# Topic: Probability and Statistics (Test 2)

Topic: Probability and Statistics
Q.1

Direction: Study the following information carefully and answer the questions that follow:

A committee of five members is to be formed out of 3 trainees, 4 professors and 6 engineers.

In how many ways this can be done if 1 trainees and 4 engineers be included in a committee

A. 45
B. 32
C. 60
D. 36
E. None of these
Answer : Option A
Explaination / Solution:

If 1 trainees and 4 engineers include in committee
3C1*6C4 = 3*15 --> 45

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Q.2
Directions: In the following questions two equations numbered I and II are given. You have to solve both the equations.

I. 8x2 + 30x + 28 = 0
II. 9y2 + 11y+2 =0
A. X>Y
B. X ≥Y
C. X
D. X≤Y
E. X = Y or the relationship cannot be established
Answer : Option C
Explaination / Solution:

I. 8x2 + 30x + 28 = 0
x = (-7/4, -2)
II. 9y2 + 11y+2 =0
y = (-1, -2/9)
So x<y

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Q.3
Directions: In the following questions two equations numbered I and II are given. You have to solve both the equations.

I. 5x2- 24x +16 = 0
II. 5y2 + 29y + 20 = 0
A. X>Y
B. X ≥Y
C. X
D. X≤Y
E. X = Y or the relationship cannot be established
Answer : Option A
Explaination / Solution:

I. 5x2- 24x +16 = 0
x = 4/5, 4
II. 5y2 + 29y + 20 = 0
y = (-4/5, -5)
So X>Y

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Q.4
Direction: In the following question, there are two equations. Solve the equations and answer accordingly:
I. 6x2 + 46x +60 = 0
II. 4y2+ 29y + 45= 0
A. X>Y
B. X ≥Y
C. X
D. X≤Y
E. X = Y or the relationship cannot be established
Answer : Option E
Explaination / Solution:

I. 6x2 + 46x +60 = 0
3x2 + 23x + 30 = 0
3x2 + 18x + 5x + 30 = 0
3x(x+6)+5(x+6)=0
(x+6)(3x+5)=0
x = -5/3, -6

II. 4y2 + 29y + 45= 0
4y2 + 20y + 9y + 45= 0
4y(y+5)+9(y+5)=0
(4y+9)(y+5)=0
y = -5, -9/4

So Relationship cannot be established

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Q.5
Directions: In the following questions two equations numbered I and II are given. You have to solve both the equations.

I. 8x – y = 40
II. 4x + y = 32
A. X>Y
B. X ≥Y
C. X
D. X≤Y
E. X = Y or the relationship cannot be established
Answer : Option C
Explaination / Solution:

After solving the both equation got values of x = 6 and y = 8 So X
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Q.6
Directions: In the following questions two equations numbered I and II are given. You have to solve both the equations.
I. 4x2 - 24x + 32 = 0
II. y2 = (12)2 – 140
A. X>Y
B. X ≥Y
C. X
D. X≤Y
E. X = Y or the relationship cannot be established
Answer : Option B
Explaination / Solution:

I. 4x2 - 24x + 32 = 0
x = (4, 2)
II. y = (+2,-2)
So X ≥Y

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Q.7
Directions: Study the given information carefully and answer the questions that follow—
A store contains 5 red, 4 blue, 5 green shirts.
If two shirts are picked at random, what is the probability that both are blue?
A. 3/91
B. 1/17
C. 6/91
D. 1/31
E. None of these
Answer : Option C
Explaination / Solution:

Probabilities if both are blue
4C2/14C2 = 6/91

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Q.8
Directions: Study the given information carefully and answer the questions that follow—
A store contains 5 red, 4 blue, 5 green shirts.
If three shirts are picked at random, what is the probability that two are blue and one is red?
A. 15/182
B. 8/143
C. 5/143
D. 17/182
E. None of these
Answer : Option A
Explaination / Solution:

Probability if two are Blue and one are Red
[(4C2*5C1)/14C3] = (6*5)/364 -->15/182

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Q.9
Directions: Study the given information carefully and answer the questions that follow—
A store contains 5 red, 4 blue, 5 green shirts.
If two shirts are picked at random, what is the probability that at least one is green?
A. 19/31
B. 55/91
C. 36/91
D. 4/11
E. None of these
Answer : Option B
Explaination / Solution:

Probability if at least one is Green
[1-(9C2/14C2)] = 55/91

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Q.10
Directions: Study the given information carefully and answer the questions that follow—
A store contains 5 red, 4 blue, 5 green shirts.
If two shirts are picked at random, what is the probability that either both are red or both are green?
A. 22/91
B. 1/13
C. 5/9
D. 20/91
E. None of these
Answer : Option D
Explaination / Solution:

Probabilities if both either are Red or either are green
(5C2 + 5C2)/ 14C2 = (10+10)/91 --> 20/91

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