Probability and Statistics | Mathematics or Quantitative Aptitude

CAT Entrance Exams Mathematics or Quantitative Aptitude # Probability and Statistics Probability Prepare / Learn

Q1.

Two coins are tossed once ,where E :no tail appears , F : no head appers. Find P(E/F).

A. 0.22

B. 0.24

C. 0

D. 0.25

Answer : Option C
Explaination / Solution:

CAT Entrance Exams Mathematics or Quantitative Aptitude # Probability and Statistics Probability and Statistics Prepare / Learn

Q2. 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

CAT Entrance Exams Mathematics or Quantitative Aptitude # Probability and Statistics Statistics Prepare / Learn

Q3. If the coefficient of variation between x and y is 0.28, covariance between x and y is 7.6, and the variance of x is 9, then the S.D. of the y series is

A. 10.05

B. 9.8

C. 9.05

D. 10.1

Answer : Option C
Explaination / Solution:

CAT Entrance Exams Mathematics or Quantitative Aptitude # Probability and Statistics Probability Prepare / Learn

Q4. Given that E and F are events such that P(E) = 0.6, P(F) = 0.3 and P(E ∩ F) = 0.2, find P (E|F) and P(F|E)

A. P (E|F) =

\frac{2}{4}

,P(F|E) =

\frac{1}{3}

B. P (E|F) =

\frac{2}{3}

,P(F|E) =

\frac{1}{4}

C. P (E|F) =

\frac{2}{5}

,P(F|E) =

\frac{1}{4}

D. P (E|F) =

\frac{2}{4}

,P(F|E) =

\frac{1}{3}

Answer : Option A
Explaination / Solution:

We have ,
P(E) = 0.6, P(F) = 0.3 and P(E ∩ F) = 0.2, then ,

P (E / F) = \frac{P (E \cap F)}{P (F)} = \frac{0.2}{0.3} = \frac{2}{3} P (F / E) = \frac{P (E \cap F)}{P (E)} = \frac{0.2}{0.6} = \frac{1}{3}

CAT Entrance Exams Mathematics or Quantitative Aptitude # Probability and Statistics Probability and Statistics Prepare / Learn

Q5. 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

CAT Entrance Exams Mathematics or Quantitative Aptitude # Probability and Statistics Statistics Prepare / Learn

Q6. The Q.D. of the daily wages (in Rs) of 7 persons given below: 12,7,15,10,17,19,25 is

A. 4.5

B. 9

C. 14.5

D. 5

Answer : Option A
Explaination / Solution:

the given data in ascending order is

$7, 10, 12, 15, 17, 19, 25$

$Q_{1} = \frac{{(n + 1)}^{t h} t e r m}{4} = 2^{n d} t e r m = 10$

$Q_{3} = \frac{3 {(n + 1)}^{t h} t e r m}{4} = 6^{t h} t e r m = 19$

$Q . D . = \frac{Q_{3} - Q_{1}}{2} = \frac{19 - 10}{2} = \frac{9}{2} = 4.5$

CAT Entrance Exams Mathematics or Quantitative Aptitude # Probability and Statistics Probability Prepare / Learn

Q7. Compute P(A|B), if P(B) = 0.5 and P (A ∩ B) = 0.32

A. P(A|B) =

\frac{16}{33}

B. P(A|B) =

\frac{16}{33}

C. P(A|B) =

\frac{15}{27}

D. P(A|B) =

\frac{16}{29}

Answer : Option A
Explaination / Solution:

We have , P(B) = 0.5 and P (A ∩ B) = 0.32

P (A / B) = \frac{P (A \cap B)}{P (B)} = \frac{0.32}{0.5} = \frac{16}{25}

CAT Entrance Exams Mathematics or Quantitative Aptitude # Probability and Statistics Probability and Statistics Prepare / Learn

Q8. 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

CAT Entrance Exams Mathematics or Quantitative Aptitude # Probability and Statistics Statistics Prepare / Learn

Q9. Which of the following is not a measure of central tendency :

A. median

B. mode

C. range

D. mean

Answer : Option C
Explaination / Solution:

meausre of central tendencies give the middle most or average value whereas range gives the difference between highest and lowest value.

CAT Entrance Exams Mathematics or Quantitative Aptitude # Probability and Statistics Probability Prepare / Learn

Q10. If P (A) = 0.8, P (B) = 0.5 and P(B|A) = 0.4, find P(A ∩ B)

A. 0.32

B. 0.29

C. 0.35

D. 0.37

Answer : Option A
Explaination / Solution:

We have ,
P (A) = 0.8, P (B) = 0.5 and P(B|A) = 0.4

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