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chetan2u
Bunuel
One number, n, is selected at random from a set of 10 integers. What is the probability that 1/2*n + 13 = 0 ?

(1) The largest integer in the set is 13.
(2) The arithmetic mean of the set is zero.


The Q can be rephrased-
\(\frac{1}{2}* n + 13 =0............n=-13*2=-26\)...............
so there are 10 integers and we have to find that what is the probability of the integer,n, being picked to be -26....
HAD the Q said been - a set of 10 DIFFERENT integers, our answer would be 1/10, but we do not know how many of 10 integers are -26...

(1) The largest integer in the set is 13.
we do not know anything about other 9 numbers
Insuff

(2) The arithmetic mean of the set is zero.

It just tells us that mean is 0, can be any variation..

Combined-
lot of variation can be made where the largest number is 13 and the mean is 0..
example 13,13,0,0,0,0,0,0,0,-26... prob = 1/10
13, 10,10,10,3,3,3,0,-26,-26..... Now prob = 2/10
Insuff
E


chetan2u,
For Statement A, I agree that the probability of choosing -26 would have been 1/10, but this would again be the case provided that it is given that -26 is a part of the set.
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adityapareshshah
chetan2u
Bunuel
One number, n, is selected at random from a set of 10 integers. What is the probability that 1/2*n + 13 = 0 ?

(1) The largest integer in the set is 13.
(2) The arithmetic mean of the set is zero.


The Q can be rephrased-
\(\frac{1}{2}* n + 13 =0............n=-13*2=-26\)...............
so there are 10 integers and we have to find that what is the probability of the integer,n, being picked to be -26....
HAD the Q said been - a set of 10 DIFFERENT integers, our answer would be 1/10, but we do not know how many of 10 integers are -26...

(1) The largest integer in the set is 13.
we do not know anything about other 9 numbers
Insuff

(2) The arithmetic mean of the set is zero.

It just tells us that mean is 0, can be any variation..

Combined-
lot of variation can be made where the largest number is 13 and the mean is 0..
example 13,13,0,0,0,0,0,0,0,-26... prob = 1/10
13, 10,10,10,3,3,3,0,-26,-26..... Now prob = 2/10
Insuff
E


chetan2u,
For Statement A, I agree that the probability of choosing -26 would have been 1/10, but this would again be the case provided that it is given that -26 is a part of the set.

I have the same question, any reasons that we can say probability would be 1/10 if in case n were categorically mentioned as dufferent integers ?

chetan2u , Bunuel - advice please ?
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