pg. 181 4.1

  1. One of the first conditional probability paradoxes was provided by Bertrand. It is called the Box Paradox. A cabinet has three drawers. In the first drawer there are two gold balls, in the second drawer there are two silver balls, and in the third drawer there is one silver and one gold ball. A drawer is picked at random and a ball chosen at random from the two balls in the drawer. Given that a gold ball was drawn, what is the probability that the drawer with the two gold balls was chosen?

For ease, we are going to name the probabilities:

P(Picking drawer w/ 2 gold) = P(A) = \(\frac{1}3\)

P(Picking drawer w/ 2 silver) = P(B) = \(\frac{1}3\)

P(Picking drawer w/ 1 gold & 1 silver) = P(C) = \(\frac{1}3\)

P(Drawing a gold ball) = P(X)

The probability of drawing a gold one from the first drawer or P(X|A)= 1.

The probability of drawing a gold one from the second drawer or P(X|B)= 0.

The probability of drawing a gold one from the third drawer or P(X|A)= \(\frac{1}2\)

Knowing this, we can use Baye’s rule to find the probability of picking the first drawer P(A) given drawing a gold ball P(X), which is P(A|G).

\(P(A|X) = \frac{P(X|A)P(A)}{P(X|A)P(A) + P(X|B)P(B) + P(X|C)P(C)} = \frac{1(\frac{1}3)}{1(\frac{1}3) + 0(\frac{1}3) + (\frac{1}2)(\frac{1}3)} = \frac{\frac{1}3}{\frac{1}2} = \frac{2}3\)