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 $\mathrm{Y}$, is :
Solution
<p>To solve this problem, we use Bayes' theorem. Let's define the events:</p>
<ul>
<li>$X$: Selecting bag $X$</li>
<li>$Y$: Selecting bag $Y$</li>
<li>$Z$: Selecting bag $Z$</li>
<li>$A$: Drawing a one-rupee coin</li>
</ul>
<p>We are given that a bag is selected at random, so the probabilities for choosing any of the bags are:</p>
<p>
<p>$P(X) = P(Y) = P(Z) = \frac{1}{3}$</p>
</p>
<p>Next, we need the probability of drawing a one-rupee coin from each bag:</p>
<ul>
<li>From bag $X$: $P(A|X) = \frac{5}{5+4} = \frac{5}{9}$</li>
<li>From bag $Y$: $P(A|Y) = \frac{4}{4+5} = \frac{4}{9}$</li>
<li>From bag $Z$: $P(A|Z) = \frac{3}{3+6} = \frac{3}{9} = \frac{1}{3}$</li>
</ul>
<p>We need to find the probability that the coin came from bag $Y$ given that a one-rupee coin was drawn, i.e., we need $P(Y|A)$. Using Bayes' theorem:</p>
<p>
<p>$P(Y|A) = \frac{P(A|Y) \cdot P(Y)}{P(A)}$</p>
</p>
<p>To find $P(A)$, the total probability of drawing a one-rupee coin can be calculated as follows:</p>
<p>
<p>$P(A) = P(A|X) \cdot P(X) + P(A|Y) \cdot P(Y) + P(A|Z) \cdot P(Z)$</p>
</p>
<p>Substituting the values:</p>
<p>
<p>$$P(A) = \left(\frac{5}{9} \cdot \frac{1}{3}\right) + \left(\frac{4}{9} \cdot \frac{1}{3}\right) + \left(\frac{1}{3} \cdot \frac{1}{3}\right)$$</p>
</p>
<p>Calculating the above, we get:</p>
<p>
<p>$P(A) = \frac{5}{27} + \frac{4}{27} + \frac{1}{9}$</p>
</p>
<p>Note that $\frac{1}{9} = \frac{3}{27}$, so:</p>
<p>
<p>$P(A) = \frac{5}{27} + \frac{4}{27} + \frac{3}{27} = \frac{12}{27} = \frac{4}{9}$</p>
</p>
<p>Now, substituting back into Bayes' theorem:</p>
<p>
<p>$$P(Y|A) = \frac{\left(\frac{4}{9}\right) \cdot \left(\frac{1}{3}\right)}{\frac{4}{9}} = \frac{4}{9} \cdot \frac{1}{3} \cdot \frac{9}{4} = \frac{1}{3}$$</p>
</p>
<p>Hence, the probability that the coin came from bag $Y$ is:</p>
<p>Option B: $\frac{1}{3}$</p>
About this question
Subject: Mathematics · Chapter: Probability · Topic: Classical and Axiomatic Probability
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