Question

Bag A contains 3 white and 2 red balls. Bag B contains only 1 white ball. A fair coin is tosse(D) If head appears then 1 ball is drawn at random from bag A and put into bag (B) However if tail appears then 2 balls are drawn at random from bag A and put into bag (B) Now one ball is drawn at random from bag (B) Given that the drawn ball from B is white, the probability that head appeared on the coin is

• A. \(\frac{23}{30}\)
• B. \(\frac{12}{23}\)
• C. \(\frac{11}{23}\)
• D. \(\frac{19}{30}\)

Hint:

In conditional probability problems, Bayes' Theorem allows us to update the probability of an event based on new information. Here, the new information is the white ball drawn from bag (B)

Answer 1

The Correct Option is B

We need to calculate the probability that the coin showed heads given that a white ball was drawn from bag (B) This is a conditional probability problem. We will use Bayes' Theorem:
\[ P(\text{Head} \mid \text{White}) = \frac{P(\text{White} \mid \text{Head}) P(\text{Head})}{P(\text{White})} \] Step 1: Calculate \( P(\text{Head}) \) and \( P(\text{Tail}) \) Since the coin is fair, \[ P(\text{Head}) = P(\text{Tail}) = \frac{1}{2} \] Step 2: Calculate \( P(\text{White} \mid \text{Head}) \) If head appears, then 1 ball is drawn from bag A and placed in bag (B) Bag A contains 3 white balls and 2 red balls. The probability of drawing a white ball is: \[ P(\text{White} \mid \text{Head}) = \frac{3}{5} \] Step 3: Calculate \( P(\text{White} \mid \text{Tail}) \) If tail appears, 2 balls are drawn from bag A and placed in bag (B) There are 3 white balls and 2 red balls in bag (A) The probability that 1 white ball and 1 red ball are drawn is: \[ P(\text{White} \mid \text{Tail}) = \frac{3}{5} \times \frac{2}{4} = \frac{6}{20} = \frac{3}{10} \] Step 4: Calculate \( P(\text{White}) \) The total probability of drawing a white ball from bag B can be found by considering both possible events (head or tail): \[ P(\text{White}) = P(\text{White} \mid \text{Head}) P(\text{Head}) + P(\text{White} \mid \text{Tail}) P(\text{Tail}) \] \[ P(\text{White}) = \left(\frac{3}{5} \times \frac{1}{2}\right) + \left(\frac{3}{10} \times \frac{1}{2}\right) \] \[ P(\text{White}) = \frac{3}{10} + \frac{3}{20} = \frac{9}{20} \] Step 5: Apply Bayes' Theorem Now, we can apply Bayes' Theorem to find \( P(\text{Head} \mid \text{White}) \): \[ P(\text{Head} \mid \text{White}) = \frac{P(\text{White} \mid \text{Head}) P(\text{Head})}{P(\text{White})} \] \[ P(\text{Head} \mid \text{White}) = \frac{\frac{3}{5} \times \frac{1}{2}}{\frac{9}{20}} = \frac{\frac{3}{10}}{\frac{9}{20}} = \frac{3}{10} \times \frac{20}{9} = \frac{12}{23} \] Thus, the probability that head appeared on the coin is \(\frac{12}{23}\).