Question:
Out of two bags, bag I contains 3 red and 4 white balls and bag II contains 8 red and 6 white balls. A die is thrown. If it shows a number less than 3 then a ball is drawn at random from bag I, otherwise a ball is drawn at random from bag II. Find the probability that the ball drawn from one of the bags is a red ball.
Out of two bags, bag I contains 3 red and 4 white balls and bag II contains 8 red and 6 white balls. A die is thrown. If it shows a number less than 3 then a ball is drawn at random from bag I, otherwise a ball is drawn at random from bag II. Find the probability that the ball drawn from one of the bags is a red ball.
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Draw a tree diagram to visualize the branching of choices for multi-stage probability problems.
Always check if the fractions can be simplified before adding to make calculations easier.
Always check if the fractions can be simplified before adding to make calculations easier.
Updated On: Sep 10, 2026
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Solution and Explanation
Concept:
• Law of Total Probability: If \( E_1, E_2 \) are mutually exclusive events, then \( P(A) = P(E_1)P(A|E_1) + P(E_2)P(A|E_2) \).
Step 1: Define the events and find probabilities of bag selection
Let \( E_1 \) be the event that Bag I is chosen (die shows \( \{1, 2\} \)). Let \( E_2 \) be the event that Bag II is chosen (die shows \( \{3, 4, 5, 6\} \)). \[ P(E_1) = \frac{2}{6} = \frac{1}{3} \] \[ P(E_2) = \frac{4}{6} = \frac{2}{3} \]
Step 2: Find the conditional probabilities of drawing a red ball
Let \( A \) be the event that the ball drawn is red. Bag I has 3 red and 4 white balls (Total 7). \[ P(A|E_1) = \frac{3}{7} \] Bag II has 8 red and 6 white balls (Total 14). \[ P(A|E_2) = \frac{8}{14} = \frac{4}{7} \]
Step 3: Calculate the total probability of drawing a red ball
Using the Total Probability Theorem:
\[ P(A) = P(E_1)P(A|E_1) + P(E_2)P(A|E_2) \] \[ P(A) = \left( \frac{1}{3} \right) \left( \frac{3}{7} \right) + \left( \frac{2}{3} \right) \left( \frac{4}{7} \right) \] \[ P(A) = \frac{3}{21} + \frac{8}{21} = \frac{11}{21} \]
• Law of Total Probability: If \( E_1, E_2 \) are mutually exclusive events, then \( P(A) = P(E_1)P(A|E_1) + P(E_2)P(A|E_2) \).
Step 1: Define the events and find probabilities of bag selection
Let \( E_1 \) be the event that Bag I is chosen (die shows \( \{1, 2\} \)). Let \( E_2 \) be the event that Bag II is chosen (die shows \( \{3, 4, 5, 6\} \)). \[ P(E_1) = \frac{2}{6} = \frac{1}{3} \] \[ P(E_2) = \frac{4}{6} = \frac{2}{3} \]
Step 2: Find the conditional probabilities of drawing a red ball
Let \( A \) be the event that the ball drawn is red. Bag I has 3 red and 4 white balls (Total 7). \[ P(A|E_1) = \frac{3}{7} \] Bag II has 8 red and 6 white balls (Total 14). \[ P(A|E_2) = \frac{8}{14} = \frac{4}{7} \]
Step 3: Calculate the total probability of drawing a red ball
Using the Total Probability Theorem:
\[ P(A) = P(E_1)P(A|E_1) + P(E_2)P(A|E_2) \] \[ P(A) = \left( \frac{1}{3} \right) \left( \frac{3}{7} \right) + \left( \frac{2}{3} \right) \left( \frac{4}{7} \right) \] \[ P(A) = \frac{3}{21} + \frac{8}{21} = \frac{11}{21} \]
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