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 :
Solutions · 1
Official solution
To solve this problem, we use Bayes' theorem. Let's define the events:
X: Selecting bag XY: Selecting bag YZ: Selecting bag ZA: Drawing a one-rupee coin
We are given that a bag is selected at random, so the probabilities for choosing any of the bags are:
\[P(X) = P(Y) = P(Z) = \frac{1}{3}\]
Next, we need the probability of drawing a one-rupee coin from each bag:
From bag X: \[P(A|X) = \frac{5}{5+4} = \frac{5}{9}\]
From bag Y: \[P(A|Y) = \frac{4}{4+5} = \frac{4}{9}\]
From bag Z: \[P(A|Z) = \frac{3}{3+6} = \frac{3}{9} = \frac{1}{3}\]
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:
\[P(Y|A) = \frac{P(A|Y) \cdot P(Y)}{P(A)}\]
To find P(A), the total probability of drawing a one-rupee coin can be calculated as follows:
\[P(A) = P(A|X) \cdot P(X) + P(A|Y) \cdot P(Y) + P(A|Z) \cdot P(Z)\]
Substituting the values:
\[P(A) = \left(\frac{5}{9} \cdot \frac{1}{3}\right) + \left(\frac{4}{9} \cdot \frac{1}{3}\right) + \left(\frac{1}{3} \cdot \frac{1}{3}\right)\]
Calculating the above, we get:
\[P(A) = \frac{5}{27} + \frac{4}{27} + \frac{1}{9}\]
Note that \[\frac{1}{9} = \frac{3}{27}\], so:
\[P(A) = \frac{5}{27} + \frac{4}{27} + \frac{3}{27} = \frac{12}{27} = \frac{4}{9}\]
Now, substituting back into Bayes' theorem:
\[P(Y|A) = \frac{\left(\frac{4}{9}\right) \cdot \left(\frac{1}{3}\right)}{\frac{4}{9}} = \frac{4}{9} \cdot \frac{1}{3} \cdot \frac{9}{4} = \frac{1}{3}\]
Hence, the probability that the coin came from bag Y is:
Option B: \[\frac{1}{3}\]