One bag contains 4 red balls and 3 black balls. A second bag contains 2 red balls and 4 black balls. One of the two bags is selected at random and then one ball is drawn at random from the chosen bag. What is the probability that the ball drawn is red?

Introduction / Context:
This question combines conditional probability with a simple selection between two bags. We choose one bag at random, then draw a ball from that bag and want the probability that the ball is red. The problem naturally leads to the use of the law of total probability over the two possible bag choices.

Given Data / Assumptions:

• Bag 1: 4 red balls, 3 black balls, total 7.
• Bag 2: 2 red balls, 4 black balls, total 6.
• One bag is selected at random, so P(Bag 1) = P(Bag 2) = 1/2.
• From the selected bag, one ball is drawn at random.
• Event of interest: the drawn ball is red.

Concept / Approach:
Let R be the event that the ball drawn is red. Let B1 be the event of choosing Bag 1 and B2 be the event of choosing Bag 2. Using total probability, P(R) = P(B1) * P(R | B1) + P(B2) * P(R | B2). We compute the conditional probabilities by simple fractions within each bag, then average them according to bag selection probabilities.

Step-by-Step Solution:
P(B1) = 1/2, P(B2) = 1/2. In Bag 1, red balls = 4 out of 7, so P(R | B1) = 4 / 7. In Bag 2, red balls = 2 out of 6, so P(R | B2) = 2 / 6 = 1 / 3. Apply total probability: P(R) = (1/2) * (4/7) + (1/2) * (1/3). P(R) = 1/2 * (4/7 + 1/3) = 1/2 * (12/21 + 7/21) = 1/2 * (19/21). P(R) = 19 / 42.

Verification / Alternative check:
We can think in terms of a two step experiment tree: first branch for bag choice, second for colour. The total probability of red equals the sum of probabilities along the two red ending branches: (1/2 * 4/7) and (1/2 * 2/6). Summing these again yields 19/42. No other branches lead to red, so this covers all possibilities.

Why Other Options Are Wrong:
23/42 and 16/39 do not reflect any natural weighting of the two bag probabilities. 7/32 uses a denominator that suggests 32 equally likely outcomes, which is not the case here.

Common Pitfalls:
Some learners average the red proportions as (4/7 + 2/6) / 2 correctly but then forget this corresponds exactly to the total probability formula. Others mistakenly treat the total balls as 13 and attempt a direct ratio 6/13, which ignores the initial random bag choice. Always condition on which bag is chosen in such problems.

Final Answer:
Hence, the probability that the ball drawn is red is 19/42.