# Would anyone check my solutions to a few probability questions (1 Viewer)

#### HeroWise

##### Active Member
Working out for 17 a , in the HSC requires this step $\bg_white ( \frac{20}{20})( \frac{1}{20})$

the $\bg_white ( \frac{20}{20})$ ensures you get any card and the $\bg_white ( \frac{1}{20})$ is any other specific card

17 b is

$\bg_white \binom{3}{2} \times ( \frac{1}{20})^2( \frac{19}{20}) \, = \, ( \frac{57}{8000})$

Uses binomials

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#### HeroWise

##### Active Member
For 18

Count all the yellows and the blue only once you should get 11 total outcomes. I understand in you thinking its from the total set but if u think about it the question is talking about "atleast 3" only.