Test: Probability- 2 - Civil Engineering (CE) MCQ

# Test: Probability- 2 - Civil Engineering (CE) MCQ

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## 15 Questions MCQ Test Engineering Mathematics - Test: Probability- 2

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

### A die is rolled three times. The probability that exactly one odd number turns up among the three outcomes is

Detailed Solution for Test: Probability- 2 - Question 1

Probability of getting an odd number in rolling of a die = 3/6 = 1/2.
Now using binomial distribution
P(Exactly one odd number among three outcomes)

Test: Probability- 2 - Question 2

### Seven (distinct) car accidents occurred in a week. What is the probability that they all occurred on the same day?

Detailed Solution for Test: Probability- 2 - Question 2

Sample space = 77
All accidents on the same day = 7 ways (all on Monday, all on Tuesday...)
So, required probability

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Test: Probability- 2 - Question 3

### Let X and Y be two exponentially distributed and independent random variables with mean α and β, respectively. If Z= min (X, Y), then the mean of Z is given by

Detailed Solution for Test: Probability- 2 - Question 3

X and Y are two independent exponentially distributed random variables. Let λ1 and λ2 parameters of X and Y respectively.

Given, Z = min (X, Y)

Since mean of exponential distribution = 1/Parameter
So,

∴ Z is random variable with parameter
Mean of Z =

Test: Probability- 2 - Question 4

A bag contains 10 blue marbles, 20 green marbles and 30 red marbles. A marble is drawn from the bag, its colour recorded and it is put back in the bag. This process is repeated 3 times. The probability that no two of the marbles drawn have the same colour is

Detailed Solution for Test: Probability- 2 - Question 4

The given condition corresponds to sampling with replacement and with order.
No 2 marbles have the same color i.e. Drawn 3 different marble.
So total number of ways for picking 3 different marbles = 3! = 6.
Probability of getting blue, green, red in order

[Since 6 ways to get the marbles]
= 1/6

Test: Probability- 2 - Question 5

In a certain town, the probability that it will rain in the afternoon is known to be 0.6. Moreover, meteorological data indicates that if the temperature at noon is less than or equal to 250C, the probability that it will rain in the afternoon is 0.4. The temperature at noon is equally likely to be above 250C, or at/below 250C.
What is the probability that it will rain in the afternoon on a day when the temperature at noon is above 250C?

Detailed Solution for Test: Probability- 2 - Question 5

Let RA : Rain in the afternoon
T > 25 : Temperature more than 250C
Let the desired probability = P(RAIT ≤ 25) = x
The tree diagram forthis problem is given below:

Given P(RA) = 0.6
by rule of total probability P(RA)

⇒ x = 0.8

Test: Probability- 2 - Question 6

Suppose a fair six-sided die is rolled once, if the value on the die is 1, 2 or 3 then die is rolled a second time. What is the probability that the sum total of values that turn up is at least 6?

Detailed Solution for Test: Probability- 2 - Question 6

If first throw is 1, 2 or 3 then sample space is only 18 possible ordered pairs. Out of this only (1, 5 ), (1, 6 ), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5) and (3,6) i e. 9 out of 18 ordered pairs gives a Sum ≥ 6.
If first throw is 4, 5 or 6 then second throw is not made and therefore the only way Sum ≥ 6 is if the throw was 6. Which is one out of 3 possible. So the tree diagram becomes as follows:

From above diagram

Test: Probability- 2 - Question 7

Suppose p is the number of cars per minute passing through a certain road junction between 5 PM and p has Poisson distribution with mean 3. What is the probability of observing fewer than 3 cars during any given minute in this interval?

Detailed Solution for Test: Probability- 2 - Question 7

Probability of observing fewer than 3 cars

Test: Probability- 2 - Question 8

A fair coin is tossed till a head appears for the first time. The probability that the number of required tosses is odd, is

Detailed Solution for Test: Probability- 2 - Question 8

P(number of tosses is odd) = P(number of tosses is 1, 3, 5, 7 ...)
P(number of toss is 1) = P(Head in first toss = 1/2
P(number of toss is 3) = P(tail in first toss, tail in second toss and head in third toss)

P(number of toss is 5) = P(T, T, T, T, H)

So P(number of tosses is odd)
Sum of infinite geometric series with

Test: Probability- 2 - Question 9

A continuous random variable Xhas a probability density function f(x) = e-x, 0 < x <∝. Then P{X> 1} is

Detailed Solution for Test: Probability- 2 - Question 9

Test: Probability- 2 - Question 10

A fair coin is tossed n times. The probability that the difference between the number of heads and tails is ( n - 3) is

Detailed Solution for Test: Probability- 2 - Question 10

when n = probability of occurrence of head
y = probability of occurrence of tail
Let number of head is P
Number of tail is q

Here required probability is zero.

Test: Probability- 2 - Question 11

Two players, A and B, alternately keep rolling a fair dice. The person to get a six first wins the game. Given that player A starts the game, the probability that A wins the game is

Detailed Solution for Test: Probability- 2 - Question 11

P(A wins) = p(6 in first throw by A) + p(A not 6, B not 6, A 6) + ...

Test: Probability- 2 - Question 12

Equation ex - 1 = 0 is required to be solved using Newton's method with an initial guess x0 = -1. Then, after one step of Newton’s method, estimate x1 of the solution will be given by

Detailed Solution for Test: Probability- 2 - Question 12

Test: Probability- 2 - Question 13

Let x2 - 117 = 0. The iterative steps for the solution using Newton-Raphson’s method is given by

Detailed Solution for Test: Probability- 2 - Question 13

Test: Probability- 2 - Question 14

Solution of the variables x1 and x2 for the following equations is to be obtained by employing the Newton-Raphson iterative method

Assuming the initial values x1 = 0.0 and x2 = 1.0, the Jacobian matrix is

Detailed Solution for Test: Probability- 2 - Question 14

Test: Probability- 2 - Question 15

When the Newton-Raphson method is applied to solve the equation f(x) = x3 + 2x - 1 = 0, the solution at the end of the first iteration with the initial guess value as x0 = 1.2 is

Detailed Solution for Test: Probability- 2 - Question 15

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