Given the following information:
- We have $2$ baskets with red and blue marbles in each basket.
- In basket 1 there are $3$ red and $4$ blue.
- In basket 2 there are $6$ red and $8$ blue.
and the following question:
You draw $3$ marbles. If you know that you are definitely drawing from basket 1, what is the probability that the $3$ marbles are red.
I thought I could solve it using Bayes' rule, namely:
\begin{align*} &P(\text{Red}\mid\text{Basket 1}) \\ &= \frac{P(\text{Basket 1}\mid\text{Red}) \cdot P(\text{Red})}{P(\text{Basket 1}\mid\text{Red}) \cdot P(\text{Red}) + P(\text{Basket 1}\mid\text{Blue}) \cdot P(\text{Blue})} \\ & = \frac{3}{7} \end{align*}
However, via combinatorics we obtain a probability of $\frac{1}{35}$. Can someone explain to me what I am doing wrong here?
There's no need to use Bayes' formula. The first red ball has probability 3/7. Subtract one from the numerator and denominator for each subsequent draw. P = (3/7)(2/6)(1/5) = 1/35