Probability and Statistics | Mathematics or Quantitative Aptitude

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Q1. The most stable measure of central tendency is

A. mode

B. median

C. mean

D. range

Answer : Option C
Explaination / Solution:

because takes into account every value in the given set of data to provide the average

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Q2. If E, F and G are events with P(G) ≠ 0 then P ((E ∪ F)|G) given by

A. P (G|E) + P (F|G) – P ((E ∩ F)|E)

B. P (E|G) + P (G|F) – P ((E ∩ F)|G)

C. P (E|G) + P (F|G) – P ((E ∩ F)|G)

D. P (E|G) + P (F|G) – P ((E ∩ F)|F)

Answer : Option C
Explaination / Solution:

P(EUF/G) = $\frac{P [(E U F) \cap G]}{P (G)}$ = $\frac{P [(E \cap G) U (F \cap G]}{P (G)}$ $\frac{P (E \cap G) + P (F \cap G) - P (E \cap G \cap F \cap G) }{P (G)}$

= $\frac{P (E \cap G) }{P (G)}$ + $\frac{P (F \cap G) }{P (G)}$ - $\frac{P (E \cap F \cap G) }{P (G)}$

= P (E|G) + P (F|G) – P ((E ∩ F)|G)

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Q3.

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 at least 2 trainees to have to be included in a committee

A. 340

B. 820

C. 560

D. 405

E. None of these

Answer : Option D
Explaination / Solution:

If at least 2 trainees include in committee

³C₂*¹⁰C₃ + ³C₃*¹⁰C₂ = 3*120+1*45

= 405

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Q4. The mean weight of a group of 10 items is 28 and that of another group of n items is 35.The mean of combined group of 10 + n items is found to be 30. The value of n is

A. 4

B. 2

C. 10

D. 12

Answer : Option A
Explaination / Solution:

sum of weights of 10 items =280

sum of weights of n items = 35n

so, sum of weights of ( 10 + n) items = 280 + 35n

so ,mean = (280 + 35 n) / ( 10 + n)

30(10 + n ) = 280 + 35n

solving we get, n = 4

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Q5. If E and F are events then P (E ∩ F) =

A. P (E) P (E|F), P (E) ≠ 0

B. P (E ∩ F) P (F|E), P (E) ≠ 0

C. P (E) P (F|E), P (E) ≠ 0

D. P (E∪F) P (F|E), P (E) ≠ 0

Answer : Option C
Explaination / Solution:

If E and F are events then P (E ∩ F) = P (E) P (F|E), P (E) ≠ 0. By the defination of conditional probability of two events

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Q6.

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 3 engineers and 2 professors or 2 trainees and 3 professor be included in a committee

A. 140

B. 160

C. 132

D. 155

E. None of these

Answer : Option C
Explaination / Solution:

If 3 engineers and 2 professors or 2 trainees and 3 professors include in committee

⁶C₃*⁴C₂ + ³C₂*⁴C₃ = 20*6 + 3*4 = 120 + 12 = 132

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Q7. The measure of variation which is least affected by extreme items is

A. range

B. mean deviation

C. quartile deviation

D. standard deviation

Answer : Option C
Explaination / Solution:

as it only depends on upper and lower quartile not on extreme values

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Q8. Two coins are tossed once ,where E : tail appears on one coin , F : one coin shows head. Find P(E/F).

A. 0.24

B. 0.33

C. 1

D. 0.23

Answer : Option C
Explaination / Solution:

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Q9.

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

# CAT Entrance Exams # Mathematics or Quantitative Aptitude # Probability and Statistics # CBSE 11th Mathematics Prepare / Learn

Q10. A group of 10 items has mean 6. If the mean of 4 of these items is 7.5, then the mean of the remaining items is

A. 5.5

B. 4.5

C. 6.5

D. 5.0

Answer : Option D
Explaination / Solution:

sum of 10 items = 60

sum of 4 items = 30

sum of remaining items = 30

their mean = 30/6=5

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