Question 1

A leap year has 366 days (52 weeks + 2 days). If a year is chosen uniformly at random among leap years, what is the probability that it contains 53 Mondays?

Accepted Answer

Introduction / Context: A leap year has 366 days, which is 52 full weeks plus 2 extra days. Exactly two weekdays will occur 53 times; the others occur 52 times. We want the chance that Monday is among the 53-times weekdays. Given Data / Assumptions: Leap year: 366 days = 52 weeks + 2 days. Start-of-year weekday is equally likely to be any of the 7 weekdays. Two consecutive weekdays occur 53 times each; all others occur 52 times. Concept / Approach: Let the first day of the year be D. Then the two 53-times weekdays are D and the weekday immediately following D. Monday is a 53-times weekday iff D is Monday or Sunday. Step-by-Step Solution: Favorable start days for “53 Mondays”: Sunday or Monday → 2 possibilities.Total possible start days = 7.Probability = 2/7. Verification / Alternative check: Compare with ordinary (365-day) years, where exactly one weekday is 53-times; then P(53 Mondays) = 1/7. The leap-year case doubles the favorable possibilities to 2/7. Why Other Options Are Wrong: 5/7 and 3/7 overstate; 1/7 is for ordinary years; “Data inadequate” is incorrect under the uniform start-day assumption. Common Pitfalls: Forgetting that in leap years, two adjacent weekdays get the extra occurrences. Final Answer: 2/7

Question 2

Two students A and B have independent failure probabilities in an exam: P(A fails) = 0.2 and P(B fails) = 0.3. What is the probability that at least one of A or B fails (i.e., A fails or B fails or both)?

Accepted Answer

Introduction / Context: This is a union-of-events problem with independence. We want P(A ∪ B) where A = “A fails” and B = “B fails.” Given Data / Assumptions: P(A) = 0.2, P(B) = 0.3. Independence: P(A ∩ B) = P(A) * P(B) = 0.06. Concept / Approach: Use inclusion–exclusion for two events: P(A ∪ B) = P(A) + P(B) − P(A ∩ B). Step-by-Step Solution: Compute the intersection: 0.2 × 0.3 = 0.06.Therefore P(A ∪ B) = 0.2 + 0.3 − 0.06 = 0.44. Verification / Alternative check: Complement method: P(neither fails) = (0.8)(0.7) = 0.56; hence P(at least one fails) = 1 − 0.56 = 0.44 (same result). Why Other Options Are Wrong: 0.38 is not obtained by any standard combination; 0.50 ignores overlap subtraction; 0.94 is near certainty and clearly too large; 0.56 is the complement (both pass), not the asked probability. Common Pitfalls: Adding probabilities without subtracting the overlap when events are not mutually exclusive. Final Answer: 0.44

Question 3

A fair coin is tossed 100 times. What is the probability that the number of tails obtained is an odd number?

Accepted Answer

Introduction / Context: This is identical in structure to the “odd heads” problem with a fair coin and an even number of tosses. By symmetry, the probability that the count of tails is odd equals 1/2. Given Data / Assumptions: n = 100 tosses (even). P(T) = 1/2, P(H) = 1/2; independence across tosses. Concept / Approach: Parity splitting for binomial with p = 1/2 and even n yields P(odd) = 1/2 and P(even) = 1/2 for the count of the tracked outcome. Step-by-Step Solution: Let X be the number of tails.Using the identities (p + q)^n and (p − q)^n with p = q = 1/2 and n even implies P(X odd) = 1/2. Verification / Alternative check: Direct pairing of outcomes by flipping the first toss (head↔tail) creates a bijection between sequences with even and odd tails. Why Other Options Are Wrong: 1/8 and 3/8 do not reflect parity symmetry; “None of these” is incorrect because 1/2 is valid; 1/4 is too small. Common Pitfalls: Forgetting the requirement that n is even and p = 1/2 for the parity split to hold exactly. Final Answer: 1/2

Question 4

An integer x is chosen uniformly at random from {1, 2, …, 100}. What is the probability that x satisfies the inequality x^2 − 13x ≤ 30?

Accepted Answer

Introduction / Context: We must count how many integers between 1 and 100 inclusive satisfy a quadratic inequality, then divide by 100 to obtain the probability. Given Data / Assumptions: Uniform choice over {1, …, 100}. Inequality: x^2 − 13x ≤ 30. Concept / Approach: Solve the inequality by finding the roots of the corresponding quadratic equation and using the sign pattern of a parabola opening upward. Step-by-Step Solution: Solve x^2 − 13x − 30 = 0.Discriminant Δ = 13^2 + 4*30 = 169 + 120 = 289; √Δ = 17.Roots: (13 ± 17)/2 → x = 15 and x = −2.Since the parabola opens upward, the solution set to x^2 − 13x − 30 ≤ 0 is −2 ≤ x ≤ 15.Within {1, …, 100}, valid x are 1 through 15 inclusive → 15 integers.Probability = 15 / 100 = 3/20. Verification / Alternative check: Quick test: x = 15 yields 225 − 195 = 30 (boundary satisfied); x = 16 yields 256 − 208 = 48 > 30 (violates), confirming the cutoff. Why Other Options Are Wrong: Fractions like 5/9, 7/9 are too large; 9/50 = 18% undercounts; 2/5 = 40% is far above the correct 15%. Common Pitfalls: Miscomputing the discriminant or forgetting that the inequality includes equality at the roots. Final Answer: 3/20

Question 5

For a student and three subjects (Mathematics M, Physics P, Chemistry C), suppose: P(M ∪ P ∪ C) = 0.75 (pass at least one), P(at least two) = 0.50, and P(exactly two) = 0.40. Which relation is correct among the options (interpreting pmc as P(M ∩ P ∩…

Accepted Answer

Introduction / Context: The statements refer to probabilities of passing subjects M, P, and C. The options mix expressions for sums and a term denoted “pmc.” Under the usual contest shorthand, “pmc” here is best read as P(M ∩ P ∩ C), the probability a student passes all three subjects, not the product p*m*c. We use standard set identities to find that value. Given Data / Assumptions: P(M ∪ P ∪ C) = 0.75. P(at least two) = 0.50. P(exactly two) = 0.40. Let S1 = P(M) + P(P) + P(C); S2 = sum of pairwise intersections; S3 = P(M ∩ P ∩ C). Concept / Approach: Recall: P(M ∪ P ∪ C) = S1 − S2 + S3. Also, P(exactly two) = S2 − 3S3, and P(at least two) = P(exactly two) + S3. Step-by-Step Solution: From P(at least two) = 0.50 and P(exactly two) = 0.40, we get S3 = 0.50 − 0.40 = 0.10.Thus pmc = P(M ∩ P ∩ C) = 0.10 = 1/10.Additionally, S2 = P(exactly two) + 3S3 = 0.40 + 0.30 = 0.70.Then 0.75 = S1 − 0.70 + 0.10 ⇒ S1 = 1.35 (not among the listed numeric options except by implication that pmc is 1/10). Verification / Alternative check: Check consistency: P(at least one) = 0.75 and P(at least two) = 0.50 implies a plausible triple-intersection S3 = 0.10. Why Other Options Are Wrong: The equalities for S1 (e.g., 19/20 or 17/20) contradict the deduced S1 = 27/20; pmc = 1/4 is too large and inconsistent with the given “exactly two.” Common Pitfalls: Misreading “pmc” as the product p*m*c; here it naturally denotes the three-way intersection probability. Final Answer: pmc = 1/10

Question 6

Integers m and n are chosen uniformly and independently from {1, 2, …, 100}. What is the probability that 7^m + 7^n is divisible by 5?

Accepted Answer

Introduction / Context: We need to evaluate 7^m + 7^n modulo 5. Working modulo 5 simplifies because powers of 7 reduce to powers of 2 (since 7 ≡ 2 mod 5) with a period of 4. Given Data / Assumptions: m and n are independently uniform on {1, …, 100}. 7 ≡ 2 (mod 5). Cycle of 2^k (mod 5) has period 4: 2, 4, 3, 1 for k ≡ 1, 2, 3, 0 (mod 4). Concept / Approach: Let f(r) be 2^r mod 5 with r in {1,2,3,0}. Then 7^m + 7^n ≡ f(m mod 4) + f(n mod 4) (mod 5). Count residue pairs that sum to 0 mod 5. Step-by-Step Solution: Residue values by exponent class: {2, 4, 3, 1}.Pairs summing to 0 mod 5 are (2,3), (3,2), (4,1), (1,4) → 4 favorable pairs.Since m mod 4 and n mod 4 are each uniform over 4 classes, total pairs = 16, favorable = 4.Probability = 4/16 = 1/4. Verification / Alternative check: Because 100 is a multiple of 4, residues are exactly balanced (25 each), justifying uniformity on residue classes. Why Other Options Are Wrong: 1/7 and 1/49 ignore the 4-cycle structure; 1/8 undercounts; 3/8 overcounts. Common Pitfalls: Forgetting that exponent classes 0..3 map to specific residues of 2^k (mod 5) and mixing the class labels. Final Answer: 1/4

Question 7

Two fair dice and two fair coins are tossed simultaneously. What is the probability that both coins show heads and the sum of the two dice is a prime number?

Accepted Answer

Introduction / Context: This is a product event: coin results are independent of dice results. We need both coins to be heads and the dice sum to be prime. Given Data / Assumptions: Two coins → P(HH) = 1/4. Two dice → 36 equally likely ordered outcomes. Prime sums possible: 2, 3, 5, 7, 11. Concept / Approach: Compute the dice probability first, then multiply by 1/4 for HH. Step-by-Step Solution: Count prime-sum outcomes: sum 2 → 1 way; sum 3 → 2 ways; sum 5 → 4 ways; sum 7 → 6 ways; sum 11 → 2 ways.Total favorable on dice = 1 + 2 + 4 + 6 + 2 = 15.P(sum is prime) = 15/36 = 5/12.P(HH and prime sum) = (1/4) * (5/12) = 5/48. Verification / Alternative check: Check symmetry: sums 2 and 12 each have 1 way; 3 and 11 have 2; 5 and 9 have 4; 7 has 6. Prime pattern count 15 is standard. Why Other Options Are Wrong: 5/72 and 13/144 reflect incorrect counts; 1/12 misses the coin factor; 7/144 is not supported by standard sum counts. Common Pitfalls: Forgetting that ordered dice pairs like (3,2) and (2,3) are distinct; or failing to multiply by the coin probability. Final Answer: 5/48

Question 8

Two reporters A and B narrate the same incident. A tells the truth in 60% of cases; B in 80% of cases. What is the probability they contradict each other (one says true while the other says false)?

Accepted Answer

Introduction / Context: They contradict when exactly one speaks truth and the other lies about the same fact. With independent truth-telling behavior, compute the two mutually exclusive ways this happens and add them. Given Data / Assumptions: P(A truthful) = 0.6 → P(A lies) = 0.4. P(B truthful) = 0.8 → P(B lies) = 0.2. Independence between A and B. Concept / Approach: P(contradict) = P(A true & B false) + P(A false & B true). Step-by-Step Solution: P(A true & B false) = 0.6 × 0.2 = 0.12.P(A false & B true) = 0.4 × 0.8 = 0.32.Total = 0.12 + 0.32 = 0.44 = 44%. Verification / Alternative check: P(agree) = P(both true) + P(both false) = (0.6×0.8) + (0.4×0.2) = 0.48 + 0.08 = 0.56, and 1 − 0.56 = 0.44 (contradict), consistent. Why Other Options Are Wrong: 70% is unrelated; 36% or 22% come from partial products; 40% cannot result from these parameters. Common Pitfalls: Multiplying 0.6 and 0.8 or adding them directly—neither operation models contradiction correctly. Final Answer: 44%

Question 9

A leap year has 366 days (52 weeks + 2 days). If a leap year is selected at random, what is the probability that it contains 53 Sundays?

Accepted Answer

Introduction / Context: In a leap year, two consecutive weekdays occur 53 times. We want the probability that Sunday is one of them. Given Data / Assumptions: 366 days = 52 weeks + 2 extra days. Start-of-year weekday is uniform over 7 days. Concept / Approach: If the year starts on Sunday, both Sunday and Monday occur 53 times. If it starts on Saturday, then Saturday and Sunday occur 53 times. Thus Sunday is 53-times if the start day is Saturday or Sunday. Step-by-Step Solution: Favorable start days = 2 (Saturday or Sunday).Total start days = 7.Probability = 2/7. Verification / Alternative check: Matches the standard leap-year weekday distribution logic; compare to ordinary years where only one weekday is 53-times (probability 1/7). Why Other Options Are Wrong: 7/366 and 26/183 are ratios unrelated to weekday-cycle logic; 1/7 holds for ordinary years, not leap years. Common Pitfalls: Assuming the extra days “belong” to fixed weekdays; they shift with the start day. Final Answer: 2/7

Question 10

From 4 children, 2 women, and 4 men (total 10 people), 4 persons are chosen uniformly at random. What is the probability that exactly 2 of the selected persons are children?

Accepted Answer

Introduction / Context: This is a hypergeometric selection: we draw a fixed-size subset without replacement from distinct groups and ask for exactly k from one group. Given Data / Assumptions: Total population N = 10 (4 children, 6 adults). Sample size n = 4. Event: exactly 2 children in the sample. Concept / Approach: Count ways to pick 2 from the 4 children and 2 from the 6 adults; divide by total ways to pick any 4 from 10. Step-by-Step Solution: Favorable = C(4, 2) * C(6, 2) = 6 * 15 = 90.Total = C(10, 4) = 210.Probability = 90/210 = 3/7 = 9/21. Verification / Alternative check: Reduce 90/210 by dividing numerator and denominator by 30 to obtain 3/7; the option list uses 21ths, so 9/21 matches. Why Other Options Are Wrong: Other fractions do not simplify to 3/7 given these counts. Common Pitfalls: Accidentally choosing 3 from one group and 1 from the other or forgetting to multiply the group combinations. Final Answer: 9/21

Question 11

A marksman hits a target with probability 3/4 on each shot, independently. He fires 5 shots. What is the probability that he hits the target at least 3 times?

Accepted Answer

Introduction / Context: We have a binomial model X ~ Binomial(n = 5, p = 3/4). The required probability is P(X ≥ 3) = P(3) + P(4) + P(5). Given Data / Assumptions: n = 5 independent trials. p = 3/4 (success = hit). Concept / Approach: Use the binomial probability mass formula: P(X = k) = C(n, k) * p^k * (1 − p)^(n − k). Sum for k = 3, 4, 5. Step-by-Step Solution: P(3) = C(5, 3) * (3/4)^3 * (1/4)^2 = 10 * 27/64 * 1/16 = 270/1024.P(4) = C(5, 4) * (3/4)^4 * (1/4) = 5 * 81/256 * 1/4 = 405/1024.P(5) = C(5, 5) * (3/4)^5 = 1 * 243/1024 = 243/1024.Sum = (270 + 405 + 243)/1024 = 918/1024 = 459/512. Verification / Alternative check: Complement: 1 − [P(0) + P(1) + P(2)] yields the same 459/512 after computation. Why Other Options Are Wrong: The other fractions do not equal the computed exact value 459/512. Common Pitfalls: Arithmetic slips in powers or forgetting coefficients C(5, k). Keeping everything over 1024 reduces mistakes. Final Answer: 459/512

Question 12

A fair die is rolled three times and the sum of the three outcomes is 15. Given this condition, what is the probability that the first roll was a 4?

Accepted Answer

Introduction / Context: This is a conditional probability on discrete uniform outcomes. We count the number of triples (x1, x2, x3) with xi ∈ {1,…,6} summing to 15, and among them count those with x1 = 4. The ratio gives the conditional probability. Given Data / Assumptions: All ordered triples of die outcomes summing to 15 are equally likely. Die faces are 1 through 6. Concept / Approach: Use bounded-integer solution counting with inclusion–exclusion. Step-by-Step Solution: Let yi = xi − 1 so yi ∈ {0,…,5}. Then y1 + y2 + y3 = 12.Unrestricted nonnegative solutions: C(12 + 3 − 1, 3 − 1) = C(14, 2) = 91.Subtract cases where some yi ≥ 6: for any i, set zi = yi − 6 ≥ 0 → z-sum = 6 gives C(8, 2) = 28. There are 3 such i, subtract 3 × 28 = 84.Add back intersections where two yi ≥ 6: z-sum = 0 → C(2, 2) = 1, with C(3, 2) = 3 such intersections.Total valid triples = 91 − 84 + 3 = 10.Now impose x1 = 4 → x2 + x3 = 11. Pairs (x2, x3) in 1..6: (5,6) and (6,5) → 2 cases.Therefore conditional probability = 2 / 10 = 1/5. Verification / Alternative check: A quick list of all 10 triples confirms two begin with 4. Why Other Options Are Wrong: 2/5 would require 4 cases out of 10; 1/6 assumes naive independence; “None of these” is false. Common Pitfalls: Forgetting that ordered triples are counted, or missing inclusion–exclusion adjustments. Final Answer: 1/5

Question 13

Form a 4-digit number without repetition using digits {1, 3, 5, 7, 9}. What is the probability that the number is divisible by 5?

Accepted Answer

Introduction / Context: A number is divisible by 5 iff its last digit is 0 or 5. Here, the available digits are {1,3,5,7,9} and each can be used at most once, so the last digit must be 5. Given Data / Assumptions: Digits: 1, 3, 5, 7, 9; no repetition. All 4-digit permutations using these digits are equally likely. Concept / Approach: Count favorable permutations with 5 in the units place; divide by total permutations of length 4 from 5 distinct digits. Step-by-Step Solution: Total 4-digit numbers = 5P4 = 5 × 4 × 3 × 2 = 120.Favorable: fix last digit = 5 → arrange remaining 4 digits in the thousands–hundreds–tens positions = 4! = 24.Probability = 24 / 120 = 1/5. Verification / Alternative check: Symmetry: each of the 5 digits is equally likely to occupy the last place under uniform random permutation; hence probability the last digit is 5 equals 1/5. Why Other Options Are Wrong: Fractions larger than 1/5 ignore the single admissible last digit; 2/3 and 1/3 are not justified by the counting. Common Pitfalls: Accidentally allowing repetition or considering numbers ending with 0 (digit 0 not available). Final Answer: 1/5

Question 14

The letters of the word CASTIGATION is arranged in different ways randomly. What is the chance that vowels occupy the even places ?

Accepted Answer

Vowels are A I A I O, C A S T I G A T I O N (O) (E) (O) (E) (O) (E) (O) (E) (O) (E) (O) So there are 5 even places in which five vowels can be arranged and in rest of 6 places 6 constants can be arranged as follows : n ( E ) = 5 P 5 2 ! * 2 ! ( A , I a r e 2 t i m e s ) * 6 P 6 2 ! ( T i s 2 t i m e s ) = 21600 n ( S ) = 11 ! 2 ! * 2 ! * 2 ! ( A , I a r e 2 t i m e s ) = 4989600 21600 4989600 = 0 . 043

Question 15

In a throw of a coin, find the probability of getting a head.

Accepted Answer

Here, s={H,T} and E={H} P(E) = n(E)/n(S) = 1/2

Question 16

A box contains 100 balls, numbered from 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability that the sum of the three numbers on the balls selected from the box will be odd?

Accepted Answer

P(odd) = P (even) = 1 2 1(because there are 50 odd and 50 even numbers) Sum or the three numbers can be odd only under the following 4 scenarios: Odd + Odd + Odd = 1 2 * 1 2 * 1 2 = 1 8 Odd + Even + Even = 1 2 * 1 2 * 1 2 = 1 8 Even + Odd + Even = 1 2 * 1 2 * 1 2 = 1 8 Even + Even + Odd = 1 2 * 1 2 * 1 2 = 1 8 Other combinations of odd and even will give even numbers. Adding up the 4 scenarios above: = 1 8 + 1 8 + 1 8 + 1 8 = 4 8 = 1 2

Question 17

A number is selected at random from the numbers 1 to 30. What is the probability that it is divisible by either 3 or 7 ?

Accepted Answer

1/30

Question 18

A fair coin is tossed 11 times. What is the probability that only the first two tosses will yield heads ?

Accepted Answer

Probability of occurrence of an event, P(E) = Number of favorable outcomes/Numeber of possible outcomes = n(E)/n(S) ⇒ Probability of getting head in one coin = ½, ⇒ Probability of not getting head in one coin = 1- ½ = ½, Hence, All the 11 tosses are independent of each other. ∴ Required probability of getting only 2 times heads = 1 2 2 × 1 2 9 = 1 2 11

Question 19

One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn find the probability the card drawn is red.

Accepted Answer

We know that: When one card is drawn from a pack of 52 cards The numbers of possible outcomes n(s) = 52 We know that there are 26 red cards in the pack of 52 cards ⇒ The numbers of favorable outcomes n(E) = 26 Probability of occurrence of an event: P(E)=Number of favorable outcomes/Numeber of possible outcomes=n(E)/n(S) ∴ required probability = 26/52 = 1/2.

Question 20

A bag contains 6 white and 4 black balls .2 balls are drawn at random. Find the probability that they are of same colour.

Accepted Answer

Let S be the sample space Then n(S) = no of ways of drawing 2 balls out of (6+4) = 10 C 2 10 = 10 * 9 2 * 1 =45 Let E = event of getting both balls of same colour Then,n(E) = no of ways (2 balls out of six) or (2 balls out of 4) = 6 C 2 + 4 C 2 = 6 * 5 2 * 1 + 4 * 3 2 * 1 = 15+6 = 21 Therefore, P(E) = n(E)/n(S) = 21/45 = 7/15

Probability Questions