CAT Exam > CAT Questions > Three unbiased dice are thrown simultaneously...
Start Learning for Free
Three unbiased dice are thrown simultaneously. What is the probability that the sum of the three numbers on them is divisible by 4?
- a)0.254
- b)0.187
- c)0.333
- d)0.286
Correct answer is option 'A'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
Three unbiased dice are thrown simultaneously. What is the probability...
When three unbiased dice are thrown, there are 63 = 216 combinations possible.
The least possible sum is 3 and the highest possible sum is 18.
The multiples of 4 in this range are 4, 8, 12 and 16
The least possible sum is 3 and the highest possible sum is 18.
The multiples of 4 in this range are 4, 8, 12 and 16
Sum = 4:
Obtained for the combination (1,1, 2).
This combination can be made in 3!/2! = 3 ways
This combination can be made in 3!/2! = 3 ways
Sum = 8:
Obtained for the combinations-
(1, 1,6)- possible in 3!/2! = 3 ways
(1,2, 5) - possible in 3! = 6 ways
(1, 3, 4) - possible in 3! = 6 ways
(2, 2, 4) - possible in 3!/2! = 3 ways
(2, 3, 3) - possible in 3!/2! = 3 ways
(1, 1,6)- possible in 3!/2! = 3 ways
(1,2, 5) - possible in 3! = 6 ways
(1, 3, 4) - possible in 3! = 6 ways
(2, 2, 4) - possible in 3!/2! = 3 ways
(2, 3, 3) - possible in 3!/2! = 3 ways
Total number of ways = 3 + 6 + 6 + 3 + 3 = 21
Sum = 12:
Obtained for the combinations-
(1, 5, 6) - possible in 3! = 6 ways
(2, 4, 6) - possible in 3! = 6 ways
(2, 5, 5) - possible in 3!/2! = 3 ways
(3, 3, 6) - possible in 3!/2! = 3 ways
(3, 4, 5) - possible in 3! = 6 ways
(4, 4, 4) - possible in 1 way
Total number of ways = 6 + 6 + 3 + 3 + 6 + 1 =25
(1, 5, 6) - possible in 3! = 6 ways
(2, 4, 6) - possible in 3! = 6 ways
(2, 5, 5) - possible in 3!/2! = 3 ways
(3, 3, 6) - possible in 3!/2! = 3 ways
(3, 4, 5) - possible in 3! = 6 ways
(4, 4, 4) - possible in 1 way
Total number of ways = 6 + 6 + 3 + 3 + 6 + 1 =25
Sum = 16:
Obtained for the combinations-
(4, 6, 6) - possible in 3!/2! = 3 ways
(5, 5, 6) - possible in 3!/2! = 3 ways
Total number of ways = 3 + 3 = 6
Overall total unmber of ways = 3 + 21 + 25 + 6 = 55
Required probability = 55/216 = 0.254 Hence, option 1.
(4, 6, 6) - possible in 3!/2! = 3 ways
(5, 5, 6) - possible in 3!/2! = 3 ways
Total number of ways = 3 + 3 = 6
Overall total unmber of ways = 3 + 21 + 25 + 6 = 55
Required probability = 55/216 = 0.254 Hence, option 1.
Most Upvoted Answer
Three unbiased dice are thrown simultaneously. What is the probability...
Introduction:
We are given three unbiased dice and we need to find the probability that the sum of the three numbers on them is divisible by 4. To solve this problem, we can use the concept of probability and combinatorics.
Approach:
To find the probability, we need to determine the favorable outcomes and the total possible outcomes.
Favorable outcomes:
For the sum to be divisible by 4, we need to consider the possible combinations of numbers on the three dice that add up to a multiple of 4. Let's list down all the possible outcomes:
- (1,1,2)
- (1,2,1)
- (2,1,1)
- (2,2,2)
- (1,1,6)
- (1,6,1)
- (6,1,1)
- (2,2,6)
- (2,6,2)
- (6,2,2)
- (3,3,4)
- (3,4,3)
- (4,3,3)
- (4,4,4)
- (3,3,8)
- (3,8,3)
- (8,3,3)
- (4,4,8)
- (4,8,4)
- (8,4,4)
- (5,5,6)
- (5,6,5)
- (6,5,5)
- (5,5,10)
- (5,10,5)
- (10,5,5)
- (6,6,8)
- (6,8,6)
- (8,6,6)
- (7,7,10)
- (7,10,7)
- (10,7,7)
- (8,8,8)
- (7,7,12)
- (7,12,7)
- (12,7,7)
- (8,8,12)
- (8,12,8)
- (12,8,8)
- (9,9,12)
- (9,12,9)
- (12,9,9)
- (10,10,12)
- (10,12,10)
- (12,10,10)
- (11,11,12)
- (11,12,11)
- (12,11,11)
- (12,12,12)
Therefore, there are 45 favorable outcomes.
Total possible outcomes:
Since each dice has 6 faces, the total possible outcomes for the three dice can be calculated as 6 * 6 * 6 = 216.
Probability:
The probability is given by the ratio of favorable outcomes to total possible outcomes.
Probability = Favorable outcomes / Total possible outcomes
Probability = 45 / 216
Probability = 0.208
Therefore, the probability that the sum of the three numbers on the dice is divisible by 4 is approximately 0.208.
Conclusion:
The probability that the sum of the numbers on the three unbiased dice is divisible by 4 is approximately 0.208.
We are given three unbiased dice and we need to find the probability that the sum of the three numbers on them is divisible by 4. To solve this problem, we can use the concept of probability and combinatorics.
Approach:
To find the probability, we need to determine the favorable outcomes and the total possible outcomes.
Favorable outcomes:
For the sum to be divisible by 4, we need to consider the possible combinations of numbers on the three dice that add up to a multiple of 4. Let's list down all the possible outcomes:
- (1,1,2)
- (1,2,1)
- (2,1,1)
- (2,2,2)
- (1,1,6)
- (1,6,1)
- (6,1,1)
- (2,2,6)
- (2,6,2)
- (6,2,2)
- (3,3,4)
- (3,4,3)
- (4,3,3)
- (4,4,4)
- (3,3,8)
- (3,8,3)
- (8,3,3)
- (4,4,8)
- (4,8,4)
- (8,4,4)
- (5,5,6)
- (5,6,5)
- (6,5,5)
- (5,5,10)
- (5,10,5)
- (10,5,5)
- (6,6,8)
- (6,8,6)
- (8,6,6)
- (7,7,10)
- (7,10,7)
- (10,7,7)
- (8,8,8)
- (7,7,12)
- (7,12,7)
- (12,7,7)
- (8,8,12)
- (8,12,8)
- (12,8,8)
- (9,9,12)
- (9,12,9)
- (12,9,9)
- (10,10,12)
- (10,12,10)
- (12,10,10)
- (11,11,12)
- (11,12,11)
- (12,11,11)
- (12,12,12)
Therefore, there are 45 favorable outcomes.
Total possible outcomes:
Since each dice has 6 faces, the total possible outcomes for the three dice can be calculated as 6 * 6 * 6 = 216.
Probability:
The probability is given by the ratio of favorable outcomes to total possible outcomes.
Probability = Favorable outcomes / Total possible outcomes
Probability = 45 / 216
Probability = 0.208
Therefore, the probability that the sum of the three numbers on the dice is divisible by 4 is approximately 0.208.
Conclusion:
The probability that the sum of the numbers on the three unbiased dice is divisible by 4 is approximately 0.208.
Attention CAT Students!
To make sure you are not studying endlessly, EduRev has designed CAT study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in CAT.
|
Explore Courses for CAT exam
|
|
Similar CAT Doubts
Top Courses for CATView all
Three unbiased dice are thrown simultaneously. What is the probability that the sum of the three numbers on them is divisible by 4?a)0.254b)0.187c)0.333d)0.286Correct answer is option 'A'. Can you explain this answer?
Question Description
Three unbiased dice are thrown simultaneously. What is the probability that the sum of the three numbers on them is divisible by 4?a)0.254b)0.187c)0.333d)0.286Correct answer is option 'A'. Can you explain this answer? for CAT 2024 is part of CAT preparation. The Question and answers have been prepared according to the CAT exam syllabus. Information about Three unbiased dice are thrown simultaneously. What is the probability that the sum of the three numbers on them is divisible by 4?a)0.254b)0.187c)0.333d)0.286Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for CAT 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Three unbiased dice are thrown simultaneously. What is the probability that the sum of the three numbers on them is divisible by 4?a)0.254b)0.187c)0.333d)0.286Correct answer is option 'A'. Can you explain this answer?.
Three unbiased dice are thrown simultaneously. What is the probability that the sum of the three numbers on them is divisible by 4?a)0.254b)0.187c)0.333d)0.286Correct answer is option 'A'. Can you explain this answer? for CAT 2024 is part of CAT preparation. The Question and answers have been prepared according to the CAT exam syllabus. Information about Three unbiased dice are thrown simultaneously. What is the probability that the sum of the three numbers on them is divisible by 4?a)0.254b)0.187c)0.333d)0.286Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for CAT 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Three unbiased dice are thrown simultaneously. What is the probability that the sum of the three numbers on them is divisible by 4?a)0.254b)0.187c)0.333d)0.286Correct answer is option 'A'. Can you explain this answer?.
Solutions for Three unbiased dice are thrown simultaneously. What is the probability that the sum of the three numbers on them is divisible by 4?a)0.254b)0.187c)0.333d)0.286Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for CAT.
Download more important topics, notes, lectures and mock test series for CAT Exam by signing up for free.
Here you can find the meaning of Three unbiased dice are thrown simultaneously. What is the probability that the sum of the three numbers on them is divisible by 4?a)0.254b)0.187c)0.333d)0.286Correct answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
Three unbiased dice are thrown simultaneously. What is the probability that the sum of the three numbers on them is divisible by 4?a)0.254b)0.187c)0.333d)0.286Correct answer is option 'A'. Can you explain this answer?, a detailed solution for Three unbiased dice are thrown simultaneously. What is the probability that the sum of the three numbers on them is divisible by 4?a)0.254b)0.187c)0.333d)0.286Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of Three unbiased dice are thrown simultaneously. What is the probability that the sum of the three numbers on them is divisible by 4?a)0.254b)0.187c)0.333d)0.286Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Three unbiased dice are thrown simultaneously. What is the probability that the sum of the three numbers on them is divisible by 4?a)0.254b)0.187c)0.333d)0.286Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice CAT tests.
|
|
Explore Courses for CAT exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup