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

A kid is carrying 15 identical but differently colored crayons in his pocket, out of these 5 are blue, 7 green and 3 yellow. If he takes out two crayons randomly from his pocket, what is the probability that they are green and yellow?

  1. 1/5

  2. 2/5

  3. 3/5

  4. 4/5

  5. 1
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

2 crayons can be drawn out of 15 in 15C2 = 105 ways.  One green crayon and one yellow crayon separately can be drawn in 7C1 = 7 and 3C1 = 3 ways, respectively. Since they are exclusive and independent events, taking one green and one yellow crayon together will make the favourable number of ways equal to 7C1 x 3C1 = 21 ways. Hence, required probability, P = 7C1 x 3C1/(15C2) = 21/105 = 1/5

Multiple choice

A shopkeeper received 3 differently sized vests to be delivered to 3 different customers. His helper delivered them randomly to the customers without worrying about their respective sizes. What is the probability that at least one of them received the right sized vest?

  1. 1/3

  2. 2/3

  3. 1

  4. 3/4

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

Let the order of correct sizes be (1, 2, 3). The vests can be delivered to the customers in the following ways: (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2) and (3, 2,1) = 6 ways. The cases (1, 2, 3), (1, 3, 2), (2, 1, 3) and (3, 2, 1) correspond to at least one right sized delivery OR 4 ways. Therefore, the required probability = 4/6 = 2/3

Multiple choice

If a student is picked randomly from a class of 40 students, what is the probability that his name starts with a vowel? Assume that there is no bias towards name starting with any letter.

  1. 5/40

  2. 1/40

  3. 5/26

  4. 5/21

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

Since there is no bias towards any letter, this is the right answer because there are 5 vowels out of 26 letters in the English.

The number of students is not a multiple of 26 but still the exact probability will be close to this.

Multiple choice

A couple has two children. What is the probability that they have one son and one daughter? Assume that there is no bias towards son or daughter.

  1. 1/3

  2. 1/2

  3. 1/4

  4. 1

  5. It depends on other factors
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

First issue can be a son or a daughter with probabilities of 1/2. Similarly the second issue too can be a son or a daughter with equal probabilities of 1/2.

So, there are four equally probable possibilities out of which two result in one son - one daughter combination. Required probability should be 2/4 or 1/2.

Multiple choice

Students in a class of strength 'p' have roll numbers 1, 2, 3, 4… p - 1, p. Take a student with roll number 'n'. What is the probability that either he or someone with a roll number higher than him is selected at random?

  1. 1/p

  2. (p - n)/p

  3. (p - n + 1)/p

  4. 1/n

  5. n/p
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

There are (p-n) students in the class with roll number greater than 'n'. Add one to this to include the student with roll number 'n'. Hence this gives the probability that the selected student has his roll number 'n' or greater than 'n'. This is the required probability.

Multiple choice

In a family of 5 persons (father, mother, a son and two daughters), 2 are to be selected randomly to go out for dinner. What is the probability that at least one of the daughters is selected for this purpose?

  1. 0.7

  2. 0.4

  3. 0.5

  4. 2/3

  5. 0.2
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

2 persons can be selected out of 5 in 5C2 = 10 ways.

Number of selections in which none of the daughters is selected = 3C2 = 3.

In remaining selections (10 – 3 = 7) at least one of the daughters will be present.

So the required probability is, 7/10 = 0.7.

Multiple choice

Four students reached late in a class and made an excuse to the teacher that they had a flat tyre while coming to the college by car. The teacher asked them to sit at four distant seats and made them to answer the same question “which tyre of the car was flat”? What is the probability that all the four give the same answer and their lie is not caught? Assume that students had told lie to the teacher and they will now select their answers randomly.

  1. 1/4

  2. 1/64

  3. 1/256

  4. 1/16

  5. 1/32
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

There are 4 tyres in a car. A person can answer any of the four tyres as flat. Thus, all the four students have four options for giving their answers. There can be 4*4*4*4 = 256 different answers. Out of these, only 4 possibilities are there in which all the four students choose the same tyre as flat. That happens when the first person answers any of the four tyres and remaining three choose the same one. So the required probability is 4/256 = 1/64.

Multiple choice

In a class of 70 students, there are 5 students who are better than others and they have double the chance of being placed at the first rank in a test in comparison to other students. Vijay is not among these five. What is the probability that Vijay gets the first rank in the test?

  1. 0

  2. 1/70

  3. 1/140

  4. 1/75

  5. 1/65
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Since, 5 students have double the chance of being first than others; you may consider each of them as 2 and compute the probability. Thus, these 5 are equivalent to 10 and remaining 65 added to this make it 75. These 5 students will have probability of being first as 2/75 and remaining 65 students will have the same probability as 1/75. Thus 65 times 1/75 and 5 times 2/75 make it 1. Vijay is one among the 65 students. This is correct.