chance of finding two balls from a pack of 300 with a twist I need to work out a probability outcome of a scenario:
I have 300 balls, 290 different and 10 red. I have two chance of picking the red balls, with an extra bonus chance if I can't find two. so it will be either 2 chances if I get both right, or 3 if I only get 1 right in the first 2.
 A: Add up: 


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*Probability of two red balls: $\dfrac{10}{300} \times \dfrac{9}{299}$  

*Probability of a red then a non-red then a red: $\dfrac{10}{300} \times \dfrac{290}{299}\times \dfrac{9}{298}$  

*Probability of a non-red then a red then a red: $\dfrac{290}{300} \times \dfrac{10}{299}\times \dfrac{9}{298}$  
A: HINT: The probability of drawing a red ball on the first draw is $\frac{10}{300}=\frac1{30}$. If you succeed, the probability of drawing a second red ball is $\frac9{299}$, so the probability of winning in the first two draws is
$$\frac1{30}\cdot\frac9{299}=\frac3{2990}\;.$$
Use the same ideas to calculate the probability of drawing red then other and other then red and add these two numbers to get the probability $p$ of drawing exactly one red ball in the first two draws. At that point there will be $9$ red balls and $298$ balls altogether, and the probability of getting a red ball on the third draw will be $\frac9{298}$. 


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*How do you combine $p$ and $\frac9{298}$ to get the probability of winning in exactly $3$ draws?


Now just add that to the $\frac3{2990}$ probability of winning in $2$ draws.
