The only way that 3 draws are needed is if the first two numbers drawn are either: \(1\)&\(2\), \(2\)&\(1\), \(1\)&\(3\) or \(3\)&\(1\) (4 pairs)

Number of ways in which two numbers can be drawn from 5: \(2*\frac{5!}{2!3!}= 20\)

Which means that the probability of there being 3 draws is: \(\frac{4}{20}\) or \(\frac{1}{5}\)

Answer D

In order to have to draw 3 times before the sum of the value exceeds 4.

The first draw cannot be 5, as it will end the sequence.
The first draw cannot be 4, as the sequence will end after the second draw.

The first draw can be 3, then the second draw must be 1, (and the third draw can be anything).
Probability for this sequence is 1/5 * 1/4 * 3/3 = 1/20

The first draw can be 2, then the second draw must be 1, (and the third draw can be anything).
Probability for this sequence is 1/5 * 1/4 * 3/3 = 1/20

The first draw can be 1, then the second draw must be either 2 or 3, (and the third draw can be anything).
Probability for this sequence is 1/5 * 2/4 * 3/3 = 2/20

Altogether, the probability is 1/20 + 1/20 + 2/20 = 4/20 = 1/5
The answer is thus (D).