<h3>
Answer: No it is not possible</h3>
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Explanation:
Page 1 is labeled with 1 and 2, which sum to 1+2 = 3
Page 2 is labeled with 3 and 4 which sum to 3+4 = 7
Page 3 is labeled with 5 and 6 which sum to 5+6 = 11
and so on until we reach
Page 96 is labeled with 191 and 192, which sum to 191+192 = 383
Note how each page has an odd page number label and an even number label (odd on the front side; even on the back side). Adding any odd number to an even number will result in an odd number. We can prove it as such
x = some odd number = 2m+1, m is any integer
y = some even number = 2n, n is an integer
x+y = 2m+1+2n = 2(m+n)+1 = some other odd integer because it is in the form 2p+1 with p = m+n as an integer
This explains why the results 3,7,11,..,383 are all odd.
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So we effectively have this set of values {3,7,11,...,383}. This is an arithmetic sequence with 3 as the first term and 4 as the common difference.
If we add two odd numbers together, we get an even number (proof is similar to one shown above)
odd + odd = even
But if we add in another odd number, then we'll go back to an odd result
odd + odd + odd = odd
If we have an odd number of odd numbers added up, then the result will be odd. In this case, we're adding 25 values from the set {3,7,11,...,383}. The value 25 is odd, so we have an odd number of values from {3,7,11,...,383} being added up. Therefore, the result Bob will get will always be odd. There is no way to get a sum of 2012 because this value is even.
, because there are 4 black balls and 1 white ball which is 5 balls in total. After picking the first ball, 4 balls remain - 3 black and 1 white, so the probability of picking a white ball is 
.