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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?
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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
Sum = 4:
Obtained for the combination (1,1, 2).
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
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
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
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
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
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
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
Most Upvoted 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
Sum = 4:
Obtained for the combination (1,1, 2).
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
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
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
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
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
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
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
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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?
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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?.
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