There are three bags X, Y and Z.

Question

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 sele...

Accepted Answer

To solve this problem, we use Bayes' theorem. Let's define the events:

$X$: Selecting bag $X$

$Y$: Selecting bag $Y$

$Z$: Selecting bag $Z$

$A$: 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:

$P(A)=(\frac{5}{9}\cdot \frac{1}{3})+(\frac{4}{9}\cdot \frac{1}{3})+(\frac{1}{3}\cdot \frac{1}{3})$

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:

$P(Y|A)=\frac{(\frac{4}{9})\cdot (\frac{1}{3})}{\frac{4}{9}}=\frac{4}{9}\cdot \frac{1}{3}\cdot \frac{9}{4}=\frac{1}{3}$

Hence, the probability that the coin came from bag $Y$ is:

Option B: $\frac{1}{3}$

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