Answer as a fraction = 53/200
Answer as a decimal = 0.265
Answer as a percent = 26.5%
Explanation
Let's say hypothetically that your first three selections are: 1,2,3 (these correspond to the white balls) and let's also say that your red ball selection is 20.
With the ticket "1,2,3,20" in mind, what is the probability of winning the white ball portion, red ball portion, or both portions?
For now let's just focus on the white balls.
Since order doesn't matter, consider the sequence 1,2,X where X is an integer from the set {3,4,5,...,9,10}
Any sequence of the form 1,2,X will have at least two white ball numbers matching.
We have 10-3+1 = 8 choices for what to replace X with.
That means there are 8 combinations of the form 1,2,X where order doesn't matter.
Similar logic will then tell us there are also 8 combos of the form 1,3,Y where Y is an integer from the set {2,4,5,...,9,10}
But since "1,2,3" is already accounted for in the previous paragraph, we only need to consider the set {4,5,6,...,9,10} which has 7 items in it.
Furthermore, we have 7 combos of the form 2,3,Z where Z is an integer from the set {4,5,...,9,10}
Note that "1" is left out since 1,2,3 is already accounted for two paragraphs prior.
We have 8+7+7 = 22 ways of winning the white ball portion.
Here are all 22 instances where we have 2 or more matches.{1,2,3} {1,2,4} {1,2,5} {1,2,6} {1,2,7} {1,2,8} {1,2,9} {1,2,10}
{1,3,4} {1,3,5} {1,3,6} {1,3,7} {1,3,8} {1,3,9} {1,3,10}
{2,3,4} {2,3,5} {2,3,6} {2,3,7} {2,3,8} {2,3,9} {2,3,10}The first line has 8 sets of curly braces, the other two lines have 7 sets each.
So that's 8+7+7 = 22 total.
I used this combinatorics calculator to generate the list
https://www.mathsisfun.com/combinatorics/combinations-permutations-calculator.html
So we have 22 ways to win the white ball portion out of 10C3 = 120 combinations.
The 10C3 refers to the nCr combination formula.
Such values are found in Pascal's Triangle. Look at the row that starts with 1,10,...
The probability of winning the white ball portion is 22/120 = (2*11)/(2*60) = 11/60.
We'll use 11/60 later.
There are 20-11+1 = 10 selections for the red ball, so the chances of winning the red ball portion is 1/10.
(11/60)*(1/10) = 11/600 is the probability of winning both white and red ball portions at the same time.
This assumes the white and red balls are independent of each other.
A = you win the white ball portion
B = you win the red ball portion
P(A or B) = P(A) + P(B) - P(A and B) ................... inclusion-exclusion principle
P(A or B) = (11/60) + (1/10) - (11/600)
P(A or B) = (110/600) + (60/600) - (11/600)
P(A or B) = (110+60-11)/600
P(A or B) = 159/600
P(A or B) = (53*3)/(200*3)
P(A or B) = 53/200
53/200 = 0.265 = 26.5%
The decimal values are exact and haven't been rounded.