Hey guys! I have been stuck on a couple of problems for a while now. I got one of them right but can't figure out the other two. It says I got the first one right. For the second one I have tried this formula: (9C2)times (19C7)/(28C5). This isn't working. Any help would be greatly appreciated.
Probability Question
- Thread starter jayyy9102
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Dr.Peterson
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Your second answer clearly is wrong, because a probability can't be greater than 1!
In your work, where did the 7 come from? I think you just let your mind wander!
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Hello, and welcome to FMH! 
For the second problem, I would first look at a particular case, such as the first and second are red and the rest are not.
[MATH]\frac{9}{28}\cdot\frac{8}{27}\cdot\frac{19}{26}\cdot\frac{18}{25}\cdot\frac{17}{24}=\frac{323}{9100}[/MATH]
Then multiply this by the number of ways to choose 2 from 5:
[MATH]P(\text{2 are red})={5 \choose 2}\cdot\frac{323}{9100}=\frac{323}{910}\approx0.354945[/MATH]
Note this is the same as:
[MATH]P(\text{2 are red})=\frac{{9 \choose 2}\cdot{19 \choose 3}}{{28 \choose 5}}=\frac{323}{910}\approx0.354945[/MATH]
For the second problem, I would first look at a particular case, such as the first and second are red and the rest are not.
[MATH]\frac{9}{28}\cdot\frac{8}{27}\cdot\frac{19}{26}\cdot\frac{18}{25}\cdot\frac{17}{24}=\frac{323}{9100}[/MATH]
Then multiply this by the number of ways to choose 2 from 5:
[MATH]P(\text{2 are red})={5 \choose 2}\cdot\frac{323}{9100}=\frac{323}{910}\approx0.354945[/MATH]
Note this is the same as:
[MATH]P(\text{2 are red})=\frac{{9 \choose 2}\cdot{19 \choose 3}}{{28 \choose 5}}=\frac{323}{910}\approx0.354945[/MATH]