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Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
Three fair 6-sided dice are rolled.
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18 Aug 2021, 09:40
18 Aug 2021, 09:40
2
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00:00
Question Stats:
12% (02:01) correct
87% (01:22) wrong based on 54 sessions
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Three fair 6-sided dice are rolled.
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Quantity A |
Quantity B |
Total number of different outcomes where the order of the dice does not matter |
56 |
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Three fair 6-sided dice are rolled.
[#permalink]
19 Aug 2021, 06:32
19 Aug 2021, 06:32
5
This question include 3 possibilities:
1. All 3 dice have same number
2. Two of the dice have same number
3. All the numbers are different
First let's calculate the combinations for 1st case. Since there are only 6 numbers we would have 6 cases (i.e. all 1s, 2s....6s).
Second, we calculate for two dice having same number such as two dice has 1s and then third die could have any number from 2-6. So 1*5 for this case. Therefore we would have 6*5 = 30 combinations of outcomes.
Third, let's consider the cases which would have all different numbers. This would be 6*5*4=120 but we need to eliminate permutation such as {1,2,3},{1,3,2},{2,1,3},{2,3,1},{3,2,1},{3,1,2} as order doesn't matter. There are 3!=6 ways that each groups of the numbers can be ordered so we need to divide 120 by 6 in order to obtain the results where order doesn't matter. Thus there are 20 combinations where all three numbers are different.
The total number of combinations are 6+30+20=56
Hence the answer is option C
1. All 3 dice have same number
2. Two of the dice have same number
3. All the numbers are different
First let's calculate the combinations for 1st case. Since there are only 6 numbers we would have 6 cases (i.e. all 1s, 2s....6s).
Second, we calculate for two dice having same number such as two dice has 1s and then third die could have any number from 2-6. So 1*5 for this case. Therefore we would have 6*5 = 30 combinations of outcomes.
Third, let's consider the cases which would have all different numbers. This would be 6*5*4=120 but we need to eliminate permutation such as {1,2,3},{1,3,2},{2,1,3},{2,3,1},{3,2,1},{3,1,2} as order doesn't matter. There are 3!=6 ways that each groups of the numbers can be ordered so we need to divide 120 by 6 in order to obtain the results where order doesn't matter. Thus there are 20 combinations where all three numbers are different.
The total number of combinations are 6+30+20=56
Hence the answer is option C
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
Re: Three fair 6-sided dice are rolled.
[#permalink]
24 Aug 2021, 10:29
24 Aug 2021, 10:29
1
Bookmarks
Replying to a PM:
Same number = S
Different number = D
Possible Outcomes;
I. All numbers same
II. 2 numbers same and 1 different
III. All 3 numbers different
I. SSS = \((^6C_1)(1)(1) = 6\)
We can pick only one number which will be written thrice
II. SSD = \((^6C_1)(1)(^5C_1) = 30\)
We have 6 option to pick first number, then we can select that one number only for the next and then we can select any of the 5 numbers left
III. DDD = \((^6C_3) = 20\)
We can pick any 3 numbers from 6 given numbers
So total cases \(= 6 + 30 + 20 = 56\)
Col. A: \(56\)
Col. B: \(56\)
Hence, option C
Same number = S
Different number = D
Possible Outcomes;
I. All numbers same
II. 2 numbers same and 1 different
III. All 3 numbers different
I. SSS = \((^6C_1)(1)(1) = 6\)
We can pick only one number which will be written thrice
II. SSD = \((^6C_1)(1)(^5C_1) = 30\)
We have 6 option to pick first number, then we can select that one number only for the next and then we can select any of the 5 numbers left
III. DDD = \((^6C_3) = 20\)
We can pick any 3 numbers from 6 given numbers
So total cases \(= 6 + 30 + 20 = 56\)
Col. A: \(56\)
Col. B: \(56\)
Hence, option C
