable = 24 out of 96, so the probability is 14.

Answer: (b) 0.25

Method. Numbers without 0 use exactly the digits 1, 2, 3, 4

Such numbers = 4! = 24, out of 96.
Required probability = 2496 = 0.25.

Answer: (c) 5648

Method. Count positive integer solutions of a + b + c + d = 6

Each die shows at least 1, and a total of 6 keeps each face at most 3, so every positive solution is a valid outcome.
Number of solutions = ^6−⁠1C₃ = ⁵C₃ = 10.
Total outcomes = 6⁴ = 1296.
Required probability = 101296 = 5648.

Answer: (d) 51296

Method. List the totals 23 and 24 (maximum 24)

Total 24: (6, 6, 6, 6) – 1 way.
Total 23: three 6's and one 5, the 5 on any of 4 dice – 4 ways.
Required probability = 51296.

Answer: (a) 221

Method. Combinations: choose 2 red out of 15 balls

P(both red) = ⁵C₂¹⁵C₂ = 10105 = 221.

No correct option. This question is defective as printed in UPSC's paper.

Defective as printed: P(passes both) = 11/⁠15 exceeds P(passes Physics) = 2/⁠3 = 10/⁠15, and no event can be more probable than an event that contains it. No option is a valid answer. (Adding blindly, P(at least one) = 23 + P(English) − 1115 gives P(English) = 1315, option (b), which other keys print; it rests on data that contradict themselves.)

Answer: (d) 12

Method. Count the sequences of 4 tosses with no HH, by the number of heads

0 heads: TTTT – 1 way.
1 head: 4 ways.
2 heads, not adjacent: HTHT, HTTH, THTH – 3 ways.
3 or more heads in 4 tosses always put two heads together – 0 ways.
Favourable = 8 out of 2⁴ = 16, so the probability is 816 = 12.

Answer: (d) 16

Method. Classify faces as prime {2, 3, 5}, composite {4, 6}, neither {1}; arrange the three types

For one die: P(prime) = 36, P(composite) = 26, P(neither) = 16.
The three types can fall on the three dice in 3! = 6 orders.
Required probability = 6 × 36 × 26 × 16 = 36216 = 16.

Answer: (d) 2549

Method. An odd sum needs one odd and one even number

50 consecutive natural numbers contain 25 odd and 25 even numbers.
Favourable pairs = 25 × 25 = 625; total pairs = ⁵⁰C₂ = 1225.
Required probability = 6251225 = 2549.

Answer: (b) 512

Method. Count the outcomes with sum 8 to 12

Number of outcomes with sum 8, 9, 10, 11, 12 = 5 + 4 + 3 + 2 + 1 = 15.
Required probability = 1536 = 512.

Answer: (a) 1033

Method. Combinations: favourable committees over all committees

Total committees = ¹¹C₆ = 462.
Exactly 3 gentlemen means 3 gentlemen and 3 ladies: ⁷C₃ × ⁴C₃ = 35 × 4 = 140.
Required probability = 140462 = 1033.

Answer: (d) 5366

Method. Complement: at most 1 lady

No lady: ⁷C₆ = 7. One lady: ⁴C₁ × ⁷C₅ = 4 × 21 = 84.
P(at most 1 lady) = 91462 = 1366.
P(at least 2 ladies) = 1 − 1366 = 5366.

Answer: (c) 115

Method. Count consecutive triples against all choices of 3 numbers

Total ways to choose 3 numbers from 10 = ¹⁰C₃ = 120.
Consecutive triples are {1, 2, 3}, {2, 3, 4}, …, {8, 9, 10}: the smallest number can be 1 to 8, so there are 8.
Required probability = 8120 = 115.

Answer: (c) 56

Method. Favourable faces over total faces

Faces showing 4 or 5 = 2 + 3 = 5 out of 6 faces.
P(4 or 5) = 56.

Answer: (a) 16

Method. Draws with replacement are independent; add over the three colours

Total balls = 12, so P(black) = 212 = 16, P(yellow) = 412 = 13, P(white) = 612 = 12.
P(all same colour) = (16)³ + (13)³ + (12)³ = 1216 + 8216 + 27216
= 36216 = 16.

Answer: (d) 135

Method. Arrangements with repeated letters; alternate pattern C V C V C V C

ZOOLOGY has 7 letters: vowels O, O, O (3) and consonants Z, L, G, Y (4).
Total arrangements = 7!3! = 840.
For alternation with 4 consonants and 3 vowels the pattern must be C V C V C V C.
Consonants fill their 4 places in 4! = 24 ways; the three O's fill theirs in 1 way.
Required probability = 24840 = 135.

Answer: (b) 4750

Method. Find the least x satisfying the inequality, then count

x² + x increases with x. For x = 6: 36 + 6 = 42 < 50; for x = 7: 49 + 7 = 56 > 50.
So x² + x > 50 exactly for x = 7, 8, …, 100, i.e. 94 numbers.
Required probability = 94100 = 4750.

Answer: (d) 1932

Method. List the sequences that end at the first HH for 2, 3, 4 and 5 tosses

The game ends at toss n when the last two tosses are HH and no earlier HH occurred.
n = 2: HH, probability 14.
n = 3: THH, probability 18.
n = 4: TTHH, HTHH, probability 216.
n = 5: TTTHH, THTHH, HTTHH, probability 332.
P(tosses < 6) = 832 + 432 + 432 + 332 = 1932.

Answer: (c) 49

Method. Both black or both white, counted with combinations

Total = ⁹C₂ = 36. Both black ⁵C₂ = 10; both white ⁴C₂ = 6.
P = 10 + 636 = 49.

Answer: (b) 14

Method. Arrangements with repeated letters; treat TT as one unit

TIRUPATI: 8 letters with T twice and I twice. Total = 8!2! 2!.
Ts together: 7 units (TT, I, I, R, U, P, A), arrangements = 7!2!.
P = 7!/⁠2!8!/⁠(2! 2!) = 28 = 14.

Answer: (c) 14

Method. Units digit of 77ⁿ is that of 7ⁿ, which repeats with period 4

Units digits of 7ⁿ for n = 1, 2, 3, 4, …: 7, 9, 3, 1, 7, 9, 3, 1, …
Exactly one value in every 4 consecutive n(n a multiple of 4) gives units digit 1.
P = 14.

Answer: (c) 165

Method. List triples {a, b, ab} with 2 ≤ a < b and ab ≤ 15

Total = ¹⁵C₃ = 455. a = 1 is useless (ab = b would repeat a number).
a = 2: b = 3, 4, 5, 6, 7 (products 6, 8, 10, 12, 14): 5 triples. a = 3: b = 4, 5 (12, 15): 2 triples. a ≥ 4 gives ab > 15.
Favourable = 7; P = 7455 = 165.

Answer: (b) 5

Method. p − q is the probability of exactly one tail

p − q = P(at least one tail) − P(at least two tails) = P(exactly one tail) = n2ⁿ.
n2ⁿ = 532 = 52⁵ ⟹ n = 5.

Answer: (a) 5732

Method. Complements with n = 5

p = 1 − 132 = 3132; q = 1 − P(0 tails) − P(1 tail) = 1 − 132 − 532 = 2632.
p + q = 5732.

Answer: (a) 17108

Method. Count outcomes for sums 4, 9, 16 (the squares between 3 and 18)

Sum 4: (1,1,2) arranged, 3 ways.
Sum 9: (1,2,6) 6, (1,3,5) 6, (1,4,4) 3, (2,2,5) 3, (2,3,4) 6, (3,3,3) 1: 25 ways.
Sum 16: (4,6,6) 3, (5,5,6) 3: 6 ways.
P = 3 + 25 + 6216 = 34216 = 17108.

Answer: (d) 136

Method. Unit digit 1 needs unit digits pairing as 1×1, 3×7 or 9×9

Pairs of distinct numbers: ⁹C₂ = 36.
A product of single digits ends in 1 only for 1 × 1, 3 × 7, 9 × 9; with distinct numbers only {3, 7} (product 21) works.
P = 136.

Answer: (b) 718

Method. List ordered pairs with |difference| 2 or 3

Difference 2: (1,3), (2,4), (3,5), (4,6) and reverses: 8.
Difference 3: (1,4), (2,5), (3,6) and reverses: 6.
P = 1436 = 718.

Answer: (c) 132

Method. The sixth toss can differ from all of the first five only if those five agree

Favourable sequences: H H H H H T and T T T T T H: 2 out of 2⁶ = 64.
P = 264 = 132.

Answer: (b) 2n

Method. Treat P and Q as one unit in a row of n seats

Total arrangements = n!.
P, Q together: (n − 1)! arrangements of the unit with the others, × 2 for the order within the unit.
P = 2(n − 1)!n! = 2n.

Shortcut. Seats in a row have n − 1 adjacent pairs out of ⁿC₂: n − 1n(n −⁠ 1)/⁠2 = 2n.

Answer: (a) 15

Method. Treat the three boys as one block

Total arrangements = 6! = 720.
Boys as a block: 4! arrangements of the block with 3 girls, × 3! inside the block = 144.
P = 144720 = 15.

Answer: (b) 110

Method. Alternate pattern BGBGBG or GBGBGB

Two patterns, each with 3! × 3! = 36 arrangements: 72 favourable.
P = 72720 = 110.

Answer: (d) 15

Method. Gap method: girls in the gaps between boys

Arrange the boys: 3! = 6. They create 4 gaps; choose 3 for the girls and arrange them: ⁴C₃ × 3! = 24.
Favourable = 6 × 24 = 144; P = 144720 = 15.

Answer: (a) 115

Method. Fix P and Q at the ends, arrange the rest

P and Q at the two ends in 2 ways; the other 4 fill the middle in 4! = 24 ways: 48 favourable.
P = 48720 = 115.

Answer: (d) 13

Method. Treat Q and U as one unit

Q, U together: 5! arrangements of the unit with the other 4, × 2 inside = 240.
P = 240720 = 13.

Answer: (b) 811024

Method. Probability of one fixed sequence of independent tosses

P(T) = 34, P(H) = 14; the sequence T T T T H has probability
(34)⁴ × 14 = 81256 × 14 = 811024.

Answer: (c) 127

Method. Each die independently shows 3 or 6

Multiples of 3 on a die: 3, 6, so P = 26 = 13 per die.
P(all three) = (13)³ = 127.

Answer: (c) 37

Method. 31 days = 4 weeks + 3 extra consecutive days

December has 31 days = 4 full weeks + 3 extra days, so there are always at least 4 Sundays.
The 3 extra days are 3 consecutive weekdays; of the 7 equally likely starting days, 3 make one of them a Sunday.
P = 37.

Answer: (b) 1725

Method. Count composites as 50 minus primes minus 1

Primes up to 50: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, which is 15 primes.
1 is neither prime nor composite, so composites = 50 − 15 − 1 = 34.
P = 3450 = 1725.

Answer: (b) 17

Method. C(n, 7) mod 7 equals the 7s digit of n in base 7 (Lucas' theorem); take n uniformly over long runs

Write n = 7k + r, 0 ≤ r ≤ 6. Among n, n − 1, …, n − 6 exactly one is a multiple of 7, namely 7k, so
7! × C(n, 7) = n(n −⁠ 1)…(n − 6) = 7k × Q, where Q is the product of the other six factors; so 6! × C(n, 7) = kQ.
The six factors in Q leave remainders 1, 2, …, 6 mod 7 in some order, so Q ≡ 6! (mod 7); as 6! is prime to 7, C(n, 7) ≡ k (mod 7).
So C(n, 7) is a multiple of 7 exactly when k = ⌊ n/⁠7 ⌋ is a multiple of 7: in every block of 49 consecutive n, 7 values qualify.
P = 749 = 17.

The question does not say how n is chosen from the infinitely many integers above 7; the probability is taken as the long-run proportion. C(n, 7) is a multiple of 7 exactly when the remainder of n on division by 49 is less than 7, which is 7 of every 49 consecutive values, giving 1/⁠7.

Answer: (b) 29

Method. Classify 1–⁠10 by remainder mod 4 and pair remainders summing to 0 mod 4

Two different numbers: ¹⁰C₂ = 45 pairs.
Remainder 0: {4, 8}; 1: {1, 5, 9}; 2: {2, 6, 10}; 3: {3, 7}.
x + y divisible by 4 needs remainders (0, 0), (2, 2) or (1, 3): ²C₂ + ³C₂ + 3 × 2 = 1 + 3 + 6 = 10.
P = 1045 = 29.

Answer: (c) (n − 1)n

Method. AM–⁠GM: x + 1x ≥ 2 with equality only at x = 1

For x > 0, x + 1x − 2 = (x − 1)²x ≥ 0, with equality only at x = 1.
So the inequality is strict for every natural number except 1: n − 1 favourable values out of n.
P = n − 1n.

Answer: (b) 118

Method. Count ordered outcomes (die 1, die 2, die 3) forming an AP with non-zero common difference

Total outcomes = 6³ = 216.
Sets of three different numbers from 1–⁠6 in AP: common difference 1 gives {1,2,3}, {2,3,4}, {3,4,5}, {4,5,6}; common difference 2 gives {1,3,5}, {2,4,6}: 6 sets.
Read in the order of the dice, each set forms an AP in two orders (increasing or decreasing): 6 × 2 = 12 outcomes.
P = 12216 = 118.

'In AP′ is read as the numbers on the first, second and third die being in arithmetic progression in that order. If the three numbers merely had to be rearrangeable into an AP the count would be 36 and the probability 1/⁠6, which is not among the options.

Answer: (a) 3132

Method. Complement of 'all five tails'

At most four tails fails only when all five are tails: P = (12)⁵ = 132.
Required probability = 1 − 132 = 3132.

Answer: (d) 554

Method. Count outcomes with totals 15, 16, 17, 18

Total 18: (6,6,6), 1 way. Total 17: (5,6,6) arranged, 3 ways.
Total 16: (4,6,6) 3 ways + (5,5,6) 3 ways = 6. Total 15: (3,6,6) 3 + (4,5,6) 6 + (5,5,5) 1 = 10.
Favourable = 1 + 3 + 6 + 10 = 20.
P = 20216 = 554.

Answer: (a) 9 : 4

Method. Addition rule for spade or ace, then odds against = unfavourable: favourable

Spades 13, aces 4, ace of spades counted twice: favourable = 13 + 4 − 1 = 16.
Unfavourable = 52 − 16 = 36.
Odds against = 36 : 16 = 9 : 4.

Answer: (b) 15

Method. A product ends in 0 only if it contains the factors 2 and 5

Two different digits are chosen: ⁵C₂ = 10 pairs.
Product ends in 0 only if 5 and an even digit are chosen: {2, 5}, {4, 5} (products 10, 20).
P = 210 = 15.

Answer: (b) 25

Method. Read the cell from the two-way table

T̅ is 'less than 3 years experience'. Graduates with less than 3 years experience = 36.
P(G ∩ ⁠T̅⁠) = 3690 = 25.

Answer: (c) 47

Method. Conditional probability restricts to the row of T̅

Applicants with less than 3 years experience = 36 + 27 = 63; of these, 36 are graduates.
P(G | ⁠T̅⁠) = 3663 = 47.

Answer: (d) 34

Method. Conditional probability restricts to the column of G̅

Non-graduates = 9 + 27 = 36; of these, 27 have less than 3 years experience.
P(⁠T̅⁠ | ⁠G̅⁠) = 2736 = 34.

Answer: (a) 0

Method. Divisibility by 3 depends only on the digit sum

Every such number uses all three digits 1, 3, 7, so its digit sum is 1 + 3 + 7 = 11.
11 is not divisible by 3, so no number formed is divisible by 3; the probability is 0.

Answer: (a) 0

Method. Discriminant test for each admissible n

n ∈ {1, 2, 3}. Real roots need discriminant 1 − 4n ≥ 0, i.e. n ≤ 14.
No admissible n satisfies this, so the probability is 0.

Answer: (a) 715

Method. Gap method: place black balls in the gaps between white balls

Only the positions of the 3 black balls matter: total ways = ¹⁰C₃ = 120.
The 7 white balls create 8 gaps (including the two ends); choosing 3 different gaps keeps the black balls apart: ⁸C₃ = 56.
P = 56120 = 715.

Answer: (c) 12

Method. Reduced sample space of ordered triples with distinct faces; complement of 'no 6'.

Distinct faces: 6 × 5 × 4 = 120 equally likely ordered outcomes.
No 6: 5 × 4 × 3 = 60.
P(a 6 appears) = 1 − 60120 = 12.

Shortcut. The three distinct faces are a random 3-subset of 6 faces, so P(6 is among them) = 36 = 12.

Answer: (b) 17

Method. A leap-year February has 4 full weeks plus one extra day.

29 days = 4 weeks + 1 day, so every weekday occurs 4 times and one weekday occurs a 5th time.
Five Sundays exactly when that extra day is a Sunday; taking the seven weekdays as equally likely for 1 February, as the exam does,
P = 17.

Answer: (a) 191

Method. Hypergeometric: all 4 from the 6 counterfeits.

Total ways = ¹⁵C₄ = 1365.
All counterfeit = ⁶C₄ = 15.
P = 151365 = 191.

Answer: (d) 12

Method. Classical probability with ⁿC_r.

Total committees = ⁴C₃ = 4.
2 boys and 1 girl: ²C₂ × ²C₁ = 2.
P = 24 = 12.

Answer: (c) 215

Method. Classical probability: both tickets from the 4 primes 2, 3, 5, 7.

Total ways = ¹⁰C₂ = 45.
Both prime: ⁴C₂ = 6.
P = 645 = 215.

Answer: (d) 0.98

Method. Complement of 'three tails' with P(T) = 14.

P(H) = 3P(T) and P(H) + P(T) = 1, so P(T) = 14.
P(3 tails) = (14)³ = 164.
P(at most two tails) = 1 − 164 = 6364 ≈ 0.98.

Answer: (a) 0.009

Method. Multiplication rule without replacement.

P = 520 × 419 × 318 = 606840 = 1114 ≈ 0.0088 ≈ 0.009.

Answer: (d) {H, TH, TTH, TTTH, …}

Method. List outcomes: the experiment stops at the first head.

The first head may come on toss 1, 2, 3, …: H, TH, TTH, TTTH, …
Sample space = {H, TH, TTH, TTTH, …}.

Answer: (a) 512

Method. Count sums that are prime among 36 outcomes.

Prime sums: 2 (1 way), 3 (2), 5 (4), 7 (6), 11 (2).
Favourable = 1 + 2 + 4 + 6 + 2 = 15.
P = 1536 = 512.

Answer: (c) 112

Method. Hypergeometric: all three chosen from the 5 failing trucks.

Total ways = ¹⁰C₃ = 120.
None meets standards: all 3 from the 5 that fail, ⁵C₃ = 10.
P = 10120 = 112.

Answer: (d) 113

Method. With replacement, every draw is a fresh draw from the full deck.

Each card is replaced, so the 14th draw is from the full deck of 52 with 4 aces.
P = 452 = 113.

Answer: (c) 512

Method. Symmetry between 'second higher' and 'first higher' after removing ties.

P(both equal) = 636 = 16.
The remaining 56 splits equally between 'second higher' and 'first higher'.
P(second higher) = 12 × 56 = 512.

Answer: (c) 14

Method. Independent events: multiply.

P(head) = 12; P(2, 4 or 6) = 36 = 12.
P(both) = 12 × 12 = 14.

Answer: (c) 518

Method. Independent dice: add P(both show colour k) over the four colours.

Face probabilities: red 26, black 26, yellow 16, white 16.
P(same) = (26)² + (26)² + (16)² + (16)² = 4 + 4 + 1 + 136 = 1036 = 518.

Answer: (b) 4

Method. Set ³C₂/⁠ⁿC₂ equal to 12 and solve for n.

³C₂ⁿC₂ = 3n(n −⁠ 1)/⁠2 = 6n(n −⁠ 1) = 12.
n(n −⁠ 1) = 12, so n = 4.

Answer: (a) 117

Method. Fix the first card; the second must match its value.

Whatever the first card, 3 of the remaining 51 cards have the same value.
P = 351 = 117.
(Check: 13 × ⁴C₂⁵²C₂ = 781326 = 117.)

Answer: (a) 14

Method. Geometric probability: ratio of areas.

For a circle of radius R, a point at distance r from the centre is closer to the centre when r < R − r, i.e. r < R2.
P = π(R/⁠2)²πR² = 14.

Answer: (a) 1332

Method. Independent selections; add the three cases by which group gives the girl.

P(girl) from the groups: 34, 12, 14; P(boy): 14, 12, 34.
Girl from group 1 only: 34 × 12 × 34 = 932.
Girl from group 2 only: 14 × 12 × 34 = 332.
Girl from group 3 only: 14 × 12 × 14 = 132.
Total = 1332.

Answer: (c) 30121

Method. Count ordered pairs (x, y) from 11 × 11 = 121 with |x − y| ≥ 6.

For x − y = d (d = 6, 7, 8, 9, 10) the number of pairs is 11 − d: 5 + 4 + 3 + 2 + 1 = 15.
The case y − x = d gives another 15.
Favourable = 30, so P = 30121.

Answer: (b) 16

Method. Count favourable outcomes among 36.

Sum 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1), 6 outcomes.
P = 636 = 16.

Answer: (a) 110

Method. Total probability over the first draw (or symmetry).

P(first is 9) = 110, then the second cannot be 9.
P(first is not 9) = 910, then P(second is 9) = 19.
P(second is 9) = 110 × 0 + 910 × 19 = 110.

Shortcut. By symmetry every chit is equally likely to be in the second position: 110.

Answer: (a) 514

Method. Classical probability with ⁿC_r.

Total selections = ⁹C₃ = 84.
Exactly 2 men: ⁴C₂ × ⁵C₁ = 6 × 5 = 30.
P = 3084 = 514.

Answer: (b) 23

Method. Classical probability: count committees with ⁿC_r.

Total committees = ⁴C₂ = 6.
Exactly one woman: ²C₁ × ²C₁ = 4.
Probability = 46 = 23.

Answer: (c) 49

Method. Loaded die: find face probabilities from total probability 1.

Let each odd face have probability k; each even face then has 2k.
3k + 3(2k) = 1, so k = 19.
Primes on a die: 2, 3, 5.
P(prime) = 29 + 19 + 19 = 49.

Answer: (d) Neither 1 nor 2

Method. Count in the sample space {0, 1, 2, …, 50} of 51 numbers.

Non-negative integers up to 50: 0, 1, …, 50, i.e. 51 numbers.
Multiples of 3: 0, 3, …, 48, i.e. 17 numbers, so P(X) = 1751 = 13 ≠ 825.
Odd numbers: 1, 3, …, 49, i.e. 25 numbers, so P(Y) = 2551 ≠ 12.
Neither statement is correct.

The sample space of non-negative integers up to 50 has 51 points, 0 included, so P(X) = 17/⁠51 = 1/⁠3 and P(Y) = 25/⁠51; the printed values 8/⁠25 and 1/⁠2 hold only for the sample space 1 to 50, under which both statements would be correct and the answer (c). The answer follows the text as set.

Answer: (c) 0.3

Method. Triangle inequality: the two shorter sticks must together exceed the longest.

Number of selections = ⁵C₃ = 10.
Any triple with the 1-ft stick fails: lengths are odd and distinct, so the other two differ by at least 2 > 1.
Without 1: (3, 5, 7): 8 > 7 yes; (3, 5, 9): 8 < 9 no; (3, 7, 9): 10 > 9 yes; (5, 7, 9): 12 > 9 yes.
Favourable = 3, probability = 310 = 0.3.

Answer: (a) 16

Method. Classical probability: count committees with ⁿC_r.

Total committees of 2 from 4 persons = ⁴C₂ = 6.
Committees with only women = ²C₂ = 1.
Required probability = 16.

Answer: (b) 14

Method. Geometric probability: ratio of areas.

Let the radius be R and the point be at distance r from the centre.
Distance to the boundary = R − r. Closer to the centre means r < R − r, i.e. r < R2.
Favourable region is the concentric circle of radius R2.
Probability = π(R/⁠2)²πR² = 14.

Answer: (a) 113

Method. Classical probability.

A deck has 4 aces among 52 cards.
P(ace) = 452 = 113.

No correct option. This question is defective as printed in UPSC's paper.

Defective as set: UPSC's paper prints P(A ∩ B) = 1/⁠2 and P(⁠A̅⁠ ∩ ⁠B̅⁠) = 1/⁠2. Then P(A ∪ B) = 1 − 1/⁠2 = 1/⁠2, and the addition theorem gives 1/⁠2 = k/⁠2 + k − 1/⁠2, so k = 2/⁠3 and P(A) = 1/⁠3 — less than P(A ∩ B) = 1/⁠2, which no two events can satisfy. No option is a valid answer.

Answer: (c) 13/⁠18

Method. Complement of the event 'sum is 5 or 7'

Sum 5: 4 outcomes; sum 7: 6 outcomes; together 10 of 36.
P(neither 5 nor 7) = 1 − 1036 = 2636 = 1318.

Answer: (d) Neither I nor II

Method. Bonferroni and Boole inequalities, and counterexamples

The true bounds are P(A ∩ B ∩ C) ≥ P(A) + P(B) + P(C) − 2 and P(A ∪ B ∪ C) ≤ P(A) + P(B) + P(C).
Both statements reverse the inequality. Take A = B = C with P(A) = 12.
I: 12 ≤ 32 − 2 = −12 is false.
II: 12 ≥ 32 is false.
Neither statement holds in general.

Answer: (a) 59

Method. Set the common ratio equal to k and use the sum = 1

P(A) = 2k, P(B) = 3k, P(C) = 5k, P(D) = 8k.
Exhaustive and mutually exclusive: 18k = 1, so k = 118.
P(A) + P(B) + P(C) = 10k = 1018 = 59.

Shortcut. P(A) + P(B) + P(C) = 1 − P(D) = 1 − 818.

Answer: (b) 1360

Method. Substitute P(A) = 2k, P(B) = 3k, P(C) = 5k, P(D) = 8k

2P(A) + 3P(B) = 4k + 9k = 13k.
4P(C) + 5P(D) = 20k + 40k = 60k.
Ratio = 1360.

Answer: (b) 15^1/⁠4

Method. Geometric mean of 2k, 3k, 5k, 8k with k = 118

G = (2k × 3k × 5k × 8k)^1/⁠4 = k(240)^1/⁠4 = (240)^1/⁠418.
9G = (240)^1/⁠42 = (24016)^1/⁠4 = 15^1/⁠4.

Answer: (a) I only

Method. De Morgan: each bracket pair is an event and its complement

A̅ ∩ B̅ ∩ C̅ is the complement of A ∪ B ∪ C, so the expression in I is ∅: an impossible event. I is correct.
A̅ ∪ B̅ ∪ C̅ is the complement of A ∩ B ∩ C, so the expression in II is also ∅: an impossible event. II is wrong.
So I only.

Answer: (b) 0.76

Method. Addition theorem P(L ∪ D) = P(L) + P(D) − P(L ∩ D)

P(laptop or desktop) = 0.68 + 0.56 − 0.48 = 0.76.

Answer: (b) 1825

Method. Inclusion–⁠exclusion on multiples of 5 and 9

Among 1 to 50: multiples of 5 = 10, multiples of 9 = 5, multiples of 45 = 1.
Divisible by 5 or 9 = 10 + 5 − 1 = 14.
Neither = 50 − 14 = 36, so the probability is 3650 = 1825.

Answer: (b) 512

Method. De Morgan: P(A^c ∩ B^c) = 1 − P(A ∪ B)

P(A ∪ B) = 13 + 12 − 14 = 4 + 6 − 312 = 712.
P(A^c ∩ B^c) = 1 − 712 = 512.

Answer: (c) 929

Method. Mutually exclusive and exhaustive events have probabilities summing to 1

P(B) = 43P(A) and P(C) = 23P(B) = 89P(A).
P(A) + P(B) + P(C) = 1 gives P(A)(1 + 43 + 89) = 1, i.e. 299P(A) = 1.
P(A) = 929.

Shortcut. Put P(A) = 9t, P(B) = 12t, P(C) = 8t; then 29t = 1.

Answer: (d) 2936

Method. Complement of the event 'sum is 9 or 10'

Sum 9: (3, 6), (4, 5), (5, 4), (6, 3) – 4 outcomes. Sum 10: (4, 6), (5, 5), (6, 4) – 3 outcomes.
P(sum is 9 or 10) = 736.
P(neither 9 nor 10) = 1 − 736 = 2936.

Answer: (d) 0.875

Method. P(A) + P(B) = P(A ∪ B) + P(A ∩ B)

P(A) + P(B) = P(A ∪ B) + P(A ∩ B) ≥ 0.75 + 0.125 = 0.875.
Both lower bounds can hold at once, so the minimum is 0.875.

Answer: (c) 1.375

Method. P(A) + P(B) = P(A ∪ B) + P(A ∩ B) with P(A ∪ B) ≤ 1

P(A) + P(B) = P(A ∪ B) + P(A ∩ B) ≤ 1 + 0.375 = 1.375.
Attained, e.g. with P(A ∪ B) = 1, P(A ∩ B) = 0.375, P(A) = P(B) = 0.6875; so the maximum is 1.375.

Answer: (b) 0.2

Method. Inclusion–⁠exclusion for three events with 0.85 ≤ P(A ∪ B ∪ C) ≤ 1

P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 0.6 + 0.4 − 0.8 = 0.2.
P(A ∪ B ∪ C) = 0.6 + 0.4 + 0.5 − 0.2 − 0.3 − P(B ∩ C) + 0.2 = 1.2 − P(B ∩ C).
0.85 ≤ 1.2 − P(B ∩ C) ≤ 1 ⟹ 0.2 ≤ P(B ∩ C) ≤ 0.35.
Minimum P(B ∩ C) = 0.2 (also forced by P(B ∩ C) ≥ P(A ∩ B ∩ C) = 0.2).

Answer: (c) 0.35

Method. Inclusion–⁠exclusion for three events with P(A ∪ B ∪ C) ≥ 0.85

P(A ∩ B) = 0.6 + 0.4 − 0.8 = 0.2.
P(A ∪ B ∪ C) = 0.6 + 0.4 + 0.5 − 0.2 − 0.3 − P(B ∩ C) + 0.2 = 1.2 − P(B ∩ C).
1.2 − P(B ∩ C) ≥ 0.85 ⟹ P(B ∩ C) ≤ 0.35; the maximum is 0.35.

Answer: (d) 30

Method. Set the common value to k; exhaustive and exclusive events sum to 1

Let 2P(A) = 3P(B) = 4P(C) = 5P(D) = k.
k(12 + 13 + 14 + 15) = 1 ⟹ k × 7760 = 1 ⟹ k = 6077.
P(A) = k2 = 3077, so 77P(A) = 30.

Answer: (b) 4750

Method. Solve the quadratic inequality n² - 50n + 50 < 0

n + 50n < 50 ⟺ n² - 50n + 50 < 0 (n > 0).
Roots 25 ± ⁠√575⁠ ≈ 25 ± 23.98, i.e. about 1.02 and 48.98.
So n = 2, 3, …, 48: 47 values (check: n = 1 gives 51, n = 49 gives 50149, n = 50 gives 51).
P = 4750.

Answer: (b) 0.7

Method. The three regions together are the complement of A ∩ B

A′ ∩ B′, A′ ∩ B and A ∩ B′ together with A ∩ B make up the whole sample space, and they do not overlap.
So the required sum = 1 − P(A ∩ B) = 1 − 0.3 = 0.7.

Answer: (c) 1 and 3 only

Method. Addition rule P(E ∪ F) = P(E) + P(F) − P(E ∩ F) with P(E ∪ F) ≤ 1

From P(E ∪ F) = P(E) + P(F) − P(E ∩ F) ≤ 1: P(E ∩ F) ≥ P(E) + P(F) − 1, so 1 is correct.
2 has + P(E ∩ F) where the rule has − P(E ∩ F); it is wrong.
Since P(E ∩ F) ≥ 0, P(E ∪ F) ≤ P(E) + P(F), so 3 is correct.

Answer: (c) 34

Method. Complement: P(E ∪ F) = 1 − P(no head)

P(E ∪ F) = 1 − P(⁠E̅⁠ ∩ ⁠F̅⁠) = 1 − 12 × 12 = 34.

Answer: (c) Both 1 and 2

Method. Extreme values of union and intersection occur when one event contains the other

P(A ∪ B) ≥ max(P(A), P(B)) = 34, with equality when B ⊂ A; so the minimum of P(A ∪ B) is 34: (1) is correct.
P(A ∩ B) ≤ min(P(A), P(B)) = 58, with equality when B ⊂ A; so the maximum of P(A ∩ B) is 58: (2) is correct.

Answer: (d) 0.3

Method. Addition rule, then complement.

P(A ∪ B) = 0.5 + 0.6 − 0.4 = 0.7.
P(⁠A̅ ∪ B̅⁠) = 1 − 0.7 = 0.3.

Answer: (b) P(A ∪ B)

Method. 'Either A or B′ is the union.

The event 'A or B occurs' is A ∪ B, so its probability is P(A ∪ B).
P(A) + P(B) equals this only when A and B are mutually exclusive.

Answer: (b) 413

Method. Mutually exclusive and exhaustive: probabilities add to 1.

P(B) = 1.5P(A), P(C) = 0.5 × 1.5P(A) = 0.75P(A).
P(A) + 1.5P(A) + 0.75P(A) = 1, so 3.25P(A) = 1.
P(A) = 13.25 = 413.

Answer: (c) Both 1 and 2

Method. Express P(A) + P(B) and P(A ∪ B) through p and q.

P(exactly one) = P(A) + P(B) − 2P(A ∩ B), so q = P(A) + P(B) − 2p and P(A) + P(B) = q + 2p.
(1) P(⁠A̅⁠) + P(⁠B̅⁠) = 2 − (q + 2p) = 2 − 2p − q: correct.
(2) P(A ∪ B) = P(A) + P(B) − p = p + q, so P(⁠A̅⁠ ∩ ⁠B̅⁠) = 1 − p − q: correct.
Both are correct.

Answer: (b) 1 and 3 only

Method. Venn-diagram identities for 'A but not B′, 'exactly one' and mutually exclusive union.

(1) If B ⊂ A, A ∩ B = B, so P(A ∩ ⁠B̅⁠) = P(A) − P(A ∩ B) = P(A) − P(B): correct.
(2) Exactly one occurs: P(A) + P(B) − 2P(A ∩ B); the printed P(A) + P(B) − P(A ∩ B) is P(A ∪ B): incorrect.
(3) Mutually exclusive, so P(A ∩ B) = 0 and P(A ∪ B) = P(A) + P(B): correct.
Correct: 1 and 3 only.

Answer: (a) 112

Method. Split B by C: B = (B ∩ C) ∪ (B ∩ C̅).

B ∩ C̅ = (A ∩ B ∩ C̅) ∪ (A̅ ∩ B ∩ C̅), disjoint, so P(B ∩ ⁠C̅⁠) = 13 + 13 = 23.
P(B ∩ C) = P(B) − P(B ∩ ⁠C̅⁠) = 34 − 23 = 112.

Answer: (c) Both 1 and 2

Method. Definition of mutually exclusive events and the addition rule.

(1) Mutually exclusive means A ∩ B = ∅, so the occurrence of one rules out the other: correct.
(2) For such events P(A ∩ B) = 0, so P(A ∪ B) = P(A) + P(B): correct.
Both statements are correct.

Answer: (c) 34

Method. Complement after counting sums 8 and 9 among 36 outcomes.

Sum 8: (2,6), (3,5), (4,4), (5,3), (6,2), 5 outcomes.
Sum 9: (3,6), (4,5), (5,4), (6,3), 4 outcomes.
P(sum is 8 or 9) = 936 = 14.
P(neither) = 1 − 14 = 34.

Answer: (d) 512

Method. De Morgan: A̅ ∩ B̅ is the complement of A ∪ B.

Mutually exclusive, so P(A ∪ B) = 13 + 14 = 712.
P(⁠A̅⁠ ∩ ⁠B̅⁠) = 1 − P(A ∪ B) = 1 − 712 = 512.

Answer: (b) 1/⁠2

Method. Conditional probability: the sample space is the 4 women

Given that a woman is selected, she is one of 4 women, each equally likely.
2 of them are senior, so the required probability = 24 = 12.

Answer: (a) 5/⁠7

Method. P(S | M) = P(M ∩ S)P(M)

P(S | M) = 0.50.7 = 57.

Answer: (b) 0.84

Method. Multiplication rule gives P(A ∩ B); independence gives P(A) and P(B)

P(A ∩ B ∩ D) = P(D | A ∩ B)P(A ∩ B), so P(A ∩ B) = 0.040.25 = 0.16.
A, B independent: P(A) × 4P(A) = 0.16, so P(A)² = 0.04, P(A) = 0.2 and P(B) = 0.8.
P(A ∪ B) = 0.2 + 0.8 − 0.16 = 0.84.

Answer: (b) 6/⁠7

Method. Set the common ratio equal to k and use the sum = 1

P(A) = 6k, P(B) = 3k, P(C) = 4k, P(D) = 2k, and 15k = 1, so k = 115.
P(A) + 2P(B)3P(C) + P(D) = 6k + 6k12k + 2k = 1214 = 67.

Answer: (b) 25⁠√3⁠

Method. Geometric mean of 6k, 3k, 4k, 2k with k = 115

G = (6k × 3k × 4k × 2k)^1/⁠4 = k(144)^1/⁠4 = k⁠√12⁠ = 2⁠√315.⁠
2⁠√315⁠ = 25⁠√3.⁠

Answer: (c) 16/⁠75

Method. Harmonic mean of 6k, 3k, 4k, 2k with k = 115

16k + 13k + 14k + 12k = 1k(2 + 4 + 3 + 612) = 1512k.
H = 415/⁠(12k) = 48k15 = 48225 = 1675.

Answer: (d) 23

Method. Addition theorem for three events with disjointness and independence

Let P(A) = x; then P(B) = 2x, P(C) = 4x.
A ∩ B = ∅, so P(A ∩ B) = P(A ∩ B ∩ C) = 0; P(A ∩ C) = 4x², P(B ∩ C) = 8x².
P(A ∪ B ∪ C) = x + 2x + 4x − 0 − 4x² − 8x² + 0 = 7x − 12x².
7x − 12x² = 5x gives 12x² = 2x, so x = 16 (x ≠ 0).
P(C) = 4x = 23.

Answer: (c) 0.216

Method. Multiplication rule for independent events

P(all three in debt) = (0.6)³ = 0.216.

Answer: (a) 310

Method. They contradict when exactly one of them speaks the truth

P(X true) = 45, P(Y true) = 56.
P(contradict) = 45 × 16 + 15 × 56 = 430 + 530 = 930 = 310.

Answer: (a) 136

Method. Independent tosses: multiply the colour probabilities in the stated order

P(black) = 36, P(white) = 26, P(red) = 16.
P(black, white, red in that order) = 36 × 26 × 16 = 6216 = 136.

Answer: (d) 0.667

Method. Multiplication rule, then the reverse conditional probability

P(A ∩ B) = P(A | B)P(B) = 0.5 × 0.4 = 0.2.
P(B | A) = P(A ∩ B)P(A) = 0.20.3 = 23 ≈ 0.667.

Answer: (d) 1112

Method. Complement: A̅ ∪ B is the complement of A ∩ B̅

P(A ∩ ⁠B̅⁠) = P(A) − P(A ∩ B) = 13 − 14 = 112.
P(⁠A̅⁠ ∪ B) = 1 − P(A ∩ ⁠B̅⁠) = 1 − 112 = 1112.