A bag contains red,orange, green, and blue gumballs. There are 8 red, 12 orange, 11 green, and 16 blue gumballs. What are the odds of reaching in the bag and randomly pulling out a gumball that is neither red or blue?

I know the total sample space is 47.

And I know that the probability of P(orange)=$\frac{12}{47}$ and P(green)=$\frac{11}{47}$. Then P(neither red or blue)= $\frac{12}{47} + \frac {11}{47} = \frac{23}{47}$ but I know it's wrong since the answer is $\frac{23}{24}$


The odds are $\frac{P(\text{neither red or blue})}{P(\text{red or blue})}=\frac{23/47}{24/47}=\frac{23}{24}$ so the answer is correct. The point is that they ask for the odds, not the probability.

  • $\begingroup$ I get it now, thanks! $\endgroup$ – Killercamin Feb 9 '17 at 16:09

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