A player picks one ball from 3 different urns. Red, green, and blue balls are to

Question

A player picks one ball from 3 different urns. Red, green, and blue balls are to be contained in these three identical urns. In this game of chance, a player will loose $1 for a green ball, $0 for a blue ball and earn $2 for a red. Urns will have the following distribution of balls: one urn will contain 3 green and 2 red, another will contain 4 greeen and 2 blue, and finally the last urn will contain 3 red, 3 blue, and 3 green.

1. What are yhe possible outcomes for the amount of money earned/

2. What are the probabilities of the different outcomes?

Explanation / Answer

1. Possible money gained:

-1 (For green)

0 (For blue)

2 (For red)

2. Probability

P(Green) : P(1st urn) *P(Green | 1st urn) + P(2nd urn) *P(Green | 2nd urn) + P(3rd urn) *P(Green | 3rd urn)

= 1/3 * 3/5 + 1/3 * 4/6 + 1/3 * 3/9 = 0.5333

P(Blue) : P(1st urn) *P(Blue | 1st urn) + P(2nd urn) *P(Blue | 2nd urn) + P(3rd urn) *P(Blue| 3rd urn)

= 1/3 * 0/5 + 1/3 * 2/6 + 1/3 * 3/9 = 0.2222

P(Red) = 1 - P(Green) - P(Blue) = 0.2445

So,

P(Gain = -1) = 0.5333

P(Gain = 0) = 0.2222

P(Gain = 2) = 0.2445

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