JEE Mathematics Probability MCQs Set 11

Answer: (d) \( \frac3{10} \)

Explanation:
1. Digits at most 9 with sum 34 (that is 36 - 2): the digit multiset is either {9, 9, 8, 8} or {9, 9, 9, 7}.
2. Counts: 4!/(2!2!) = 6 and 4!/3! = 4, so 10 numbers in F.
3. Even ones must end in 8: arrangements of {9, 9, 8} in the first three places = 3.
4. P = 3/10.

Teacher's Note:
1. The maximum digit sum is 36, so a sum of 34 means the digits fall short by only 2 in total.

Answer: (a) \( \frac14 \)

Explanation:
1. The number of sequences with k tails is C(100, k).
2. Sum of C(100,k) over odd k from 1 to 99 is 2^99.
3. Since C(100,k) = C(100,100-k) and 100 - k is odd when k is odd, the odd k up to 49 contribute half of this: 2^98.
4. P = 2^98/2^100 = 1/4.

Teacher's Note:
1. Use the symmetry of binomial coefficients to halve the sum over all odd k.

Answer: (c) 3

Explanation:
1. Total pairs = C(25,2) = 300.
2. Pairs with difference d at least 10: for d = 10, ..., 24 there are 25 - d pairs, giving 15 + 14 + ... + 1 = 120.
3. Pairs with difference less than 10 = 300 - 120 = 180, so P = 3/5.
4. 5P = 3.

Teacher's Note:
1. It is easier to count pairs with a large difference using 25 - d for each difference d.

Answer: (b) \( \frac19 \)

Explanation:
1. The limit is of the form 1^infinity and equals e raised to lim (2/x)((a^x + b^x)/2 - 1) = e^(ln a + ln b) = ab.
2. So ab = 6: (a, b) = (1,6), (6,1), (2,3), (3,2).
3. P = 4/36 = 1/9.

Teacher's Note:
1. For a 1^infinity limit use exp of the limit of (base - 1) times the exponent. The limit simplifies to the product ab.

Answer: (a) 5 or 10

Explanation:
1. Let B have x outcomes. Independence needs P(A and B) = P(A)P(B) = (4/10)(x/10), so n(A and B) = 4x/10 = 2x/5.
2. This must be a whole number, so x is a multiple of 5: x = 5 or 10.
3. Both values work (n(A and B) = 2 or 4, which does not exceed 4).

Teacher's Note:
1. Independence means n(A and B) = n(A) n(B)/n(S). This has to be an integer.

Answer: (a) \( \frac{24}{29} \)

Explanation:
1. P(correct) = (1/3)(1/4) + (1/6)(1/8) + (1/2)(1) = 1/12 + 1/48 + 1/2 = 29/48.
2. P(computed and correct) = 1/2 = 24/48.
3. Bayes: P = 24/29.

Teacher's Note:
1. Combine the three routes for the total probability, then divide the 'computed' route by it.

Answer: (b) \( \frac17 \)

Explanation:
1. P(on time
2. mode) = 7/9, 8/9, 5/9, 8/9.
3. Numerators: (1/7)(7/9) = 7/63; (3/7)(8/9) = 24/63; (2/7)(5/9) = 10/63; (1/7)(8/9) = 8/63.
4. Total = 49/63.
5. P(car
6. on time) = 7/49 = 1/7.

Teacher's Note:
1. Convert the 'late' probabilities to 'on time' probabilities first. Bayes then uses these.

Answer: (c) \( \frac{43}{216} \)

Explanation:
1. Real roots need b^2 at least 4ac.
2. b = 1: none. b = 2: ac at most 1 gives 1 case. b = 3: ac at most 2 gives 3 cases (1,1),(1,2),(2,1). b = 4: ac at most 4 gives 8 cases. b = 5: ac at most 6 gives 14 cases. b = 6: ac at most 9 gives 17 cases.
3. Total = 0 + 1 + 3 + 8 + 14 + 17 = 43.
4. P = 43/216.

Teacher's Note:
1. Fix b and count the pairs (a, c) with ac at most b^2/4. Add the cases.

Answer: (d) \( \frac{13}{20} \)

Explanation:
1. The parabola opens upward, so it is strictly above the x-axis when D < 0.
2. 64(a+5)^2 + 64(7a+5) < 0, which gives a^2 + 17a + 30 < 0, i.e. (a+15)(a+2) < 0, so -15 < a < -2.
3. Favourable length = 13 out of 20.
4. P = 13/20.

Teacher's Note:
1. For an upward parabola, the discriminant must be negative. Compare the length of the valid interval with 20.

Answer: (b) \( \frac{31}{50} \)

Explanation:
1. Real roots need p^2 at least 4q.
2. Number of q for each p: p = 1: 0; p = 2: 1; p = 3: 2; p = 4: 4; p = 5: 6; p = 6: 9; p = 7 to 10: 10 each.
3. Total = 0 + 1 + 2 + 4 + 6 + 9 + 40 = 62.
4. P = 62/100 = 31/50.

Teacher's Note:
1. Count the values of q (1 to 10) with q at most p^2/4 for each p, remembering q cannot exceed 10.

Answer: (c) \( \frac7{222} \)

Explanation:
1. Total distributions: solutions of x1 + x2 + x3 = 35 in non-negative integers = C(37, 2) = 666.
2. Each patient at least 10: put 10 aside to each, leaving 5 to share: C(7, 2) = 21.
3. P = 21/666 = 7/222.

Teacher's Note:
1. Use stars and bars for both counts. Setting aside the minimum requirement first is the usual trick.

Answer: (d) \( \frac4{35} \)

Explanation:
1. P4 reaches the final only if it is in a group of four whose other three members are all weaker (numbered above 4).
2. The three companions in its half come from P5, P6, P7, P8: C(4,3) = 4 ways.
3. Total ways to choose the 3 companions from the other 7 players: C(7,3) = 35.
4. P = 4/35.

Teacher's Note:
1. P4 loses only to P1, P2 or P3. So none of them may be in its half of the draw.

Answer: (a) \( \left(\frac{11}{20}\right)^6 \)

Explanation:
1. Subtraction needs no borrowing at a decimal place when the digit of the first number is at least that of the second.
2. For one place, the favourable pairs (x, y) with x at least y: 10 + 9 + ... + 1 = 55 out of 100, so 11/20.
3. There are six independent places: (11/20)^6.

Teacher's Note:
1. Compute the probability for one digit position and multiply across the six independent positions.

Answer: (a) \( \frac9{50} \)

Explanation:
1. Total three-digit numbers = 900.
2. Pattern aab (a from 1 to 9, b from 0 to 9, b not a): 9 x 9 = 81.
3. Pattern abb (a from 1 to 9, b not a): 81.
4. Total favourable = 162.
5. P = 162/900 = 9/50.

Teacher's Note:
1. Count the two patterns aab and abb. Excluding b = a avoids counting the case aaa.

Answer: (d) 0.0875

Explanation:
1. There are 4 matches and a maximum of 8 points.
2. P(8 points) = (0.50)^4 = 0.0625.
3. P(7 points) = 4 x (0.50)^3 x 0.05 = 0.025.
4. P(at least 7) = 0.0625 + 0.025 = 0.0875.

Teacher's Note:
1. With four matches, 7 points means three 2-point wins and one 1-point result in any order.

Answer: (a) \( \frac{9m}{8N+m} \)

Explanation:
1. P(HT
2. fair) = 1/4 and P(HT
3. biased) = (2/3)(1/3) = 2/9.
4. P(fair and HT) = (m/N)(1/4) and P(biased and HT) = ((N-m)/N)(2/9).
5. P(fair
6. HT) = (m/4)/(m/4 + 2(N-m)/9) = 9m/(9m + 8N - 8m) = 9m/(8N + m).

Teacher's Note:
1. Compute the probability of the observed sequence under each type of coin, then apply Bayes' theorem.

Answer: (a) \( \frac{369}{625} \)

Explanation:
1. A product divisible by 10 is also divisible by 5, so the event is 'divisible by 5'.
2. The product is not divisible by 5 only if no number ends in 0 or 5: (8/10)^4 = 4096/10000 = 256/625.
3. P = 1 - 256/625 = 369/625.

Teacher's Note:
1. Divisible by 10 already implies divisible by 5. Use the complement.

Answer: (c) \( \frac{20}{23} \)

Explanation:
1. P(B shows green
2. original green) = (3/4)(3/4) + (1/4)(1/4) = 10/16. So P(original green and B green) = (4/5)(10/16) = 1/2.
3. P(B shows green
4. original red) = (3/4)(1/4) + (1/4)(3/4) = 6/16, so that part is (1/5)(6/16) = 3/40.
5. P(B green) = 1/2 + 3/40 = 23/40.
6. P = (1/2)/(23/40) = 20/23.

Teacher's Note:
1. Green at B arises with two errors or none if the original was green, and with exactly one error if it was red.

Answer: (b) \( \frac29 \)

Explanation:
1. The sum of three cube roots of unity is zero only if they are 1, w and w^2 in some order (one from each residue of r modulo 3).
2. Each residue class modulo 3 (0, 1, 2) has 2 numbers on the die.
3. Ordered triples: 3! x 2 x 2 x 2 = 48.
4. P = 48/216 = 2/9.

Teacher's Note:
1. The three terms must be 1, w, w^2 in some order. Each residue class mod 3 has two faces.

Answer: (c) \( \frac{15}{92} \)

Explanation:
1. The 10 vacant places are chosen out of the other 24 places: C(24, 10).
2. If both neighbours are vacant, the remaining 8 vacant places are chosen out of 22: C(22, 8).
3. P = C(22,8)/C(24,10) = 15/92.

Teacher's Note:
1. Fix the two neighbouring places as vacant and count the arrangements of the remaining vacancies.

Answer: (a) \( \frac{35}{324} \)

Explanation:
1. Total outcomes = 6^4 = 1296.
2. The number of ways to get sum 13 with four dice is the coefficient of x^13 in (x + x^2 + ... + x^6)^4, which is 140.
3. P = 140/1296 = 35/324.

Teacher's Note:
1. Use the generating function or subtract the minimum 1 from each die and count solutions with upper bounds.