When a die that has one of six consecutive integers on each

I think it is C

We know it is consecutive so it can be many diffrent things
a-1 2 3 4 5 6
b-2 3 4 5 6 7
c-3 4 5 6 7 8
ect.

statment 1 tells us that 8 is not there.
so it could be a or b or something else.

statment 2 tells us the 7 is one of the numbers.
because the probability not to get it is 5/6*5/6

so if 7 have to be and 8 cannot be there
it got to be 2 3 4 5 6 7 ( since they are consecutive integers)

Using the info from 2)

the only possible integers on the die are

2,3,4,5,6,7
3,4,5,6,7,8
4,5,6,7,8,9
5,6,7,8,9,10
6,7,8,9,10,11
7,8,9,10,11,12

In all of those cases the Probability of rolling number 1 is zero.

agree with answer choice B

P(7) not equal to 0 hence 7 is on the die -> P(0) = 0.

1) Indicates that 8 is definitely not on the faces. That takes us nowhere because then the numbers could have been 0-5,1-6,2-7,9-14, . . . .
2)The fact that the probability of not getting a seven is not unity, this means that the number 7 is on the die. This automatically eliminates the presence of 1 on any face of the die.

A Dice will have only values from 1 to 6 right?

Can someone please explain it in detail...i am sorry, i could not understand the above explinations....

No, die will have consecutive integers on its sides: {1, 2, 3 4, 5, 6} or {7, 8, 9, 10, 11, 12} or {-7, -6, -5, -4, -3, -2}, ... Basically it can have ANY 6 consecutive integers on its sides.

Check this: when-a-die-that-has-one-of-six-consecutive-integers-on-each-11954.html#p1060734

Hope it helps.

Nice Question.... Keep posting Good questions.

Bunuel,
Thank you for the good explanations that you give.
However I have a question.
I have always had a hard time with DS.
Since we cannot get a 7 and 1 on same dice how exactly did we answer this question.
The question is what is the probability of getting a 1?
so which answer did we give
ie
did statement 2 help us answer a yes or a no
as in
statement 2 helped us answer "yes we cannot establish the probability of 1"
or "no we cannot establish the probability of 1"

There are two kinds of data sufficient questions: YES/NO DS questions and DS questions which ask to find a value.

In a Yes/No Data Sufficiency questions, statement is sufficient if the answer is “always yes” or “always no” while a statement is insufficient if the answer is "sometimes yes" and "sometimes no".

When a DS question asks about the value of some variable, then the statement is sufficient ONLY if you can get the single numerical value of this variable.

Now, our original question is asking about the value (the probability of getting the number 1 on both rolls). From (2) we get that there is no 1 on the die, thus the probability of getting the number 1 on both rolls is 0.

Hope this helps.

Bunuel,
Now its clear.
Thank you.

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.


When a die that has one of six consecutive integers on each of its sides is rolled twice, what is the probability of getting the number 1 on both rolls?

(1) the probability of NOT getting an eight is 1.
(2) the probability of NOT getting a seven is 25/36

In the original condition, there are 6 consecutive integers and 1 variable(you only need to know the first number of the consecutive integers), which should match with the number of equation. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer.
In 1), there are 2 cases; {1,2,3,4,5,6} and {2,3,4,5,6,7}. the probability of getting the number 1 from the both front sides is 1/36, but from the both back sides is 0, which is not unique and not sufficient.
In 2), there are a number of cases like {2,3,4,5,6,7}, {3,4,5,6,7,8}........ In all cases, the probability of getting the number 1 from the both sides is 0, which is unique and sufficient. Therefore, the answer is B.


-> For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.

I don't suppose I am being overly microscopic here if I were to ask what about instances where we have consecutive odds?

1,3,5,7,9,11?

When we see "consecutive integers" it ALWAYS means integers that follow each other in order with common difference of 1: ... x-3, x-2, x-1, x, x+1, x+2, .... So, 1, 3, 5, 7, 9, 11 are NOT consecutive integers, they are consecutive odd integers.

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