A 5 digit number is formed by using digits 1, 2, 3, 4 and 5 with no two digits same. What is the probability that the formed number is divisible by 4?

**(A)** 1/5

**(B)** 4/5

**(C)** 3/5

**(D)** 1/4

**Answer:** **(A)** **Explanation:** To be divisible by 4, the number has to end with the last two digits forming a number divisible by 4. Since there are only 5 digits: 1, 2, 3, 4, 5.

So, possible last two digits according to divisibility by 4:

12

24

32

52

For each of them we now have to put the remaining numbers in front: 3x2x1, therefore 6 combinations for each number set = 6×4 (we have 4 sets from above) = 24.

So, required probability is:

P = Number of desired outcomes / number of possible outcomes

= 24 /(5*4*3*2*1)

= 24 / (5!)

= 24 / 120

= 1/5

So, option (A) is correct.

Quiz of this Question

## Recommended Posts:

- Aptitude | Probability | Question 1
- Aptitude | Probability | Question 2
- Aptitude | Probability | Question 3
- Aptitude | Probability | Question 4
- Aptitude | Probability | Question 5
- Aptitude | Probability | Question 6
- Aptitude | Probability | Question 7
- Aptitude | Probability | Question 9
- Aptitude | Probability | Question 10
- Aptitude | Probability | Question 1
- QA - Placement Quizzes | Probability | Question 11
- QA - Placement Quizzes | Probability | Question 12
- QA - Placement Quizzes | Probability | Question 13
- QA - Placement Quizzes | Probability | Question 14
- QA - Placement Quizzes | Probability | Question 15
- Probability of Knight to remain in the chessboard
- Probability of a random pair being the maximum weighted pair
- Probability of choosing a random pair with maximum sum in an array
- Select a Random Node from a tree with equal probability
- Probability of reaching a point with 2 or 3 steps at a time