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 $\frac{23}{24}$
Best Answer
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.