There are 3 urns containing coins. Urn I contains 7 gold coins, urn II contains 3 gold coins and 3 silver coins, and urn III contains 3 silver coins. An urn is selected and a coin is drawn from the urn. If the coin is silver, what is the probability that urn III was selected?
The correct answer and explanation is :
We can solve this problem using Bayes’ Theorem, which helps us find the probability of an event given prior knowledge of related conditions.
Step 1: Define the Events
Let:
- ( U_1 ), ( U_2 ), and ( U_3 ) be the events of selecting urn I, II, and III, respectively.
- ( S ) be the event that a silver coin is drawn.
The problem provides:
- Urn I (( U_1 )) contains 7 gold coins, so ( P(S | U_1) = 0 ).
- Urn II (( U_2 )) contains 3 gold and 3 silver coins, so ( P(S | U_2) = \frac{3}{6} = \frac{1}{2} ).
- Urn III (( U_3 )) contains 3 silver coins, so ( P(S | U_3) = 1 ).
Since an urn is selected at random, the probability of choosing any urn is:
[
P(U_1) = P(U_2) = P(U_3) = \frac{1}{3}
]
Step 2: Apply the Law of Total Probability
The probability of drawing a silver coin (( P(S) )) is given by:
[
P(S) = P(S | U_1) P(U_1) + P(S | U_2) P(U_2) + P(S | U_3) P(U_3)
]
[
P(S) = (0 \times \frac{1}{3}) + \left(\frac{1}{2} \times \frac{1}{3}\right) + \left(1 \times \frac{1}{3}\right)
]
[
P(S) = 0 + \frac{1}{6} + \frac{1}{3} = \frac{1}{6} + \frac{2}{6} = \frac{3}{6} = \frac{1}{2}
]
Step 3: Apply Bayes’ Theorem
We want to find ( P(U_3 | S) ):
[
P(U_3 | S) = \frac{P(S | U_3) P(U_3)}{P(S)}
]
[
P(U_3 | S) = \frac{1 \times \frac{1}{3}}{\frac{1}{2}} = \frac{\frac{1}{3}}{\frac{1}{2}} = \frac{2}{3}
]
Final Answer:
[
\boxed{\frac{2}{3}}
]
Explanation (Summary)
- We used Bayes’ Theorem to calculate the probability of selecting urn III given that a silver coin was drawn.
- We determined the probabilities of drawing a silver coin from each urn and the total probability of drawing a silver coin.
- Applying Bayes’ Theorem, we found that the probability that urn III was chosen given that a silver coin was drawn is ( \frac{2}{3} ).