What is the probability that the sum of the numbers drawn is $10$ at any point in the process? The question is...
"bag contains five chips numbered 2 through 6. Danya draws chips from the bag one at a time and sets them aside. After each draw, she totals the numbers on all the chips she has already drawn. What is the probability that at any point in this process her total will equal 10?"
and the official answer is...
"Danya can get a total of 10 with two or three chips if the first two chips drawn are 4 and 6, which can occur in 2 ways, or if the first three chips drawn are 2, 3 and 5, which can occur in 6 ways. There are 5 × 4 = 20 ways to randomly select two chips, and there are 5 × 4 × 3 = 60 ways to randomly select three chips. The probability that Danya’s total will equal 10 at some point is 2/20 + 6/60 = 1/10 + 1/10 = 2/10 = 0.2".
The question said "at any point" which is to me when for example drew one chips or 2 chips or 3 chips or 4 chips ... etc.
But the way the question is answered is by assuming that she drew 2 or 3 chips only. without tacking into consideration that she might drew any number of cards rather than 2 or 3
So is this the correct answer? and if it is, why I am wrong?
 A: The probability that the total is $10$ at some point is the probability that the total is $10$ after one draw OR $10$ after two draws OR $\ldots$ OR $10$ after five draws.
These five events are mutually exclusive since there is no ball numbered $0$.
So the desired probability is the sum of the individual probabilities of these events. Now the probability that the sum is $10$ after one draw (or four, or five) is $0$. And the probabilities that it is $10$ after two or three draws are (hopefully, clearly this time) respectively $\frac{2}{20}$ and  $\frac{6}{60}$.
A: Explicitly, the value of the five chips are as follows:  $$\{2, 3, 4, 5, 6\}.$$
The statement of the problem clearly implies that draws are performed without replacement.  Therefore, if $4$ chips are drawn, the least possible value their sum could be is $2 + 3 + 4 + 5 > 10$.  So if a sum of exactly $10$ is achieved at some point during the drawing process, it must occur before the fourth draw.
I trust that it should be obvious to the reader that one draw is insufficient to obtain a total of $10$.
