RankLab
Back to Questions
MathsMediumMCQ2021 · 16 Mar Shift 2

Q80.Let A denote the event that a 6 -digit integer formed by 0, 1, 2, 3, 4, 5, 6 without repetitions, be divisible by 3 . Then probability of event A is equal to : (1) 9 (2) 4 56 9 (3) 3 (4) 11 7 27

What This Question Tests

This problem combines permutation principles with the divisibility rule for 3 to calculate the probability of forming a 6-digit number, without repetition, that is divisible by 3.

Concepts Tested

PermutationsDivisibility rule for 3Probability calculation

Formulas Used

P(A) = n(A)/n(S)

Divisibility rule for 3: sum of digits is divisible by 3

Number of permutations P(n,k) = n!/(n-k)!

📚 NCERT Sections This Tests

1.28Calculate The Mass Percentage Of Aspirin (C9H8O4) In Acetonitrile (Ch3Cn) When

Chemistry Class 11 · Chapter 1

70% match

1.28 Calculate the mass percentage of aspirin (C9H8O4) in acetonitrile (CH3CN) when 6.5 g of C9H8O4 is dissolved in 450 g of CH3CN.

12.5A Hydrogen Atom Initially In The Ground Level Absorbs A Photon,

Physics Class 12 · Chapter 12

70% match

12.5 A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of photon.

3.26The Decomposition Of Hydrocarbon Follows The Equation

Chemistry Class 11 · Chapter 3

70% match

3.26 The decomposition of hydrocarbon follows the equation k = (4.5 × 1011s–1) e-28000K/T Calculate Ea. 87 Chemical Kinetics Reprint 2025-26

📋 Question Details

Chapter
Probability
Topic
Probability of an event with permutations
Year
2021
Shift
16 Mar Shift 2
Q Number
Q80
Type
MCQ
NCERT Ref
Class 12 Mathematics Ch 13: Probability | Class 11 Mathematics Ch 7: Permutation & Combination

More from this Chapter

Q90.One ticket is selected at random from 50 tickets numbered 00, 01, 02, … , 49. Then the probability that the sum of the digits on the selected ticket is 8 , given that the product of these digits is zero, equals (1) 1 (2) 1 14 7 (3) 5 (4) 1 14 50 JEE Main 2009 JEE Main Previous Year Paper

2009
Medium

Q89.Four numbers are chosen at random (without replacement) from the set {1, 2, 3, … . , 20} . Statement-1: The probability that the chosen numbers when arranged in some order will form an AP is 1 . Statement-2: If the 85 four chosen numbers from an AP, then the set of all possible values of common difference is {±1, ±2, ±3, ±4, ±5}. (1) Statement-1 is true, Statement-2 is true; (2) Statement-1 is true, Statement-2 is false Statement-2 is not the correct explanation for Statement-1 (3) Statement-1 is false, Statement-2 is true (4) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1

2010
Hard

Q90.An urn contains nine balls of which three are red, four are blue and two are green. Three balls are drawn at random without replacement from the urn. The probability that the three balls have different colour is (1) 2 (2) 1 7 21 (3) 2 (4) 1 23 3 JEE Main 2010 JEE Main Previous Year Paper

2010
Medium

Q89.Consider 5 independent Bernoulli's trials each with probability of success p . If the probability of at least one failure is greater than or equal to 31 , then p lies in the interval 32 (1) ( 34 , 1112 ] (2) [0, 12 ] (3) ( 1112 , 1] (4) ( 12 , 34 ]

2011
Medium
More Mathematics questions