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Probability

Q.No: 1
Test Name : CAT Paper 2008

Five horses, Red, White, Grey, Black and Spotted participated in a race. As per the rules of the race, the persons betting on the winning horse get four times the bet amount and those betting on the horse that came in second get thrice the bet amount. Moreover, the bet amount is returned to those betting on the horse that came in third, and the rest lose the bet amount. Raju bets Rs. 3000, Rs. 2000 and Rs. 1000 on Red, White and Black horses respectively and ends up with no profit and no loss.


Which of the following cannot be true?

A
At least two horses finished before Spotted
B
Red finished last
C
There were three horses between Black and Spotted
D
There were three horses between White and Red
E
Grey came in second
Answer key/Solution
Solution:
Q.No: 2
Test Name : CAT Paper 2008

Five horses, Red, White, Grey, Black and Spotted participated in a race. As per the rules of the race, the persons betting on the winning horse get four times the bet amount and those betting on the horse that came in second get thrice the bet amount. Moreover, the bet amount is returned to those betting on the horse that came in third, and the rest lose the bet amount. Raju bets Rs. 3000, Rs. 2000 and Rs. 1000 on Red, White and Black horses respectively and ends up with no profit and no loss.


Suppose, in addition, it is known that Grey came in fourth. Then which of the following cannot be true?

A
Spotted came in first
B
Red finished last
C
White came in second
D
Black came in second
E
There was one horse between Black and White
Answer key/Solution
Solution:
Q.No: 3
Test Name : CAT Paper 1991
Q 61 to 100 : Choose the best answer choice from those provided

Q.75 and 76 are based on the given data:
There were a hundred schools in a town. Of these, the number of schools having a play − ground was 30, and these schools had neither a library nor a laboratory. The number of schools having a laboratory alone was twice the number of those having a library only. The number of schools having a laboratory as well as a library was one fourth the number of those having a laboratory alone. The number of schools having either a laboratory or a library or both was 35.

What was the ratio of schools having laboratory to those having library?

A
1 : 2
B
5 : 3
C
2 : 1
D
2 : 3
E
NA
Answer key/Solution
Solution:
Q.No: 4
Test Name : CAT Paper 1993
Q58 to 100 : Choose the appropriate answer choice.

There are ten 50 paise coins placed on a table. Six of these show tails, four show heads. A coin is chosen at random and flipped over (not tossed). This operation is performed seven times. One of the coins is then covered. Of the remaining nine coins, five show tails and four show heads. The covered coin shows

A
a head
B
a tail
C
more likely a head
D
more likely a tail
Answer key/Solution
Solution:
The initial reading for 10 coins is : 6 Tails and 4 Heads After repeating the process of flipping one coin at random for 7 times, the final reading for 9 coins is: 5 Tails and 4 Heads.
Therefore, possible final reading for 10 coins is: 6 Tails and 4 Heads or 5 Tails and 5 Heads.
If the final reading is 6T and 4H, it is same as the initial one. However, this is not possible as the process of flipping a coin has taken place an odd number of times, so there has to be atleast one change in the final reading.
Therefore, the final reading is 5T and 5H.
So the covered coin will certainly be a Head.
Q.No: 5
Test Name : CAT Paper 1993
Q58 to 100 : Choose the appropriate answer choice.

A box contains 6 red balls, 7 green balls and 5 blue balls. Each ball is of a different size. The probability that the red ball selected is the smallest red ball, is

A
1/18
B
1/3
C
1/6
D
2/3
Answer key/Solution
Solution:
Q.No: 6
Test Name : CAT Paper 1994
Q51 – 90 : Choose the best alternative.

Q81 – 85 : Data is provided followed by two statements – I and II – both resulting in a value, say I and II. As your answer,
Mark (a) if I > II.
Mark (b) if I < II.
Mark (c) if I = II.
Mark (d) if nothing can be said.

I. The probability of encountering 54 Sundays in a leap year.
II. The probability of encountering 53 Sundays in a non-leap year.

A
a
B
b
C
c
D
d
Answer key/Solution
Solution:
53 Sundays can occur in a non-leap year, if 1st January is either a Saturday or a Sunday. But 54 Sundays can never occur. Hence, I < II.
Q.No: 7
Test Name : CAT Paper 1995

Direction for questions 58 to 87: Answer the questions independently.

If a 4 digit number is formed with digits 1, 2, 3 and 5. What is the probability that the number is divisible by 25, if repetition of digits is not allowed?

A
B
C
D
None of these
Answer key/Solution
Solution:
Q.No: 8
Test Name : CAT Paper 2003 (L)
DIRECTIONS for Questions 114 to 120: Answer the questions independently of each other.

Twenty-seven persons attend a party. Which one of the following statements can never be true?

A
There is a person in the party who is acquainted with all the twenty-six others.
B
Each person in the party has a different number of acquaintances.
C
There is a person in the party who has an odd number of acquaintances.
D
In the party, there is no set of three mutual acquaintances.
Answer key/Solution
Solution:
Q.No: 9
Test Name : CAT Paper 2003 (L)
DIRECTIONS for Questions 114 to 120: Answer the questions independently of each other.

Let g(x) = max(5 – x, x + 2). The smallest possible value of g(x) is

A
4.0
B
4.5
C
1.5
D
None of the above
Answer key/Solution
Solution:
We can see that x + 2 is an increasing function and 5 – x is a decreasing function. This system of equation will have smallest value at the point of intersection of the two. i.e. 5 – x = x + 2 or x = 1.5. Thus smallest value of g(x) = 3.5
Q.No: 10
Test Name : CAT Paper 2003 (L)
DIRECTIONS for Questions 114 to 120: Answer the questions independently of each other.

The function f(x) = |x – 2| + |2.5 – x| + |3.6 – x|, where x is a real number, attains a minimum at

A
x = 2.3
B
x = 2.5
C
x = 2.7
D
None of the above
Answer key/Solution
Solution:
Q.No: 11
Test Name : CAT Paper 2003 (L)
DIRECTIONS for Questions 126 to 150: Answer the questions independently of each other.

There are 6 boxes numbered 1,2,… 6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is

A
5
B
21
C
33
D
60
Answer key/Solution
Solution:
GRRRRR, RGRRRR, RRGRRR, RRRGRR, RRRRGR,
RRRRRG
GGRRRR, RGGRRR, RRGGRR, RRRGGR, RRRRGG
GGGRRR, RGGGRR, RRGGGR, RRRGGG
GGGGRR, RGGGGR, RRGGGG
GGGGGR, RGGGGG
GGGGGG
Hence 21 ways.
Q.No: 12
Test Name : CAT Paper 2003 (R)
Directions for questions 60 to 93: Answer the following questions independently.

In a coastal village, every year floods destroy exactly half of the huts. After the flood water recedes, twice the number of huts destroyed are rebuilt. The floods occurred consecutively in the last three years — 2001, 2002 and 2003. If floods are expected again in 2004, the number of huts expected to be destroyed is

A
less than the number of huts existing at the beginning of 2001
B
less than the total number of huts destroyed by floods in 2001 and 2003
C
less than the total number of huts destroyed by floods in 2002 and 2003
D
more than the total number of huts built in 2001 and 2002
Answer key/Solution
Solution:
Solution:

Solution:

Solution:

Solution:
The initial reading for 10 coins is : 6 Tails and 4 Heads After repeating the process of flipping one coin at random for 7 times, the final reading for 9 coins is: 5 Tails and 4 Heads.
Therefore, possible final reading for 10 coins is: 6 Tails and 4 Heads or 5 Tails and 5 Heads.
If the final reading is 6T and 4H, it is same as the initial one. However, this is not possible as the process of flipping a coin has taken place an odd number of times, so there has to be atleast one change in the final reading.
Therefore, the final reading is 5T and 5H.
So the covered coin will certainly be a Head.

Solution:

Solution:
53 Sundays can occur in a non-leap year, if 1st January is either a Saturday or a Sunday. But 54 Sundays can never occur. Hence, I < II.

Solution:

Solution:

Solution:
We can see that x + 2 is an increasing function and 5 – x is a decreasing function. This system of equation will have smallest value at the point of intersection of the two. i.e. 5 – x = x + 2 or x = 1.5. Thus smallest value of g(x) = 3.5

Solution:

Solution:
GRRRRR, RGRRRR, RRGRRR, RRRGRR, RRRRGR,
RRRRRG
GGRRRR, RGGRRR, RRGGRR, RRRGGR, RRRRGG
GGGRRR, RGGGRR, RRGGGR, RRRGGG
GGGGRR, RGGGGR, RRGGGG
GGGGGR, RGGGGG
GGGGGG
Hence 21 ways.

Solution:

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