Test: Probability & Statistics- 1 - Civil Engineering (CE) MCQ

Here the sample space S = 6 

Therefore  P(odd number ) = 3/6 = 1/2

and  P (even number )  = 3/6 = 1/2

since events are independent, 

 therefore, P (odd / even ) 

Probability of defective computer supplied by Y = 
(Case when Y produces defective)/(All cases of producing defective product)
Case when Y produces defective = (0.3)(0.02) = 0.006
All cases of producing defective product= (0.6x0.01)+(0.3x0.02)
(0.1x0.03)= 0.006+0.006+0.003=0.015

Probability = 0.006/0.015=0.4

Let A be  the event that ‘failed in paper 1’.  

B be the event that ‘failed  in paper 2’. 

Given P(A) = 0.3, P(B) = 0.2. 

Let A be the event that first toss is head  

And B be the event that second toss is head. 

By the given condition rest all 8 tosses should be tail

∴ The probability of getting head in first two cases 

T being the random variable of f(t). 

Let A = the event that the birth month of first friend  

And B= that of second friend.

∴ P( A )= 1, as 1st friend can born in any month

and P(B) = 1/12, by the condition.

∴ Probability of two friends share same birth-month 

Here the possible combination of picking up three balls without replacement is  BBR,  BRB,  RBB.

(B = Black ball,  R = Red balls) 

9

∴ Probability of getting two black balls and one red ball is (5/28)x3   =  15/28.

Here sample space = 6 × 6 = 36 

Here, there are five such points whose sum is 8. They are (2,6), (3,5), (4,4), (5,3), (6,2). 

The correct option is a) 0.7
Given data:

λ=0.1 per minute,

Service rate, μ=0.33 per minute

ρ=0.1/0.33=0.3

Probability that an arrival does not have to wait before service or probability that a system is idle,

P₀=1−ρ=1−0.3=0.7

The probability of drawing two red balls without replacement 

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 

The probability of defective  items = 20/100

Therefore the probability of first two defective items without replacement 

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

y =ax+b − (i) and x = cy + d − (ii)

If a bell curve is assumed, the probability of a “six sigma” event is on the order of one ten millionth of a percent.

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A = First drawn screw non-defective.

B = Second drawn screw non-defective.

P(A) = 7/10

Because 7 out of 10 screws are non-defective and we sample at random,

P(B) = 7/10

The second drawing is the same as the beginning.

P(A Intersection B) = P(A) * P(B) = 0.7 * 0.7 = 0.49

It is Converted to percentage = 0.49 * 100 = 49%

The probability of two screws are defective is 49%.