# Modern Math - Probability - Previous Year CAT/MBA Questions

You can practice all previous year CAT questions from the topic Modern Math - Probability. This will help you understand the type of questions asked in CAT. It would be best if you clear your concepts before you practice previous year CAT questions.

**CAT 1995 QA | Modern Math - Probability**

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.
$\frac{1}{12}$

- B.
$\frac{1}{4}$

- C.
$\frac{1}{6}$

- D.
None of these

Answer: Option A

**Explanation** :

Total number of four-digit numbers that can be formed = 4!.

If the number is divisible by 25, then the last two digits are 25.

So the first two digits can be arranged in 2! ways.

Hence, required probability = $\frac{2!}{4!}=\frac{1}{12}.$

Workspace:

**CAT 1994 QA | Modern Math - Probability**

**Data is provided followed by two statements – I and II – both resulting in a value, say I and II. **

As your answer,

Type 1, if I > II.

Type 2, if I < II.

Type 3, if I = II.

Type 4, 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.

Answer: 2

**Explanation** :

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.

Workspace:

**CAT 1993 QA | Modern Math - Probability**

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: Option C

**Explanation** :

Since there are 6 red balls and all six of them are of different sizes, probability of choosing the smallest among them is $\frac{1}{6}$.

Workspace:

**CAT 1991 QA | Modern Math - Probability**

A player rolls a die and receives the same number of rupees as the number of dots on the face that turns up. What should the player pay for each roll if he wants to make a profit of one rupee per throw of the die in the long run?

- A.
Rs. 2.50

- B.
Rs. 2

- C.
Rs. 3.50

- D.
Rs. 4

Answer: Option A

**Explanation** :

Since in the long run the probability of each number appearing is the same, we can say in ‘n’ throws one can get 1, 2, 3, 4, 5 and 6, n/6 times each.

Hence he would earn (1 + 2 + 3 + 4 + 5 + 6)n/6 = Rs. 7n/2 = Rs. 3.5n.

∴ His average earning for each through is 3.5n/n = Rs. 3.5

In order to make a profit of 1 Re. per throw his cost for the each through should be = 3.5 – 1 = Rs. 2.50

Hence, option (a).

Workspace:

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