Math Help Please?

something like Keno with 15 numbered balls and drawing
4 without replacement.

I agree that X has to be 4 to 15

for 4 to be the highest the other 3 must be {1,2,3}
only 1 combination for that to happen

there are a total of 15 choose 4 = 1365 combinations

the distribution is interesting:
4: 1/1365
5: 4/1365
6: 10/1365
7: 20/1365
8: 35/1365
9: 56/1365
10: 84/1365
11: 120/1365
12: 165/1365
13: 220/1365
14: 286/1365
15: 364/1365

the numerators come from Pascal's triangle column 3
in other words
from 3 choose 3 to 14 choose 3

you did not ask how to do the math for the probabilities

X=4 , 5 , ...,15
P(X=4) = Pr(a 4 and 3 balls less than 4) = 6/(15*14*13*12)=..
P(X=5) = Pr(a 5 and 3 balls less than 5) = (1/15)*C(4,3)/C(14,3)=...calculator
...

there are 15 balls, and we have to choose 4. 4 times 15 is 60 possibilities. 1/60

.

C(15, 4) = 1365
4 ≤ X ≤ 15

( i )
possible values of X = { 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 }

( ii )
P(X = 4) = C(4,4) / C(15,4) = 1/1365
P(X = 5) = [ C(5, 4 ) - C( 4, 4) ] / C(15,4) = 4 / 1365
P(X = 6) = [ C(6, 4 ) - C( 5, 4) ] / C(15,4) = 10/1365 (reduced to 2/273 )
P(X = 7) = [ C(7, 4 ) - C( 6, 4) ] / C(15,4) = 20/1365 (reduced to 4/273 )
P(X = 8) = [ C(8, 4 ) - C( 7, 4) ] / C(15,4) = 35/1365 (reduced to 1/39 )
P(X = 9) = [ C(9, 4 ) - C( 8, 4) ] / C(15,4) = 56/1365 (reduced to 8/195 )
P(X=10) = [ C(10,4) - C( 9, 4) ] / C(15,4) = 84/1365 (reduced to 4/65 )
P(X=11) = [ C(11,4) - C(10,4) ] / C(15,4) = 120/1365 (reduced to 8/91 )
P(X=12) = [ C(12,4) - C(11,4) ] / C(15,4) = 165/1365 (reduced to 11/91 )
P(X=13) = [ C(13,4) - C(12,4) ] / C(15,4) = 220/1365 (reduced to 44/273 )
P(X=14) = [ C(14,4) - C(13,4) ] / C(15,4) = 286/1365 (reduced to 22/105 )
P(X=15) = [ C(15,4) - C(14,4) ] / C(15,4) = 364/1365 (reduced to 128/455 )

Interestingly the probability is also related to X as follows;
where 4 ≤ X ≤ 15
P(X) = (1/6) (1/1365) ( X³ - 6X² + 11 X - 6 )
P(X) = 1/8190 ( X³ - 6X² + 11 X - 6 )
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