I don't understand it:
If the odds for pulling a prize out of the box are 3:4, what is the probability of not pulling the prize out of the box? Express your answer as a common fraction.
Thanks in advance!
+6248 \(\text{odds of doing X} = \dfrac{P[\text{doing X}]}{P[\text{Not doing X}]}=\dfrac{P[\text{doing X}]}{1-P[\text{doing X}]}\)
\(\text{odds of pulling prize }=\dfrac{3}{4} \Rightarrow P[\text{pulling prize}] = \dfrac{3}{3+4}= \dfrac 3 7\)
\(P[\text{not pulling prize}]=1-P[\text{pulling prize}] = \\ 1 - \dfrac 3 7 = \dfrac 4 7\)
.+6248 \(\text{odds of doing X} = \dfrac{P[\text{doing X}]}{P[\text{Not doing X}]}=\dfrac{P[\text{doing X}]}{1-P[\text{doing X}]}\)
\(\text{odds of pulling prize }=\dfrac{3}{4} \Rightarrow P[\text{pulling prize}] = \dfrac{3}{3+4}= \dfrac 3 7\)
\(P[\text{not pulling prize}]=1-P[\text{pulling prize}] = \\ 1 - \dfrac 3 7 = \dfrac 4 7\)
