Three fair six-sided dice, each numbered 1 to 6, are rolled once. What

Hi Bunuel,
If I am not wrong, the OA suggests that the sequencing of the digits coming up on the dices is important too. Implying that only 1,2,3 can be considered but not 1,3,2. However, the question does not explicitly suggest whether we need to follow a sequence while looking at the dice outcomes.
Please let me know if my interpretation is incorrect.

Is the OA incorrect? Isn't it 1/36 as GMATist1 pointed out?

Bunuel
chetan2u
KarishmaB

What is the right answer to this question? Help plz...

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Given: Three fair six-sided dice, each numbered 1 to 6, are rolled once.

Asked: What is the probability that resulting three numbers are in arithmetic progression (an arithmetic progression is a sequence of numbers such that the difference between the consecutive terms is constant)?

Total ways = 6*6*6 = 6^3

Favorable ways =

d=1
{x,x+1,x+2} & {x+2,x+1,x}: {1,2,3},{2,3,4},{3,4,5},{4,5,6} & {3,2,1},{4,3,2},{5,4,3},{6,5,4} : 8 cases

d=2
{x,x+2,x+4} & {x+4,x+2,x}: {1,3,5},{2,4,6} & {5,3,1},{6,4,2} : 4 cases

d=3 onwards
{x,x+3,x+6}: 0 cases since x is minimum 1.

Favorable ways = 8+4 = 12

Probability that resulting three numbers are in arithmetic progression = 12/6^3 = 1/18

IMO B

Total outcomes = 6*6*6 = 216

Favourable outcomes = (1,2,3)(1,3,5)(2,3,4)(2,4,6)(3,4,5)(4,5,6)

Now, we CANNOT arrange each of these terms in 3! ways because then they would not be in AP

But there are 6 other favourable outcomes that we have to consider which are

= (3,2,1)(5,3,1)(4,3,2)(6,4,2)(5,4,3)(6,5,4)

These are the reverse of the 6 favourable outcomes chosen earlier and these are also valid because AP can consist of negative common difference

So total outcomes = 216
Favourable outcomes = 12

Probability = 12/216 = 1/18

Answer - B

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Hi Bunuel, Cant an AP have a common difference of Zero as well? So should we not take the sequence (1,1,1) (2,2,2) etc into account as well?

When Common difference = 1 or -1, we have 4*2 = 8 cases -- > (1,2,3), (2,3,4), (3,4,5), (4,5,6) + the same cases taken in reverse order
When Common difference = 2 or -2, we have 2*2 = 4 cases -- > (1,3,5), (2,4,6) + the same cases taken in reverse order
When Common difference = 0 , we have 6 cases -- > (1,1,1), (2,2,2) ................ (6,6,6)

In total = 18 cases
Total cases = 6*6*6 = 216 Cases

Probability = 18/216 = 1/12

Deleted

Had I got this question in the exam, I would have flipped between (D) and (E). Under no circumstances would I have thought that (B) is possible. We are given that the three dice (say red, blue and yellow) are rolled once. It seems to imply that they are rolled together once (not one after the other in sequence). The three numbers obtained would form an AP in case of any these results
R - 1, Y - 2, B - 3
R - 2, Y - 1, B - 3
etc.
This would give me 36 cases (as shown by others above)

If the dice were thrown one after another, I would assume that they would mention that the dice were thrown in a sequence one after another.

I would also wonder if {1, 1, 1}, {2, 2, 2} ... {6, 6, 6} i.e. another 6 cases are acceptable too. They should be since an AP can have all identical numbers. Then by total number of acceptable cases would become 36 + 6 = 42

Probability = 42/216 = 7/36

Completely agree with you!! Although I missed out on the part of considering {1, 1, 1}, {2, 2, 2} ... {6, 6, 6} as APs.

{1,2,3,4,5,6}
AP can happen if we get {1,2,3} {1,3,5} , {2,3,4}, {2,4,6}, {3,4,5},{4,5,6} 3! times or {x,x,x} (6 times)

P= (3!*6+6)/6^6
=7/36

­There are 6^3 total combinations = 216 possible outcomes

For numbers 1 to 6 below are the possible combinations in arithmetic progression

Combinations with difference = 1 
123 (or 321 or 312 or 213 or 231 or 132) (all 6 can be

arranged in the order 123) (diference = 1)

234 ((can be arranged in 3! or  6 ways)) 

345 ((can be arranged in 3! or  6 ways))

456 (can be arranged in 3! or  6 ways) 

here we have 6* 4 = 24 combinations

Combinations with difference = 2
246 (can be arranged in 3! or  6 ways)

1 3 5 (can be arranged in 3! or 6 ways )

here we have 6 + 6 =12 combinations

Combinations with difference = 0
1 1 1 

2 2 2

3 3 3

4 4 4

5 5 5

6 6 6

here we have 6 combinations

Adding all of the above combinations 12 + 24+ 6 = 42

P(arithmetic series arrangement) = 42/216 = 21 / 108 = 7/ 36