There are three bags \(X\), \(Y\), and \(Z\). Bag \(X\) contains 5 one-rupee coins and 4 five-rupee coins; Bag \(Y\) contains 4 one-rupee coins and 5 five-rupee coins, and Bag \(Z\) contains 3 one-rupee coins and 6 five-rupee coins. A bag is selected at random and a coin drawn from it at random is found to be a one-rupee coin. Then the probability, that it came from bag \(Y\), is:
- \(\frac{1}{3}\)
- \(\frac{1}{2}\)
- \(\frac{1}{4}\)
- \(\frac{5}{12}\)
The Correct Option is A
Approach Solution - 1
Let's solve this problem using the concept of conditional probability. We are given three bags \(X\), \(Y\), and \(Z\) containing different numbers of one-rupee and five-rupee coins. We need to find the probability that the chosen coin came from bag \(Y\) given that it is a one-rupee coin.
- Understanding the Contents:
- Bag \(X\) contains 5 one-rupee coins and 4 five-rupee coins. Total coins = 9.
- Bag \(Y\) contains 4 one-rupee coins and 5 five-rupee coins. Total coins = 9.
- Bag \(Z\) contains 3 one-rupee coins and 6 five-rupee coins. Total coins = 9.
- Probability of Selecting Each Bag:
Each bag is equally likely to be selected. Therefore, the probability of selecting any one bag is:
\(\frac{1}{3}\)
- Probability of Drawing a One-Rupee Coin Given the Selected Bag:
- From Bag \(X\): Probability = \(\frac{5}{9}\) (since 5 out of 9 coins are one-rupee coins).
- From Bag \(Y\): Probability = \(\frac{4}{9}\) (since 4 out of 9 coins are one-rupee coins).
- From Bag \(Z\): Probability = \(\frac{3}{9}\) (since 3 out of 9 coins are one-rupee coins).
- Total Probability of Drawing a One-Rupee Coin:
The total probability that a randomly drawn coin is a one-rupee coin is given by:
\(P(\text{One-Rupee}) = \frac{1}{3} \cdot \frac{5}{9} + \frac{1}{3} \cdot \frac{4}{9} + \frac{1}{3} \cdot \frac{3}{9}\)
Calculating, we get:
\(P(\text{One-Rupee}) = \frac{5}{27} + \frac{4}{27} + \frac{3}{27} = \frac{12}{27} = \frac{4}{9}\)
- Using Bayes' Theorem to find the Required Probability:
We need to find the probability that the coin came from bag \(Y\) given that it is a one-rupee coin, which is represented by:
\(P(Y | \text{One-Rupee}) = \frac{P(\text{One-Rupee | } Y) \cdot P(Y)}{P(\text{One-Rupee})}\)
Substituting values,
\(P(Y | \text{One-Rupee}) = \frac{\frac{4}{9} \cdot \frac{1}{3}}{\frac{4}{9}}\)
Simplify:
\(P(Y | \text{One-Rupee}) = \frac{4}{27} \times \frac{9}{4} = \frac{1}{3}\)
Therefore, the probability that the coin came from bag \(Y\), given that it is a one-rupee coin, is \(\frac{1}{3}\). Thus, the correct answer is:
Option B: \(\frac{1}{3}\)
Approach Solution -2
Let the events \( E_X, E_Y, \) and \( E_Z \) denote the selection of bags \( X, Y, \) and \( Z \) respectively. Let the event \( A \) denote drawing a one-rupee coin. We are required to find the conditional probability: \[ P(E_Y|A) = \frac{P(E_Y) \times P(A|E_Y)}{P(A)}. \]
The probabilities of selecting each bag are: \[ P(E_X) = P(E_Y) = P(E_Z) = \frac{1}{3}. \]
The probability of drawing a one-rupee coin from each bag is given by: \[ P(A|E_X) = \frac{5}{9}, \quad P(A|E_Y) = \frac{4}{9}, \quad P(A|E_Z) = \frac{3}{9}. \]
The total probability of drawing a one-rupee coin, using the law of total probability: \[ P(A) = P(E_X) \times P(A|E_X) + P(E_Y) \times P(A|E_Y) + P(E_Z) \times P(A|E_Z). \]
Substituting the values: \[ P(A) = \frac{1}{3} \times \frac{5}{9} + \frac{1}{3} \times \frac{4}{9} + \frac{1}{3} \times \frac{3}{9}, \] \[ P(A) = \frac{5}{27} + \frac{4}{27} + \frac{3}{27} = \frac{12}{27} = \frac{4}{9}. \]
Now, the conditional probability that the coin came from bag \( Y \) given that it is a one-rupee coin is: \[ P(E_Y|A) = \frac{P(E_Y) \times P(A|E_Y)}{P(A)}, \] \[ P(E_Y|A) = \frac{\frac{1}{3} \times \frac{4}{9}}{\frac{4}{9}} = \frac{\frac{4}{27}}{\frac{4}{9}} = \frac{1}{3}. \]
Therefore: \[ \frac{1}{3}. \]
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