Question

What is the probability that he selects a red cap?

Answer 1

Concept: To find the total probability of an event that can happen through multiple mutually exclusive paths, we use the Law of Total Probability. Let \(E\) be the event of selecting a red cap. If \(B_1, B_2, B_3\) represent the mutually exclusive events of choosing Box I, Box II, and Box III respectively, then the total probability \(P(E)\) is given by: \[ P(E) = P(B_1) \cdot P(E|B_1) + P(B_2) \cdot P(E|B_2) + P(B_3) \cdot P(E|B_3) \] Where:
• \(P(B_i)\) is the structural probability of selecting Box \(i\) and picking a cap out, given as \(\frac{i}{6}\) for \(i = 1, 2, 3\).
• \(P(E|B_i)\) is the conditional probability of drawing a red cap specifically from Box \(i\).

Step 1: Extracting individual box selection probabilities.
The problem states that the probability that a Box \(i\) is selected and a cap is picked out is explicitly defined by the mathematical expression \(\frac{i}{6}\) for \(i = 1, 2, 3\). Let us write these probabilities down clearly:
• For Box I (\(i = 1\)): \[ P(B_1) = \frac{1}{6} \]
• For Box II (\(i = 2\)): \[ P(B_2) = \frac{2}{6} = \frac{1}{3} \]
• For Box III (\(i = 3\)): \[ P(B_3) = \frac{3}{6} = \frac{1}{2} \] Let's quickly verify that the sum of these probabilities is valid: \[ P(B_1) + P(B_2) + P(B_3) = \frac{1}{6} + \frac{2}{6} + \frac{3}{6} = \frac{6}{6} = 1 \] This forms a complete set of exhaustive and mutually exclusive events.

Step 2: Determining conditional probabilities of drawing a red cap from each box.
Now, let us analyze the composition of each separate box to find the probability of drawing a red cap, which we denote as event \(R\):
• Box I: Contains exactly 30 red caps and 0 green caps. The total number of caps in Box I is \(30 + 0 = 30\). \[ P(R|B_1) = \frac{\text{Number of red caps in Box I}}{\text{Total caps in Box I}} = \frac{30}{30} = 1 \]
• Box II: Contains exactly 20 red caps and 10 green caps. The total number of caps in Box II is \(20 + 10 = 30\). \[ P(R|B_2) = \frac{\text{Number of red caps in Box II}}{\text{Total caps in Box II}} = \frac{20}{30} = \frac{2}{3} \]
• Box III: Contains exactly 0 red caps and 30 green caps. The total number of caps in Box III is \(0 + 30 = 30\). \[ P(R|B_3) = \frac{\text{Number of red caps in Box III}}{\text{Total caps in Box III}} = \frac{0}{30} = 0 \]

Step 3: Applying the Law of Total Probability to find the final probability.
By substituting our derived fractional individual components into the Law of Total Probability formula, we compute: \[ P(R) = \left[ P(B_1) \cdot P(R|B_1) \right] + \left[ P(B_2) \cdot P(R|B_2) \right] + \left[ P(B_3) \cdot P(R|B_3) \right] \] Substitute the exact values calculated in Step 1 and
Step 2: \[ P(R) = \left( \frac{1}{6} \cdot 1 \right) + \left( \frac{2}{6} \cdot \frac{2}{3} \right) + \left( \frac{3}{6} \cdot 0 \right) \] Now evaluate each term carefully step-by-step:
• First term: \(\frac{1}{6} \cdot 1 = \frac{1}{6}\)
• Second term: \(\frac{2}{6} \cdot \frac{2}{3} = \frac{4}{18} = \frac{2}{9}\)
• Third term: \(\frac{3}{6} \cdot 0 = 0\) Add these non-zero terms together by finding a common denominator for 6 and 9, which is 18: \[ P(R) = \frac{1}{6} + \frac{2}{9} = \frac{1 \cdot 3}{6 \cdot 3} + \frac{2 \cdot 2}{9 \cdot 2} = \frac{3}{18} + \frac{4}{18} \] \[ P(R) = \frac{3 + 4}{18} = \frac{7}{18} \text{ ??? Wait, let's re-add carefully using original fractions: } \] Let's check the values again: \[ P(R) = \frac{1}{6} + \frac{4}{18} = \frac{3}{18} + \frac{4}{18} = \frac{7}{18} \] Wait! Let's re-verify the option values or math fractions. Ah, let's re-read the options: (B) option indicates \(\frac{4}{9}\). Let's check why: \[ P(R) = \frac{1}{6}\cdot 1 + \frac{2}{6}\cdot \frac{2}{3} + 0 = \frac{1}{6} + \frac{4}{18} = \frac{3}{18} + \frac{4}{18} = \frac{7}{18} \] Let's double-check if any information was misread. Box I has 30 red caps. Box II has 20 red, 10 green. Box III has 30 green. Probabilities are \(\frac{1}{6}\), \(\frac{2}{6}\), \(\frac{3}{6}\). Total probability is exactly \(\frac{7}{18}\). Therefore option (A) is \(\frac{7}{18}\). Let's fix the correct answer choice according to correct arithmetic, matching option (A).

Answer 2

Concept: This question requires us to compute a conditional reverse probability, which is solved using Bayes' Theorem. Let \(G\) be the event that a green cap is selected. We want to find the conditional probability \(P(B_2|G)\). According to Bayes' Theorem: \[ P(B_2|G) = \frac{P(B_2) \cdot P(G|B_2)}{P(G)} \] Where the total probability of selecting a green cap, \(P(G)\), can be computed using the law of total probability: \[ P(G) = P(B_1) \cdot P(G|B_1) + P(B_2) \cdot P(G|B_2) + P(B_3) \cdot P(G|B_3) \] Alternatively, since a cap must be either red or green, we can use the complementary relationship: \[ P(G) = 1 - P(R) \]

Step 1: Computing the total probability of selecting a green cap, \(P(G)\).
From the previous question, we already explicitly calculated the total probability of selecting a red cap to be \(P(R) = \frac{7}{18}\). Since a cap can only be either red or green, these two outcomes are completely complementary. Therefore: \[ P(G) = 1 - P(R) = 1 - \frac{7}{18} = \frac{18 - 7}{18} = \frac{11}{18} \] Let us independently verify this total probability using the composition of green caps across all boxes to ensure complete structural reliability:
• \(P(G|B_1) = \frac{0}{30} = 0\)
• \(P(G|B_2) = \frac{10}{30} = \frac{1}{3}\)
• \(P(G|B_3) = \frac{30}{30} = 1\) Now, expanding via the law of total probability: \[ P(G) = \left( \frac{1}{6} \cdot 0 \right) + \left( \frac{2}{6} \cdot \frac{1}{3} \right) + \left( \frac{3}{6} \cdot 1 \right) \] \[ P(G) = 0 + \frac{2}{18} + \frac{3}{6} = \frac{2}{18} + \frac{9}{18} = \frac{11}{18} \] Both methods yield identical results, confirming \(P(G) = \frac{11}{18}\).

Step 2: Calculating the numerator for Bayes' Theorem.
The numerator represents the probability that Box II is chosen and a green cap is drawn from it: \[ \text{Numerator} = P(B_2) \cdot P(G|B_2) = \frac{2}{6} \cdot \frac{1}{3} = \frac{2}{18} \]

Step 3: Evaluating the final conditional probability using Bayes' Theorem.
Now, substitute our calculated numerator value and the total denominator value into the equation: \[ P(B_2|G) = \frac{\frac{2}{18}}{\frac{11}{18}} \] Since the denominators of both the numerator and the denominator fractions are equal to 18, they cancel each other out completely: \[ P(B_2|G) = \frac{2}{11} \] Thus, the posterior probability that the green cap originated from Box II is exactly \(\frac{2}{11}\), which perfectly matches Option (B).