Probability Problem: Drawing a Ball from an Urn with Known Remaining Color

(a) In order to calculate the probability that the ball she drew is also red, we can use Bayes' theorem. Let R be the event that the ball she drew is red, and let U1, U2, U3, U4, and U5 be the events that she picked urn 1, urn 2, urn 3, urn 4, and urn 5 respectively. We want to calculate P(R | R has been drawn).

According to Bayes' theorem: P(R | R has been drawn) = (P(R has been drawn | R) * P(R)) / P(R has been drawn)

P(R has been drawn | R) = 1, as the experimenter has already told us that the remaining ball in the urn she drew from is red.

P(R) = P(R | U1) * P(U1) + P(R | U2) * P(U2) + P(R | U3) * P(U3) + P(R | U4) * P(U4) + P(R | U5) * P(U5) = 1 * (1/5) + 1/2 * (1/5) + 1/2 * (1/5) + 1/2 * (1/5) + 1/2 * (1/5) = 5/10 = 1/2

P(R has been drawn) = P(R has been drawn | R) * P(R) + P(R has been drawn | not R) * P(not R) = 1 * (1/2) + 1/2 * (1/2) = 1/2 + 1/4 = 3/4

Therefore, the probability that the ball she drew is also red is: P(R | R has been drawn) = (1 * (1/2)) / (3/4) = 2/3

(b) The probability that the ball she drew is green can be calculated by subtracting the probability that the ball she drew is red from 1.

P(G | R has been drawn) = 1 - P(R | R has been drawn) = 1 - 2/3 = 1/3

Therefore, the probability that the ball she drew is green is 1/3.