Comprehension
An NGO organises a charity event in which they decide to distribute woollen caps to protect children from winter. The caps to be distributed are in three separate boxes, Box I has 30 red caps, Box II has 20 red and 10 green caps, and Box III has 30 green caps. The probability that a Box \(i\) is selected and a cap picked out is \(i/6\), where \(i = 1, 2, 3\).
Question: 1
What is the probability that he selects a red cap ?
What is the probability that he selects a red cap ?
Show Hint
• Check that the sum of box probabilities is 1: \(1/6 + 2/6 + 3/6 = 1\).
• Total probability problems involve "branching" calculations—multiply along paths and add the final results.
• Total probability problems involve "branching" calculations—multiply along paths and add the final results.
Updated On: Sep 10, 2026
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Solution and Explanation
Concept:
• Let \(B_1, B_2, B_3\) be the events of selecting Box I, II, and III respectively.
• Based on the problem description, the probability of choosing Box \(i\) is \(P(B_i) = i/6\).
• Let \(R\) be the event of selecting a red cap.
• The Law of Total Probability states: \(P(R) = \sum P(B_i) \cdot P(R|B_i)\).
Step 1: Identify probabilities for each box
Box I (\(B_1\)): 30 Red, 0 Green. Total = 30. \[ P(B_1) = 1/6, \quad P(R|B_1) = 30/30 = 1 \] Box II (\(B_2\)): 20 Red, 10 Green. Total = 30. \[ P(B_2) = 2/6 = 1/3, \quad P(R|B_2) = 20/30 = 2/3 \] Box III (\(B_3\)): 0 Red, 30 Green. Total = 30. \[ P(B_3) = 3/6 = 1/2, \quad P(R|B_3) = 0/30 = 0 \]
Step 2: Apply the Law of Total Probability
\[ P(R) = P(B_1)P(R|B_1) + P(B_2)P(R|B_2) + P(B_3)P(R|B_3) \] \[ P(R) = \left( \frac{1}{6} \times 1 \right) + \left( \frac{1}{3} \times \frac{2}{3} \right) + \left( \frac{1}{2} \times 0 \right) \] \[ P(R) = \frac{1}{6} + \frac{2}{9} + 0 \]
Step 3: Calculate the final sum
Find the common denominator (18): \[ P(R) = \frac{3}{18} + \frac{4}{18} = \frac{7}{18} \]
• Let \(B_1, B_2, B_3\) be the events of selecting Box I, II, and III respectively.
• Based on the problem description, the probability of choosing Box \(i\) is \(P(B_i) = i/6\).
• Let \(R\) be the event of selecting a red cap.
• The Law of Total Probability states: \(P(R) = \sum P(B_i) \cdot P(R|B_i)\).
Step 1: Identify probabilities for each box
Box I (\(B_1\)): 30 Red, 0 Green. Total = 30. \[ P(B_1) = 1/6, \quad P(R|B_1) = 30/30 = 1 \] Box II (\(B_2\)): 20 Red, 10 Green. Total = 30. \[ P(B_2) = 2/6 = 1/3, \quad P(R|B_2) = 20/30 = 2/3 \] Box III (\(B_3\)): 0 Red, 30 Green. Total = 30. \[ P(B_3) = 3/6 = 1/2, \quad P(R|B_3) = 0/30 = 0 \]
Step 2: Apply the Law of Total Probability
\[ P(R) = P(B_1)P(R|B_1) + P(B_2)P(R|B_2) + P(B_3)P(R|B_3) \] \[ P(R) = \left( \frac{1}{6} \times 1 \right) + \left( \frac{1}{3} \times \frac{2}{3} \right) + \left( \frac{1}{2} \times 0 \right) \] \[ P(R) = \frac{1}{6} + \frac{2}{9} + 0 \]
Step 3: Calculate the final sum
Find the common denominator (18): \[ P(R) = \frac{3}{18} + \frac{4}{18} = \frac{7}{18} \]
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Question: 2
If he selects a green cap, what is the probability that the cap has come from Box II ?
If he selects a green cap, what is the probability that the cap has come from Box II ?
Show Hint
• If you've already found the probability of the complement (Red), subtraction is much faster for finding \(P(G)\).
• Bayes' Theorem essentially finds the ratio of "success in one specific way" over "total successful ways".
• Bayes' Theorem essentially finds the ratio of "success in one specific way" over "total successful ways".
Updated On: Sep 10, 2026
Show Solution
Verified By Collegedunia
Solution and Explanation
Concept:
• This is a problem of reverse probability, solved using Bayes' Theorem.
• Let \(G\) be the event of selecting a green cap.
• Bayes' Theorem: \(P(B_2|G) = \frac{P(B_2) \cdot P(G|B_2)}{P(G)}\).
Step 1: Calculate the total probability of selecting a green cap \(P(G)\)
We can find this using \(P(G) = 1 - P(R)\) from the previous part: \[ P(G) = 1 - \frac{7}{18} = \frac{11}{18} \] Alternatively, using Law of Total Probability: \[ P(G) = P(B_1)P(G|B_1) + P(B_2)P(G|B_2) + P(B_3)P(G|B_3) \] \[ P(G) = \left(\frac{1}{6} \times 0\right) + \left(\frac{1}{3} \times \frac{10}{30}\right) + \left(\frac{1}{2} \times 1\right) = \frac{1}{9} + \frac{1}{2} = \frac{11}{18} \]
Step 2: Identify the components for the numerator
For Box II (\(B_2\)): \[ P(B_2) = 1/3, \quad P(G|B_2) = 10/30 = 1/3 \] The joint probability \(P(B_2 \cap G) = P(B_2) \cdot P(G|B_2) = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9} \).
Step 3: Apply Bayes' Theorem
\[ P(B_2|G) = \frac{1/9}{11/18} \] \[ P(B_2|G) = \frac{1}{9} \times \frac{18}{11} = \frac{2}{11} \] The probability that the green cap came from Box II is \(2/11\).
• This is a problem of reverse probability, solved using Bayes' Theorem.
• Let \(G\) be the event of selecting a green cap.
• Bayes' Theorem: \(P(B_2|G) = \frac{P(B_2) \cdot P(G|B_2)}{P(G)}\).
Step 1: Calculate the total probability of selecting a green cap \(P(G)\)
We can find this using \(P(G) = 1 - P(R)\) from the previous part: \[ P(G) = 1 - \frac{7}{18} = \frac{11}{18} \] Alternatively, using Law of Total Probability: \[ P(G) = P(B_1)P(G|B_1) + P(B_2)P(G|B_2) + P(B_3)P(G|B_3) \] \[ P(G) = \left(\frac{1}{6} \times 0\right) + \left(\frac{1}{3} \times \frac{10}{30}\right) + \left(\frac{1}{2} \times 1\right) = \frac{1}{9} + \frac{1}{2} = \frac{11}{18} \]
Step 2: Identify the components for the numerator
For Box II (\(B_2\)): \[ P(B_2) = 1/3, \quad P(G|B_2) = 10/30 = 1/3 \] The joint probability \(P(B_2 \cap G) = P(B_2) \cdot P(G|B_2) = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9} \).
Step 3: Apply Bayes' Theorem
\[ P(B_2|G) = \frac{1/9}{11/18} \] \[ P(B_2|G) = \frac{1}{9} \times \frac{18}{11} = \frac{2}{11} \] The probability that the green cap came from Box II is \(2/11\).
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