Mathematics

Probability Distributions

457 Questions

Probability distributions describe the likelihood of different outcomes in a statistical experiment. Key topics include calculating confidence intervals, understanding Type II errors, and working with the standard normal distribution. These concepts are frequently tested in mathematics and statistics sections of various competitive exams.

Standard normal distributionType II error probabilityConfidence interval interpretationPoisson distributionProbability density function

Probability Distributions Questions

Multiple choice

Two distinct numbers a and b are chosen randomly from the set {2, 22, 23, 24, ------ 225} (25 numbers are there). Find the probability that logab (log b to the base a) is an integer.

  1. 2/25

  2. 31/150

  3. 31/300

  4. 37/300

  5. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For Logab to be integer, b>a. So, obviously b cannot be 2 because a and b are distinct. So, if b=2, a should be greater than that, and then it will not be an integer. So if a=2 ,then possible values of b are {22,23,24,------224,225} There are 24 possible values. Similarly, if a=22, then possible values of b are {24,26,28,------222,224}, i.e. there are 11 possible values. Similarly, if a=23, then possible values of b are {26,29,212,------221,224}, i.e. there are 7 possible values. Similarly, if a=24, then possible values of b are {28,212,216,------220,224}, i.e. there are 5 possible values. Similarly, if a=25, then possible values of b are {210,215,220,------225}, i.e. there are 4 possible values. Similarly, if a=26,then possible values of b are {212,218,224}, i.e. there are 3 possible values. Similalry, if a=27, then possible values of b are {214,221}, i.e. there are 2 possible values. Similalry, if a=28, then possible values of b are {216,224}, i.e. there are 2 possible values. Similarly, if a=29, then possible values of b are {218), i.e. only 1 possible value. For a=210,211,212, the possible values are 220,222,224 respectively. After that, there is no case possible. So, total possibilities are (24+11+7+5+4+3+2+2+1+1+1+1)= 62 cases Total number of possibilities = a can be 25 numbers and b can be 24 numbers (As a and b should be distinct) Total possibilities are 25*24. Probability = 62/(25*24) = 31/300 (Correct Answer)

Multiple choice
If one number is chosen at random from the first 900 positive integers, what is the probability that the number chosen is a multiple of both 3 and 5?

  1. 3/125

  2. 1/9

  3. 3/45

  4. 9/75
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Any number that is a multiple of both 3 and 5 must be a multiple of 15. The number of integers from 1 and 900 which are multiples of 15 are 15 × 1, 15 × 2, 15 × 3, ..., 15 × 60. These are 60 in number. So, the probability that one among the first 900 numbers chosen at random being a multiple of 15 is 60/900 = 6/90 = 1/15 = 3/45

Multiple choice

A box contains 20 electric bulbs, out of which 5 are defective. Two bulbs are chosen at random from this box. The probability that at least one of these is defective is

  1. 1/4

  2. 17/38

  3. 21/38

  4. 3/38

  5. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

P (none of the balls is defective) = 15C2/20C2

                                                         = 21/38

P(at least one ball is defective) = 1 - 21/38

                                                    = 17/38

 

Multiple choice

Which of the following algorithms is used to simulate complex systems in statistical physics and provides the method for obtaining a sequence of random samples from a probability distribution?

  1. Minimax algorithm

  2. Breadth first search algorithm

  3. Simulated annealing algorithm

  4. A* search algorithm

  5. Metropolis hastings algorithm
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

This algorithm is used to simulate complex systems in statistical physics and provides the method for obtaining a sequence of random samples from a probability distribution.

Multiple choice

A candidate is to be selected for an interview for 2 posts. The number of candidates for the first post is 5 and for the second post is 6. Find the probability of getting at least one job.

  1. 5/11

  2. 6/11

  3. 2/11

  4. 1/3

  5. 1/11
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The calculation of probability is as follows:
 
Probability that the candidate does not get an offer from the first interview is 4/5.
Probability that the candidate does not get an offer from the second interview is 5/6.
Total probability of not getting an offer is 4/5 * 5/6 = 2/3
Hence, the probability of getting at least one job is 1 - 2/3 = 1/3  
 
 

Multiple choice

A hash table has space for 100 records. What is the probability of collision before the table is 10% full?

  1. 0.45

  2. 0.5

  3. 0.3

  4. 0.34
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

This is a variation of the birthday paradox. The probability of no collision for n items in m slots is (m/m) * ((m-1)/m) * ... * ((m-n+1)/m). For 10 records in 100 slots, the probability of no collision is approx 0.63, meaning the probability of at least one collision is 1 - 0.63 = 0.37, which is closest to 0.34.