Q6:

Suppose we have four boxes A,B,C and D containing coloured marbles as given below: 

One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box A?, box B?, box C?

Step 1: Defining Events and Prior Probabilities

Let $E$ be the event that the marble drawn is red. Let $A, B, C,$ and $D$ be the events of selecting boxes A, B, C, and D, respectively. Since there are four boxes and one is selected at random:

$P(A) = P(B) = P(C) = P(D) = \frac{1}{4}$ [By the principle of equal likelihood]

Step 2: Conditional Probabilities of Drawing a Red Marble

Based on the provided distribution (A: 1 Red, 6 White; B: 2 Red, 2 White; C: 3 Red, 1 White; D: 4 Red, 0 White), we calculate the probability of drawing a red marble given a specific box selection:

• $P(E|A) = \frac{1}{1+6} = \frac{1}{7}$
• $P(E|B) = \frac{2}{2+2} = \frac{2}{4} = \frac{1}{2}$
• $P(E|C) = \frac{3}{3+1} = \frac{3}{4}$
• $P(E|D) = \frac{4}{4+0} = \frac{4}{4} = 1$

Step 3: Calculating Total Probability $P(E)$

Per the Law of Total Probability: $P(E) = P(A)P(E|A) + P(B)P(E|B) + P(C)P(E|C) + P(D)P(E|D)$

$P(E) = \left( \frac{1}{4} \times \frac{1}{7} \right) + \left( \frac{1}{4} \times \frac{1}{2} \right) + \left( \frac{1}{4} \times \frac{3}{4} \right) + \left( \frac{1}{4} \times 1 \right)$

$P(E) = \frac{1}{4} \left( \frac{1}{7} + \frac{1}{2} + \frac{3}{4} + 1 \right) = \frac{1}{4} \left( \frac{8 + 28 + 42 + 56}{56} \right) = \frac{1}{4} \left( \frac{134}{56} \right) = \frac{67}{112}$

Step 4: Applying Bayes' Theorem

To find the probability that the marble was drawn from a specific box given it is red, we use: $P(X|E) = \frac{P(X)P(E|X)}{P(E)}$ where $X \in \{A, B, C, D\}$.

Final Solution: The probabilities that the red marble was drawn from boxes A, B, C, and D are $\frac{4}{67}, \frac{14}{67}, \frac{21}{67},$ and $\frac{28}{67}$ respectively.