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"GMAT - Statistics and Probability Problems with Answers"

Graduate Management Admission Test - Statistics and Probability Problems and Answers with Explanation. Practice Questions with Solutions presented to you by MBA Rendezvous.
GMAT - Statistics and Probability Problems with Answers

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Home » GMAT » QA » GMAT - Statistics and Probability Problems with Answers

Statistics and Probability

1. Which of the following is not a measure of central tendency?
    A. mean
    B. mode
    C. median
    D. variance
    E. None of these

2. Consider the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. If 1 is added to each number, the variance of the numbers so obtained is

    A. 3.87
    B. 8.25
    C. 1.87
    D. 2.87
    E. Both A and B

3. If the mean of x1, x2, …., xn is  , then mean of n observations 2x1 + 3, 2x2 + 3,....,2xn + 3, is
    A.  
    B.  
    C.  
    D.  
    E. None of these

4. In an arranged data of n observations (n being an even number),the median is the value of

A.  (n+1/2)th item

B.  n/2th item

C.  1/2{(n/2)th item + (n/2 + 1)th item}

D.  (n+2)th item

E. None of these

5. The median of a set of 9 observations is 20.5. If each of the largest 4 observations is increased by 2, then the median of the new set is

    A. remains unchanged

    B. is increased by 2

    C. is decreased by 2

    D. Both B and C

    E. None of these

6. Three digit numbers are formed using the digits 0, 2, 4, 6, 8. A number is chosen at random out of these numbers. The probability that this number has the same digits is

A.  1/25

B. 1/16

C. 16/25

D. 1/645

E. None of these

7. One mapping (function is selected at random from all the mappings of the set A = {1, 2, 3, …, n} into itself. The chance that the selected mapping is one-one is

A. 1/n

B.  
C.  

D. none of these

8. Without repetition of the numbers, four digited numbers are formed with the numbers 0, 2, 3, 5. The chance of such a number being divisible by 5 is

A. 1/5

B. 4/5

C. 1/30

D. 5/9

E. None of these

9. If odds against an event E are m : n, then P (E) =

A.  m/n

B.  n/m

C. m/m+n

D.  n/(m+n)

E. None of these

10. If the probability for A to fail in an examination is 0.2 and that for B is 0.3, then the probability that either A or B fails is

A. > 0.5

B. < 0.5

C. 0.5

D. 0

E. None of these

11. Three squares of chess board are selected at random. The probability of getting 2 squares of one colour and the third of the other colour is

A. 16/21

B. 8/21

C. 3/32

D. 3/8 
E. None of these

12. A coin is tossed again and again until it shows up a head or it has been tossed three times. The probability that the coin is tossed three times is

A. 1/8

B. 1/4

C. 1/2

D. 1/3

E. None of these

13. There are four letters and four directed envelopes, Letters are put into the envelope without taking care of the addresses on them. The probability that none of the letters is put into the right envelope is

A. 23/24

B. 1/24

C. 9/16

D. 9/24

E. None of these

14. If A and B are any two events associated with an experiment, such that

P(A) = 1/2, P(B) = 2/3 then

A.

B .

C.

D.

E. None of these

15. Three numbers are chosen from 1 to 30. The probability that they are consecutive is

A.  142/145

B.  144/145

C.  143/145

D.  140/145

E. None of these

Solutions

1. D
    Variance is not a measure of central tendency. It is a measure of dispersion.

2. B

S.D. of a data is not changed, if each item is increased or decreased by the same quantity.

3. B

4. C

Median is 1/2 {(n/2)th item + (n/2 + 1)th item}  when n is even.

5. A
Since median is (n + 1/2) th i.e. 5th item, therefore in this case the median will remain the same.

6. A

Note that 0 cannot be placed in hundres’s place.

Total number of 3-digit numbers that can be formed = 4 × 5 × 5 = 100.

Out of these, only four numbers 222, 444, 666 and 888 have the same digits.

7. C

Total number of functions that can be defined from A to A is nn as an element of A can be associated to an element of A in n ways.

Out of those nn functions, the number of one-one (and hence onto also) mappins is equal to the number of permutations of these n numbers taken all at a time,

which is equal to 

Hence, the required probability =

8. D

Note that 0 cannot be placed in thousand’s place.

So thousand’s place can be filled in 3 ways only.

Remaining three places can be filled in with remaining three digits in 3P3 ways.

Total number of 4-digited numbers = 3 × 3P3 = 3 ×   = 18.

The numbers which are divisible by 5 have either 0 or 5 in the unit’s place.

Number of numbers divisible by 5


 = 6 + 2 × 2 = 10.

9.  D

10. B

11. A

3 squares out of 64 can be selected in 64C3 ways

2 white and 1 black or 1 white and 2 black squares can be selected in 32C2 × 32C1 + 32C1 × 32C2 ways.

Required probability = 

12. B

In this case the same space is

S = {H, TH, TTH, TTT}

Clearly, 

and P(TTT) = P(TTH) = 1/8.

P (coin is tossed three times) = P({TTT, TTH}) = P(TTT) + P(TTH) =  1/8 + 1/8 = 1/4

13. D
Four letters can be put into four envelopes in    ways. The number of ways in which each of four letters is put into wrong envelope = number of dearrangements of 4 distinct objects
    = 4P4 – 4P3 + 4P2 – 4P1 + 4P0
    
    = 24 – 24 + 12 – 4 + 1 = 9
    (If n distinct objects are arranged in a row, then the number of their dearrangements such that no object remains in its original position = nPnnPn–1 + nPn–2 – ….. + (–1)nnP0)

14. A

As 

therefore, 

Similarly, 

and 

15. D

Three numbers out of 30 can be chosen in  30C3 ways.These numbers can be consecutive in 28 ways as they can be

Required probability 

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