E
ele
Guest
Consider an urn of unknown composition containing 5 balls that can be white or black, but not all of the same color. Calculate the probability that the urn contains 3 white balls, assumed that in 3 draws without return you get 3 white balls.
My thoughts:
There are 4 different ways the urn can be:
- 4 White 1 Black
- 3 White 2 Black
- 2 White 3 Black
- 1 White 4 Black
Let the random variable \(\displaystyle A = \text{"The urn contains 3 white balls"}\) and \(\displaystyle B = \text{"3 white balls are drawn with 3 draws without return"}\), we need to find \(\displaystyle P(A|B)\).
Using Bayes' rule we need to find \(\displaystyle P(A|B) = \frac{P(B|A)P(A)}{P(B)} = \frac{{\frac{1}{{4}}{\left(\frac{3}{{5}}\cdot\frac{2}{{4}}\cdot\frac{1}{{3}}\right)}}}{{\frac{1}{{4}}{\left(\frac{4}{{5}}\cdot\frac{3}{{4}}\cdot\frac{2}{{3}}\right)}+\frac{1}{{4}}{\left(\frac{3}{{5}}\cdot\frac{2}{{4}}\cdot\frac{1}{{3}}\right)}}} = \frac{1}{5}\)
Is my method correct?
-Elle
My thoughts:
There are 4 different ways the urn can be:
- 4 White 1 Black
- 3 White 2 Black
- 2 White 3 Black
- 1 White 4 Black
Let the random variable \(\displaystyle A = \text{"The urn contains 3 white balls"}\) and \(\displaystyle B = \text{"3 white balls are drawn with 3 draws without return"}\), we need to find \(\displaystyle P(A|B)\).
Using Bayes' rule we need to find \(\displaystyle P(A|B) = \frac{P(B|A)P(A)}{P(B)} = \frac{{\frac{1}{{4}}{\left(\frac{3}{{5}}\cdot\frac{2}{{4}}\cdot\frac{1}{{3}}\right)}}}{{\frac{1}{{4}}{\left(\frac{4}{{5}}\cdot\frac{3}{{4}}\cdot\frac{2}{{3}}\right)}+\frac{1}{{4}}{\left(\frac{3}{{5}}\cdot\frac{2}{{4}}\cdot\frac{1}{{3}}\right)}}} = \frac{1}{5}\)
Is my method correct?
-Elle