- #1

Darkmisc

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- 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?

1/30,012

Thanks

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 |

^{7}C_{5}x^{2}C_{1}x^{36}C_{1}/^{45}C_{7}=1/30,012

Division 7 | 3 Winning Numbers + 1 Supplementary | 1 : 87 |

^{7}C_{3}x^{2}C_{1}x^{36}C_{3}/^{45}C_{7}= 1/91Thanks