Combinatorics: calculating Oz Lotto odds for divisions

[Darkmisc]

Homework Statement

In Oz Lotto, balls are numbered 1 to 45. Nine are selected, seven of which are winning numbers and two being supplementary numbers. Players select seven numbers.

The odds of winning can be found here: https://www.lottoland.com.au/magazine/oz-lotto-everything-there-is-to-know.html

I tried calculating the odds, and get all of them right except for div 4 and 7. Could somebody please explain what I've done wrong?

Relevant Equations

[SUP]7[/SUP]C[SUB]5[/SUB] x [SUP]2[/SUP]C[SUB]1[/SUB] x [SUP]36[/SUP]C[SUB]1[/SUB]/ [SUP]45[/SUP]C[SUB]7[/SUB] =

1/30,012

In Oz Lotto, balls are numbered 1 to 45. Nine are selected, seven of which are winning numbers and two being supplementary numbers. Players select seven numbers.

The odds of winning can be found here: https://www.lottoland.com.au/magazine/oz-lotto-everything-there-is-to-know.html

I tried calculating the odds, and get all of them right except for Div 4 and 7. Could somebody please explain what I've done wrong?

Division 4

5 Winning Numbers + 1 Supplementary Number

1 : 29,602

⁷C₅ x ²C₁ x ³⁶C₁/ ⁴⁵C₇ =

1/30,012

Division 7

3 Winning Numbers + 1 Supplementary

1 : 87

⁷C₃ x ²C₁ x ³⁶C₃/ ⁴⁵C₇ = 1/91

Thanks

 

[PeroK]

I think I agree with your answers!

 

[PeroK]

PS alternative calculation for division 4:
$$p = 7 \cdot 6 \cdot P(WWWWWSX) = 42 \cdot \frac{7}{45} \cdot \frac{6}{44} \dots \frac{2}{40} \cdot \frac{36}{39}$$