JEE Mathematics Probability MCQs Set 10

Answer: (c) 50/99

Explanation:
1. Place the tallest boy in a group.
2. Of the other 99 boys, 50 can be placed in the other group (49 places remain in his own group).
3. Required probability = 50/99.

Teacher's Note:
1. Fix one object and count the places left for the other. This avoids long combination counts.

Answer: (d) 1/4

Explanation:
1. Units digits of 3, 9, 27, 81, 243, ... repeat in the cycle 3, 9, 7, 1.
2. Each of the four digits is equally likely.
3. P(units digit 3) = 1/4.

Teacher's Note:
1. Powers of 3 have a repeating units digit cycle of length 4. Equal weight to each position gives 1/4.

Answer: (a) 11/15

Explanation:
1. The group is B1 G1, B2 G2, B3 G3, B4 G4, B5, B6, with the four pairs being siblings.
2. Total selections = C(10,3) = 120.
3. Selections containing a sibling pair: choose the pair in 4 ways and the third child in 8 ways = 32 (no selection can contain two pairs).
4. P = 1 - 32/120 = 11/15.

Teacher's Note:
1. Use the complement: count groups that contain a related pair. Three members cannot hold two pairs, so there is no double counting.

Answer: (a) 71/324

Explanation:
1. P(even) = 5/6 and P(odd) = 1/6, so each even face has probability 5/18 and each odd face 1/18.
2. Prime sums possible: 2, 3, 5, 7, 11.
3. Sum 2: (1,1). Sum 3: (1,2),(2,1). Sum 5: (1,4),(4,1),(2,3),(3,2). Sum 7: (1,6),(6,1),(2,5),(5,2),(3,4),(4,3). Sum 11: (5,6),(6,5).
4. Adding the probabilities of all these pairs (each pair has probability (1/18)^2 times 1, 5 or 25 according to the number of even faces) gives 71/324.

Teacher's Note:
1. List the prime sums and classify each pair by the number of even faces. A face with probability 1/18 or 5/18 gives pair products 1, 5 or 25 over 324.

Answer: (d) \( \frac5{12} \)

Explanation:
1. A hexagon has 6C2 - 6 = 9 diagonals, so two can be chosen in 9C2 = 36 ways.
2. Two diagonals meet inside exactly when their four endpoints are distinct vertices: 6C4 = 15 ways (each four vertices give one crossing pair).
3. P = 15/36 = 5/12.

Teacher's Note:
1. Every set of four vertices of a convex polygon gives exactly one pair of crossing diagonals. This is the standard count.

Answer: (d) \( \frac{71}{100} \)

Explanation:
1. Consider the paise part p = 0, 1, ..., 99 (100 equally likely values).
2. The rounding error is p paise if p is up to 49 and (100 - p) paise if p is 50 or more.
3. Error at least 15 means p lies from 15 to 85, which is 71 values.
4. P = 71/100.

Teacher's Note:
1. Treat the paise part as one of 100 equally likely values. Count those whose distance from the nearest rupee is at least 15.

Answer: (b) \( \frac14 \)

Explanation:
1. Powers of 7 leave remainders 2, 4, 3, 1 on division by 5 in a cycle of 4.
2. For 100 values of m, each remainder occurs 25 times.
3. The sum is divisible by 5 only for remainder pairs (2,3), (3,2), (4,1), (1,4).
4. P = 4 x (25 x 25) / (100 x 100) = 1/4.

Teacher's Note:
1. Work with remainders mod 5 and the cycle of length 4. Count the complementary pairs of remainders.

Answer: (d) \( \frac12 \)

Explanation:
1. The parabola opens downward, so it lies strictly below the x-axis when the discriminant is negative.
2. D = 4(a+4)^2 - 4(3a+40) < 0 gives a^2 + 5a - 24 < 0, so (a+8)(a-3) < 0, i.e. -8 < a < 3.
3. Within [-6, 12], the favourable values are -6 < a < 3, length 9, out of total length 18.
4. P = 1/2.

Teacher's Note:
1. For a downward parabola to stay below the axis, the discriminant must be negative. Intersect the result with the allowed interval.

Answer: (d) \( \left(\frac37\right)^4-\left(\frac27\right)^4 \)

Explanation:
1. Least number at least 5 means every draw is 5, 6 or 7: probability (3/7)^4.
2. Least number at least 6 means every draw is 6 or 7: probability (2/7)^4.
3. Least number exactly 5 = (3/7)^4 - (2/7)^4.

Teacher's Note:
1. Exactly-minimum problems are solved as (all at least k) minus (all at least k+1).

Answer: (b) \( \frac{16}{625} \)

Explanation:
1. The last digit of the product depends only on the last digits of the numbers.
2. The product ends in 1, 3, 7 or 9 only when each number ends in 1, 3, 7 or 9.
3. Probability = (4/10)^4 = 256/10000 = 16/625.

Teacher's Note:
1. An even number or a multiple of 5 among the factors makes the product even or a multiple of 5. So all four factors must have an odd last digit other than 5.

Answer: (d) \( \frac{33}{1000} \)

Explanation:
1. Since [t] is at most t, the left side is at most x/2 + x/3 + x/5 = 31x/30.
2. Equality needs x/2, x/3 and x/5 all to be integers, i.e. x is a multiple of 30.
3. Multiples of 30 up to 1000: 33.
4. P = 33/1000.

Teacher's Note:
1. The floor of a number equals the number only when it is an integer. So the condition reduces to divisibility by 2, 3 and 5.

Answer: (a) \( \frac5{12}\le P(B)\le\frac34 \)

Explanation:
1. Odds against A are 2:1, so P(A) = 1/3.
2. Odds in favour of A union B are 3:1, so P(A union B) = 3/4.
3. P(B) is at least P(A union B) - P(A) = 3/4 - 1/3 = 5/12 and at most P(A union B) = 3/4.

Teacher's Note:
1. Use P(A union B) = P(A) + P(B) - P(A and B) with P(A and B) between 0 and P(A). It gives bounds on P(B).

Answer: (c) \( 1-\left(\frac{43}{45}\right)^3 \)

Explanation:
1. Two-digit numbers with digit product 12: 26, 62, 34, 43.
2. P(one card has product 12) = 4/90 = 2/45.
3. P(none of three) = (43/45)^3, so P(at least once) = 1 - (43/45)^3.

Teacher's Note:
1. List the favourable numbers carefully (digits 1 and 12 are impossible). Then use the complement for 'at least once'.

Answer: (c) \( \frac{12}{13} \)

Explanation:
1. P(asleep) = (4/5)(3/4) + (1/5)(1/4) = 3/5 + 1/20 = 13/20.
2. By Bayes' theorem, P(TV
3. asleep) = (3/5)/(13/20) = 12/13.

Teacher's Note:
1. Total probability gives the denominator and Bayes' theorem gives the posterior.

Answer: (d) \( \frac1{n-1} \)

Explanation:
1. z/(z+1) must be an n-th root of unity other than 1, so there are n - 1 solutions.
2. z is real only when z/(z+1) is real, i.e. equals -1 (possible for even n), giving z = -1/2.
3. Exactly one of the n - 1 roots is real, so P = 1/(n-1).

Teacher's Note:
1. Write z/(z+1) = w with w^n = 1 and w not 1. A real z gives a real w, and the only real option is -1.

Answer: (d) 0

Explanation:
1. f'(x) = 3x^2 = 0 gives the only critical point x = 0.
2. f' does not change sign at 0, so there is a point of inflection and no extremum.
3. P = 0.

Teacher's Note:
1. A critical point is an extremum only if the derivative changes sign. Here it does not.

Answer: (c) \( \frac23 \)

Explanation:
1. The semi-major axis equals the radius a of the circle.
2. b^2 = a^2(1 - e^2) = a^2(1 - 8/9) = a^2/9, so b = a/3.
3. Area of ellipse = pi a b = pi a^2/3.
4. Probability outside = 1 - (pi a^2/3)/(pi a^2) = 2/3.

Teacher's Note:
1. The ellipse inscribed in a circle has its major axis as a diameter. Geometric probability is a ratio of areas.

Answer: (a) \( \frac59 \)

Explanation:
1. Let the arrival times be x and y minutes after 11, in a square of side 60.
2. They meet if |x - y| is at most 20.
3. The non-meeting region consists of two right triangles with legs 40, total area 40 x 40 = 1600.
4. P = 1 - 1600/3600 = 5/9.

Teacher's Note:
1. Draw the square of side 60 and cut off the corner triangles where the gap exceeds 20. Take the ratio of areas.

Answer: (c) \( \frac13 \)

Explanation:
1. Since a^3 is congruent to a modulo 3, x^3 + y^3 is divisible by 3 exactly when x + y is.
2. Among 3n numbers, n leave remainder 0, n leave 1, n leave 2.
3. Favourable pairs: both remainder 0: C(n,2); one remainder 1 and one remainder 2: n^2. Total = n^2 + n(n-1)/2 = n(3n-1)/2.
4. Total pairs = 3n(3n-1)/2.
5. P = 1/3.

Teacher's Note:
1. Cube and number leave the same remainder on division by 3. Then count remainder classes.

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

Explanation:
1. Only the units digits matter. Squares of digits 0 to 9 end in 0,1,4,9,6,5,6,9,4,1.
2. Digit-ending counts: 0 (1 digit), 5 (1 digit), 1 (2 digits: 1, 9), 9 (2: 3, 7), 4 (2: 2, 8), 6 (2: 4, 6).
3. Pairs of endings adding to 10 or 0: (0,0) = 1; (5,5) = 1; (1,9) in either order = 2 x 2 x 2 = 8; (4,6) in either order = 8. Total 18 of 100.
4. P = 18/100 = 9/50.

Teacher's Note:
1. Reduce the problem to the units digits of the two numbers (100 equally likely pairs). Then check which pairs of square-endings add to a multiple of 10.