The probability of drawing #1 is p(1)=6/21=2/7, so q(1)=1-2/7=5/7 is the probability of not drawing #1 from the set of numbers. The question doesn't say how many times the exercise is to be repeated. It's assumed that after each draw the number is replaced. Each time a draw is made q(1)=5/7. For this to happen twice in a row is (5/7)2=25/49 (almost ½). That is the probability of failing to draw #1 on two successive occasions.
Let's suppose that 4 draws are made and that #1 is drawn last, then the combined probability would be q3p=250/2401 (about 0.104 or about 10.4%) if each number were to be replaced after drawing. In general, if n draws took place, this would be qn-1p. If n=10, this reduces to about 0.014 or about 1.4%.
If the same number of draws were to be repeated and #1 were to be drawn last again the probability would be q2(n-1)p2. So for n=4, this would be 0.01 or 1%.