Discrete probability distribution: Basic

Fill in the values in the table below to give a legitimate probability distribution for the discrete random variable x , whose possible values are 2 ,3 , 4, 5, and 6.
I obviously cannot provide a table but I will provide the sides. There are two sides one called
value x of X which has the following numbers going down
2
3
4
5
6
and then the other side which is called
P(X=x) with the following numbers going down
0.20
(blank)
(blank)
0.24
0.19

I need help with filling in the blanks thank you very much!!!

Since the problem asks for "a legitimate probability distribution," it doesn't matter which values we fill in as long as the sum of all the probabilities is 1. Thus, first note that the sum of all the probabilities BUT the two we want to fill in is
0.20+0.24+0.19=0.63
So we have remaining probability
1-0.63=0.37
to distribute as we see fit among the remaining two values of X. For example, one solution would be
0.19 and 0.18
since 0.19+0.18=0.37.

A valid distribution must have total probability of 1.
.20 + .24 + .19 = 0.63 so we have 0.37 left to distribute to P(3) and P(4)

There are infinitely many ways to this, e.g.:
0, .37
.10, .27
etc.

I might try this distribution.
X -- Pr(X = x)
2 .20
3 .20
4 .17
5 .24
6 .19

As long as the total probability is 1 you're OK