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Test: Probability Part:- 2 - Mathematics MCQ


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20 Questions MCQ Test - Test: Probability Part:- 2

Test: Probability Part:- 2 for Mathematics 2024 is part of Mathematics preparation. The Test: Probability Part:- 2 questions and answers have been prepared according to the Mathematics exam syllabus.The Test: Probability Part:- 2 MCQs are made for Mathematics 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Probability Part:- 2 below.
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Test: Probability Part:- 2 - Question 1

A lot has 10% defective items. Ten items are chosen randomly from this lot. The probability that exactly 2 of the chosen items are defective is 

Detailed Solution for Test: Probability Part:- 2 - Question 1

Let A be the event that items are defective and B be the event that items are non-defective. 
∴ P(A) = 0.1 and P(B) = 0.9
∴ Probability that exactly two of those items are defective

Test: Probability Part:- 2 - Question 2

A single die is thrown twice. What is the sum is neither 8 nor 9?

Detailed Solution for Test: Probability Part:- 2 - Question 2

Here sample space = 36
Total No. of way in which sum is either 8 or 9 are (2,6), (3,5),(3,6),(4,4),(4,5),(5,3),(5,4),(6,2),(6,3)
So probability of getting sum 8 or 9 = 9/36 = 1/4
So the probability of not getting sum 8 or 9  

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Test: Probability Part:- 2 - Question 3

A box contains 20 defective items and 80 non-defective items. If two items are selected at random without replacement, what will be the probability that both items are defective?

Detailed Solution for Test: Probability Part:- 2 - Question 3

The probability of defective  items = 20/100.
Therefore the probability of first two defective items without replacement 

Test: Probability Part:- 2 - Question 4

A coin is tossed 4 times. What is the probability of getting heads exactly 3 times?

Detailed Solution for Test: Probability Part:- 2 - Question 4

Probability of getting exactly three heads

Test: Probability Part:- 2 - Question 5

If three coins are tossed simultaneously, the probability of getting at least one head is

Detailed Solution for Test: Probability Part:- 2 - Question 5

Here the sample space = S = 23 = 8.
No. of ways to get all tails = 1.
∴ probability to get all tails = 1/8
∴ Probability to get at least one head is = 1 - 1/8 = 7/8

Test: Probability Part:- 2 - Question 6

A box contains 2 washers, 3 nuts and 4 bolts. Items are drawn from the box at random one at a time without replacement. The probability of drawing 2 washers first followed by 3 nuts and subsequently the 4 bolts is

Detailed Solution for Test: Probability Part:- 2 - Question 6

Here sample space = 9
The required probability of drawing 2 washers, 3 nuts and 4 bolts respectively without replacement

= 1/1260

Test: Probability Part:- 2 - Question 7

If 20 per cent managers are technocrats, the probability that a random committee of 5 managers consists of exactly 2 technocrats is

Detailed Solution for Test: Probability Part:- 2 - Question 7

The probability of technocrats manager = 20/100
= 1/5
∴ Probability of non technocrats manager = 4/5
Now the require probability 

Test: Probability Part:- 2 - Question 8

Analysis of variance is concerned with:

Detailed Solution for Test: Probability Part:- 2 - Question 8

Analysis of variance is used in comparing two or more populations, e.g. Different types of manures for yelding a single crop. 

Test: Probability Part:- 2 - Question 9

Four arbitrary point (x1,y1), (x2,y2),(x3,y3), (x4,y4), are given in the x, y - Plane Using the method of least squares, if, regressing y upon x gives the fitted line y = ax + b; and regressing y upon x given the fitted line x = cy + d then

Detailed Solution for Test: Probability Part:- 2 - Question 9

y = ax + b -(i) and x = cy + d - (ii)
From (ii) wee get x - d = cy ⇒ y = 1/c x - d/c - (iii)
comparing (i) and (ii), a = 1/c and b = -d/c
 

Test: Probability Part:- 2 - Question 10

A  regression model is used to express a variable Y as a function of another variable X. This implies that

Test: Probability Part:- 2 - Question 11

Let X and Y be two independent random variables. Which one of the relations between expectation (E), variance (Var) and covariance (Cov) given below is FALSE?

Test: Probability Part:- 2 - Question 12

A class of first year B. Tech. Students is composed of four batches A, B, C and D, each consisting of 30 students. It is found that the sessional marks of students in Engineering Drawing in batch C have a mean of 6.6 and standard deviation of 2.3. The mean and standard deviation of the marks for the entire class are 5.5 and 4.2, respectively. It is decided by the course instructor to normalize the marks of the students of all batches to have the same mean and standard deviation as that of the entire class. Due to this, the marks of a student in batch C are changed from 8.5 to

Detailed Solution for Test: Probability Part:- 2 - Question 12

Let mean and stander deviation of batch C be μc and σc respectively and mean and standard deviation of entire class of 1st year students be μ and σ respectively.
Given μc = 6.6 and σc = 2.3
and μ = 5.5 and σ = 4.2
In order to normalize batch F  to entire class, the normalized score must be equated 
Since
 

Test: Probability Part:- 2 - Question 13

Three values of x and y are to be fitted in a straight line in the form y = a + bx by the method of least squares. Given ∑x = 6, ∑y = 21, ∑x2 = 14 and ∑xy  = 46, the values of a and b are respectively.

Detailed Solution for Test: Probability Part:- 2 - Question 13

y = a + bx
Given
n = 3, Σx = 6, Σy = 21, Σx2 = 14 
And
Σxy = 46 


Substituting, we get

∴ a = 3 and b = 2 

Test: Probability Part:- 2 - Question 14

A box contains 10 screws, 3 of which are defective. Two screws are drawn at random with replacement. The probability that none of the two screws is defective will be

Detailed Solution for Test: Probability Part:- 2 - Question 14

Non defective screws = 7
∴ Probability of the two screws are non defective

Test: Probability Part:- 2 - Question 15

A hydraulic structure has four gates which operate independently. The probability of failure off each gate is 0.2. Given that gate 1 has failed, the probability that both gates 2 and 3 will fail is

Detailed Solution for Test: Probability Part:- 2 - Question 15

P(gate to and gate 3/gate 1 failed)
= P(gate 2 and gate 3)
[∴ all three gates are independent corresponding to each other]

Test: Probability Part:- 2 - Question 16

Which one of the following statements is NOT true?

Detailed Solution for Test: Probability Part:- 2 - Question 16

is not true since in a negatively skewed distribution, mode > median > mean 

Test: Probability Part:- 2 - Question 17

There are 25 calculators in a box. Two of them are defective. Suppose 5 calculators are randomly picked for inspection (i.e., each has the same chance of being selected), what is the probability that only one of the defective calculators will be included in the inspection?

Detailed Solution for Test: Probability Part:- 2 - Question 17

Probability of only one is defective out of 5 calculators 

Test: Probability Part:- 2 - Question 18

If the standard deviation of the spot speed of vehicles in a highway is 8.8 kmph and the mean speed of the vehicles is 33 kmph, the coefficient of variation in speed is

Detailed Solution for Test: Probability Part:- 2 - Question 18

Test: Probability Part:- 2 - Question 19

If probability density functions of a random variable X is f(x) = x2 for -1 ≤ x ≤​ 1, and  = for any other value of x 
Then, the percentage probability  is

Detailed Solution for Test: Probability Part:- 2 - Question 19


∴ Percentage probability = 2/81 x 100 = 2.47% 

Test: Probability Part:- 2 - Question 20

A person on a trip has a choice between private car and public transport. The probability of using a private car is 0.45. While using the public transport, further choices available are bus and metro, out of which the probability of commuting by a bus is 0.55. In such a situation, the probability (rounded up to two decimals) of using a car, bus and metro, respectively would be

Detailed Solution for Test: Probability Part:- 2 - Question 20

p(private car) = 0.45
p(bus / public transport) = 0.55 
Since a person has a choice between private car and public transport 
p (public transport)= 1 – p(private car)
= 1 – 0.4 5 = 0.55
p (bus) = p(bus ∩ public transport)
= p(bus | public transport)
× p( public transport)
= 0.55 × 0.55 
= 0.3025 = 0.30
Now p (metro) = 1 – [p(private car) + p(bus)]
= 1 – (0.45 + 0.30) = 0.25
∴ p (private car) = 0.45 
p (bus) = 0.30
and p(metro) = 0.25

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