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Probability and Statistics
- Online Test
Probability and Statistics
TEST : 3
# Probability and Statistics
# General Aptitude
Prepare / Learn
Q1.
Directions:
In the following questions two equations numbered I and II are given. You have to solve both the equations.
I. 8x
2
+ 30x + 28 = 0
II. 9y
2
+ 11y+2 =0
A.
X>Y
B.
X ≥Y
C.
X
D.
X≤Y
E.
X = Y or the relationship cannot be established
View Solution
Add Work Space
Answer :
Option C
Explaination / Solution:
I. 8x
2
+ 30x + 28 = 0
x = (-7/4, -2)
II. 9y
2
+ 11y+2 =0
y = (-1, -2/9)
So x<y
# Probability and Statistics
# General Aptitude
Prepare / Learn
Q2.
Directions:
In the following questions two equations numbered I and II are given. You have to solve both the equations.
I. 5x
2
- 24x +16 = 0
II. 5y
2
+ 29y + 20 = 0
A.
X>Y
B.
X ≥Y
C.
X
D.
X≤Y
E.
X = Y or the relationship cannot be established
View Solution
Add Work Space
Answer :
Option A
Explaination / Solution:
I. 5x
2
- 24x +16 = 0
x = 4/5, 4
II. 5y
2
+ 29y + 20 = 0
y = (-4/5, -5)
So X>Y
# Probability and Statistics
# General Aptitude
Prepare / Learn
Q3.
Direction: In the following question, there are two equations. Solve the equations and answer accordingly:
I. 6x
2
+ 46x +60 = 0
II. 4y
2
+ 29y + 45= 0
A.
X>Y
B.
X ≥Y
C.
X
D.
X≤Y
E.
X = Y or the relationship cannot be established
View Solution
Add Work Space
Answer :
Option E
Explaination / Solution:
I. 6x
2
+ 46x +60 = 0
3x
2
+ 23x + 30 = 0
3x
2
+ 18x + 5x + 30 = 0
3x(x+6)+5(x+6)=0
(x+6)(3x+5)=0
x = -5/3, -6
II. 4y
2
+ 29y + 45= 0
4y
2
+ 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
# Probability and Statistics
# General Aptitude
Prepare / Learn
Q4.
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
View Solution
Add Work Space
Answer :
Option C
Explaination / Solution:
After solving the both equation got values of x = 6 and y = 8 So X
# Probability and Statistics
# General Aptitude
Prepare / Learn
Q5.
Directions:
In the following questions two equations numbered I and II are given. You have to solve both the equations.
I. 4x
2
- 24x + 32 = 0
II. y
2
= (12)
2
– 140
A.
X>Y
B.
X ≥Y
C.
X
D.
X≤Y
E.
X = Y or the relationship cannot be established
View Solution
Add Work Space
Answer :
Option B
Explaination / Solution:
I. 4x
2
- 24x + 32 = 0
x = (4, 2)
II. y = (+2,-2)
So X ≥Y
# Probability and Statistics
# General Aptitude
Prepare / Learn
Q6.
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
View Solution
Add Work Space
Answer :
Option C
Explaination / Solution:
Probabilities if both are blue
4
C
2
/
14
C
2
= 6/91
# Probability and Statistics
# General Aptitude
Prepare / Learn
Q7.
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
View Solution
Add Work Space
Answer :
Option A
Explaination / Solution:
Probability if two are Blue and one are Red
[(
4
C
2
*
5
C
1
)/
14
C
3
] = (6*5)/364 -->15/182
# Probability and Statistics
# General Aptitude
Prepare / Learn
Q8.
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
View Solution
Add Work Space
Answer :
Option B
Explaination / Solution:
Probability if at least one is Green
[1-(
9
C
2
/
14
C
2
)] = 55/91
# Probability and Statistics
# General Aptitude
Prepare / Learn
Q9.
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
View Solution
Add Work Space
Answer :
Option D
Explaination / Solution:
Probabilities if both either are Red or either are green
(
5
C
2
+
5
C
2
)/
14
C
2
= (10+10)/91 --> 20/91
# Probability and Statistics
# General Aptitude
Prepare / Learn
Q10.
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 most one is blue?
A.
15/91
B.
85/91
C.
25/29
D.
75/91
E.
None of these
View Solution
Add Work Space
Answer :
Option B
Explaination / Solution:
Probabilities if at most one is blue =
[(
4
C
0
*
10
C
2
+
4
C
1
*
10
C
1
)/
14
C
2
] = (1*45 + 4*10)/( 91) = 85/91
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