Vase 1 has 4 red marbles and 3 green marble. Vase 2 has 5 red marbles and 2 gree
ID: 3230679 • Letter: V
Question
Vase 1 has 4 red marbles and 3 green marble. Vase 2 has 5 red marbles and 2 green marbles. Vase 3 has 6 red and 1 green marble. A four-sided die is rolled, with Vase 1 being selected if 1 comes up, Vase 2 if 2 comes up, and Vase 3 if 3 or 4 comes up. Then a marble is selected from the given vase. Find each of the following probabilities, leaving all answers as fractions. a) P(a red marble being selected) b) P(Vase 3 was selected, given a red marble was selected) c) What theorem was used to find the answer in part b?Explanation / Answer
P(Vase 1 being selected) = 1 / 4 (since vase 1 will be selected if the 4 sided die shows 1) P(Vase 2 being selected) = 1 / 4 (since vase 1 will be selected if the 4 sided die shows 2) P(Vase 3 being selected) = 1 / 4 + 1/4 = 1/2 (since vase 1 will be selected if the 4 sided die shows 3 or 4) a) P(a red marble is being selected) = P(selecting Vase 1 ) x P(selecting red marble from Vase 1) + P(selecting Vase 2 ) x P(selecting red marble from Vase 2) + P(selecting Vase 3 ) x P(selecting red marble from Vase 3) P(a red marble is being selected) = (1/4 x 4/7) + (1/4 x 5/7) + (1/2 x 6/7) (7 is the total number of marbles in each vase) = 3 / 4 P(a red marble is being selected) = 3 / 4 b) P(Vase 3 was selected given a red marble was selected) = P(Vase 3 was selected / a red marble was selected) = P(Vase 3 was selected and a red marble was selected) / P( a red marble was selected) From (a) we know P(a red marble was selected ) = 3 / 4 P(Vase 3 was selected and a red marble was selected) = 1/2 x 6/7 = 3/7 P(Vase 3 was selected given a red marble was selected) = (3 / 7) / (3 / 4) = 4 / 7 P(Vase 3 was selected given a red marble was selected) = 4 / 7 c) To find the answer in Part b, the Conditional Probability Theorem was used It is also a derived version of the Baye's theorem on Conditional Probability P(A/B) = P(AB) / P(B)
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