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" are arranged at random in all possible distinct ways. What is the probability that all the vowels occupy the even positions in the arrangement?

Accepted Answer

Introduction: This is a challenging permutations and probability question involving repeated letters. You are asked to find the probability that all vowels of the word "CASTIGATION" appear only in the even positions of a randomly arranged permutation of its letters. The difficulty arises because some letters (including vowels and consonants) are repeated, so we must use permutations of multisets rather than simple factorial counts. Given Data / Assumptions: The word is "CASTIGATION". Total letters = 11. Vowels: A, A, I, I, O → 5 vowels (A and I each repeated twice, O once). Consonants: C, S, T, G, T, N → 6 consonants (T repeated twice). Even positions in an 11-letter word are positions 2, 4, 6, 8 and 10 (5 positions). We want the probability that all vowels occupy these even positions and consonants occupy the odd positions. Concept / Approach: The probability is (number of favourable arrangements) divided by (total number of distinct arrangements). Because some letters repeat, we count permutations using factorials divided by factorials of repeats. For a valid arrangement, the 5 vowels must fill the 5 even spots, and the 6 consonants must fill the 6 odd spots. We count arrangements of vowels and consonants separately and then divide by the total permutations of the 11-letter multiset. Step-by-Step Solution: Step 1: Total number of distinct permutations of "CASTIGATION".There are 11 letters, with A repeated twice, I repeated twice, and T repeated twice.Total permutations = 11! / (2! * 2! * 2!).Step 2: Count favourable permutations.Even positions (2, 4, 6, 8, 10) must contain the 5 vowels A, A, I, I, O.Number of ways to arrange vowels = 5! / (2! * 2!) (because A and I each repeat twice).Odd positions (1, 3, 5, 7, 9, 11) must contain the 6 consonants C, S, T, G, T, N.Number of ways to arrange consonants = 6! / 2! (because T repeats twice).Total favourable arrangements = [5! / (2! * 2!)] * [6! / 2!].Step 3: Form the probability.P = favourable / total = { [5! / (2! * 2!)] * [6! / 2!] } / [11! / (2! * 2! * 2!) ].Notice that 2! * 2! * 2! cancels, leaving:P = (5! * 6!) / 11!.Compute: 5! = 120, 6! = 720, so numerator = 86400.11! = 39916800.So P = 86400 / 39916800.Simplify this fraction: divide numerator and denominator step by step to get P = 1/462. Verification / Alternative check: We can simplify algebraically before plugging numbers: P = (5! * 6!) / 11! = (1 * 2 * 3 * 4 * 5 * 1 * 2 * 3 * 4 * 5 * 6) / (1 * 2 * ... * 11). Cancelling common factors systematically leads to 1/462. Numerical simplification of 86400/39916800 using common divisors also confirms that the reduced fraction is 1/462 ≈ 0.00216, which is a small but reasonable probability. Why Other Options Are Wrong: 1/231, 1/154, 1/308 and 1/420 are larger probabilities that would correspond to fewer total permutations or more favourable arrangements than actually exist. Only 1/462 matches the exact combinatorial counts derived from the repeated-letter permutations. Common Pitfalls: Typical mistakes include ignoring repeated letters and using plain 11! instead of 11!/(2! * 2! * 2!), or treating vowels and consonants as all distinct. Another pitfall is to misinterpret the requirement and allow vowels in odd positions. Careful counting of both vowels and consonants with their repetitions is essential to get the correct probability. Final Answer: The required probability that all vowels occupy the even positions is 1/462.

Question 15

A fair coin is tossed once. What is the probability that the outcome is a head?

Accepted Answer

Introduction: This is one of the simplest probability questions and is a foundational example in elementary probability theory. It involves a single toss of a fair coin and asks for the probability of getting a head. Understanding this example clearly helps build intuition for more complex probability problems. Given Data / Assumptions: The coin is fair, meaning it is not biased. The coin has two possible faces: Head (H) and Tail (T). The coin is tossed once. Each outcome (H or T) is equally likely. We must find P(Head). Concept / Approach: In a fair coin toss, the sample space (set of possible outcomes) is S = {H, T}. Each outcome has equal probability. For equally likely outcomes, the probability of an event is given by: P(Event) = (Number of favourable outcomes) / (Total number of outcomes).Here, the event "Head appears" has exactly one favourable outcome: H. Step-by-Step Solution: Step 1: Write the sample space.S = {H, T}.Step 2: Count total outcomes.Total number of equally likely outcomes = 2.Step 3: Identify favourable outcomes.Event A = "Getting a head" = {H}.Number of favourable outcomes = 1.Step 4: Compute the probability.P(Head) = 1 / 2. Verification / Alternative check: Another way to think about it is symmetry: for a fair coin, there is no reason for head to be more or less likely than tail. The total probability must be 1, and P(H) = P(T). So 2 * P(H) = 1, which again gives P(H) = 1/2. Why Other Options Are Wrong: Probability 1 would mean a head is guaranteed every time, which is not true for a fair coin. 1/4 and 0.1 are too small and would be appropriate only in experiments with more outcomes. Probability 0 would mean head can never appear, which contradicts the coin having a head side. Only 1/2 correctly reflects the equal likelihood of head and tail. Common Pitfalls: There are usually no major pitfalls in this basic problem, but occasionally learners confuse the probability of "at least one head" in multiple tosses with the single-toss case. Always pay attention to how many times the coin is tossed and exactly what event is being asked about. Final Answer: The probability of getting a head in one fair coin toss is 1/2.

Question 16

A box contains balls numbered from 1 to 100. Three balls are selected at random with replacement. What is the probability that the sum of the three selected numbers is odd?

Accepted Answer

Introduction: This probability problem focuses on the parity (odd or even) of the sum of numbers chosen from a set. Instead of analysing all specific values from 1 to 100, we can exploit symmetry between odd and even numbers and use rules about when sums are odd or even. This greatly simplifies the calculation. Given Data / Assumptions: Balls are numbered 1 to 100. Each draw is random and with replacement (so the distribution remains the same for each draw). We draw three balls: say numbers X₁, X₂, X₃. We need the probability that X₁ + X₂ + X₃ is odd. Concept / Approach: Parity rules: Even + Even = Even, Odd + Odd = Even, Odd + Even = Odd.For the sum of three numbers to be odd, the total number of odd addends must itself be odd (i.e., 1 or 3). Also, from 1 to 100, there are 50 odd numbers and 50 even numbers, so each draw has probability 1/2 of being odd and 1/2 of being even. We treat each draw as an independent Bernoulli trial with "odd" as success with probability 1/2. Step-by-Step Solution: Let p = probability that a single draw is odd.From 1 to 100, half the numbers are odd, so p = 50/100 = 1/2.Probability of even = 1 − p = 1/2.We want sum to be odd. This happens when the number of odd draws among the three is 1 or 3.Case 1: Exactly one odd and two even numbers.Probability = C(3, 1) * (1/2)¹ * (1/2)² = 3 * (1/8) = 3/8.Case 2: All three numbers are odd.Probability = (1/2)³ = 1/8.Total probability of an odd sum = 3/8 + 1/8 = 4/8 = 1/2. Verification / Alternative check: Another neat observation is that if each draw has equal chance of being odd or even, then the distribution of the sum’s parity is symmetric: exactly half of all possible parity combinations (odd-odd-odd, odd-even-even, etc.) yield an odd sum. This symmetry argument also leads to the result 1/2 without detailed counting. Why Other Options Are Wrong: 1/6, 1/3 and 1/4 are too small and would correspond to far fewer favourable combinations than actually exist. 2/3 is too large, implying that odd sums are significantly more likely than even sums, which contradicts the symmetry in odd and even probabilities. Only 1/2 correctly reflects the equal likelihood of odd and even sums under these conditions. Common Pitfalls: Common mistakes include trying to work with actual numbers from 1 to 100 instead of just using odd/even counts, or forgetting that the draws are with replacement, which keeps probabilities constant for each draw. Another error is to consider only the all-odd case and forget the one-odd-two-even case, or vice versa. Final Answer: The probability that the sum of the three selected numbers is odd is 1/2.

Question 17

A number is chosen at random from the integers 1 to 30. What is the probability that the chosen number is divisible by 3 or by 7?

Accepted Answer

Introduction: This question is about basic probability using divisibility and counting. You must find how many numbers between 1 and 30 are divisible by 3 or by 7, and then divide by the total number of possibilities. It also illustrates the principle of inclusion and exclusion when dealing with "or" conditions in counting problems. Given Data / Assumptions: We consider integers from 1 to 30 inclusive. Each integer is equally likely to be chosen. We want the probability that the chosen number is divisible by 3 or 7 (or both). Total number of possible outcomes = 30. Concept / Approach: Let A be the event "number divisible by 3" and B be the event "number divisible by 7". We want P(A ∪ B). Use inclusion–exclusion: P(A ∪ B) = P(A) + P(B) − P(A ∩ B).In counting form, we first count multiples of 3, then multiples of 7, then subtract multiples common to both (i.e., multiples of lcm(3, 7) = 21), and finally divide by 30. Step-by-Step Solution: Step 1: Count numbers divisible by 3.Multiples of 3 up to 30: 3, 6, 9, ..., 30.Number of such multiples = 30 ÷ 3 = 10.Step 2: Count numbers divisible by 7.Multiples of 7 up to 30: 7, 14, 21, 28.Number of such multiples = 4.Step 3: Count numbers divisible by both 3 and 7.Numbers divisible by both 3 and 7 are multiples of lcm(3, 7) = 21.Multiples of 21 up to 30: only 21.So there is 1 such number.Step 4: Apply inclusion–exclusion to count numbers divisible by 3 or 7.Count(A ∪ B) = Count(A) + Count(B) − Count(A ∩ B).= 10 + 4 − 1 = 13.Step 5: Convert to probability.Total possibilities = 30 numbers.P(divisible by 3 or 7) = 13 / 30. Verification / Alternative check: We can explicitly list the 13 numbers: divisible by 3 → 3, 6, 9, 12, 15, 18, 21, 24, 27, 30; divisible by 7 → 7, 14, 21, 28. The union is {3, 6, 7, 9, 12, 14, 15, 18, 21, 24, 27, 28, 30}, which clearly has 13 elements, matching our count. Why Other Options Are Wrong: 7/15 = 14/30, 2/5 = 12/30, 11/30 and 3/10 = 9/30 all correspond to incorrect counts of numbers satisfying the condition. Only 13/30 captures the correct 13 favourable outcomes out of 30 possible choices. Common Pitfalls: Common errors include double-counting multiples of both 3 and 7 or forgetting to subtract the overlap for 21. Some students only consider one of the sets or add the probabilities directly without adjusting for intersection. Always remember to use inclusion–exclusion when working with "or" events that can overlap. Final Answer: The probability that a randomly chosen number from 1 to 30 is divisible by 3 or by 7 is 13/30.

Question 18

A fair coin is tossed 11 times in succession. What is the probability that only the first two tosses show heads and all the remaining nine tosses show tails?

Accepted Answer

Introduction: This probability question involves repeated independent tosses of a fair coin. You are asked to find the probability of a very specific outcome sequence: heads on the first two tosses and tails on all remaining tosses. Unlike problems that ask for "exactly two heads anywhere", this problem fixes the positions of heads and tails, leading to a straightforward calculation. Given Data / Assumptions: The coin is fair, so P(Head) = 1/2 and P(Tail) = 1/2 on each toss. The coin is tossed 11 times. We require the event: toss 1 = H, toss 2 = H, tosses 3 to 11 = T. All tosses are independent. Concept / Approach: When outcomes are fixed in a specific order and the tosses are independent, the probability of the combined event is the product of the probabilities of each individual event. Since each toss has probability 1/2 of matching the required outcome (either H or T), the probability of any particular 11-toss sequence is (1/2)^11. Step-by-Step Solution: We want the specific sequence: H, H, T, T, T, T, T, T, T, T, T.Probability that toss 1 is H = 1/2.Probability that toss 2 is H = 1/2.For each of tosses 3 to 11, we need T, each with probability 1/2.There are 2 heads required and 9 tails required.Total number of tosses = 11.Because the tosses are independent, multiply probabilities:P(required sequence) = (1/2) * (1/2) * (1/2) * ... (11 factors) = (1/2)^11. Verification / Alternative check: Note that every exact 11-toss sequence (like HHTTTTTTTTT, THTHTHTHTHT, etc.) has the same probability (1/2)^11, because for each of the 11 tosses, there are 2 equally likely outcomes. The sequence specified in the question is just one of those many sequences, so its probability must be (1/2)^11, consistent with this reasoning. Why Other Options Are Wrong: 9(1/2) is not even a probability (it is greater than 1). (11C2)(1/2)^9 would correspond to exactly two heads anywhere among 11 tosses, without fixing their positions, which is a different event with many more sequences. 1/2 and (1/2)^9 also refer to wrong events or incorrect exponent counts and do not represent the probability of this specific 11-toss sequence. Common Pitfalls: A common mistake is to treat this as a problem of "exactly two heads in 11 tosses" and introduce combinations (11C2), which overcounts the allowed sequences. Here the heads must occur specifically on the first two tosses, so there is exactly one favourable sequence. Always read carefully to see whether positions are fixed or not. Final Answer: The probability that only the first two tosses show heads and the remaining nine are tails is (1/2)^11.

Question 19

In a standard deck of 52 playing cards, one card is drawn at random. Assuming that each of the 52 cards is equally likely to be drawn, what is the probability that the card drawn is red?

Accepted Answer

Introduction: This question tests a very important basic concept in probability related to a standard deck of playing cards. You are asked to calculate the probability that a randomly drawn card is red when all 52 cards are equally likely to be drawn. Understanding this kind of question builds a foundation for more complex probability problems that involve conditional events and combinations. Given Data / Assumptions: We are using a standard deck of 52 playing cards. The deck has no jokers or missing cards. Each one of the 52 cards is equally likely to be drawn. We want the probability that the card drawn is a red card. Red cards in a standard deck are hearts and diamonds. Concept / Approach: In basic probability, when all outcomes are equally likely, the probability of an event is given by: Probability of event = (Number of favorable outcomes) / (Total number of possible outcomes) Here, the event is "the card drawn is red." Therefore, we need to count how many red cards are in the deck and divide that by the total number of cards in the deck. A standard deck has two red suits: hearts and diamonds, and each suit has 13 cards. Step-by-Step Solution: Step 1: Find the total number of possible outcomes. There are 52 distinct cards in a standard deck, so the total number of possible outcomes is 52. Step 2: Count the number of favorable outcomes. There are 2 red suits (hearts and diamonds), and each suit has 13 cards. So the total number of red cards is: Number of red cards = 13 (hearts) + 13 (diamonds) = 26. Step 3: Apply the probability formula. Probability(card is red) = Number of red cards / Total number of cards = 26 / 52. Step 4: Simplify the fraction. Divide numerator and denominator by 26: 26 / 52 = 1 / 2. So the probability that the card drawn is red is 1/2. Verification / Alternative check: Another way to think about this is that in a standard deck, exactly half of the cards are red and half are black. Since there are only two colors (red and black) and each color has the same number of cards (26), the chance of drawing either color is equally likely. Therefore, the probability of drawing a red card is 1 out of 2, which again gives 1/2. This confirms our calculation and provides an intuitive check on the answer. Why Other Options Are Wrong: 2/3: This would suggest that two thirds of the deck is red, which is not true because only half the deck is red and the other half is black. 1/52: This would be the probability of drawing one specific card, such as the ace of hearts, not any one of the 26 red cards. 13/51: This fraction does not correspond to any natural count in a full 52 card deck for this event. It might appear in other card problems but not here. 1/4: This would suggest that only 13 of 52 cards are red, which is incorrect because there are 26 red cards in total. Common Pitfalls: A common mistake is to forget the composition of a standard deck and miscount the number of red cards. Some learners confuse the probability of drawing a specific card (1 out of 52) with the probability of drawing any card of a certain color or suit. Others may incorrectly assume that each suit is a color category on its own. Always remember: a standard deck has 4 suits with 13 cards each, and exactly 2 of these suits (hearts and diamonds) are red, giving 26 red cards. Using the basic probability formula correctly avoids these errors. Final Answer: The probability that the card drawn from a standard 52 card deck is red is 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

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