Theory of Probability: Miscellaneous Exercise for Class 11 NCERT Maths

Unpacking the Miscellaneous Exercise

The miscellaneous exercise in the Theory of Probability chapter serves as a crucial checkpoint for Class 11 students. Unlike standard exercises that often focus on a single concept, this section presents problems that integrate various probability principles learned throughout the chapter. You'll encounter scenarios requiring you to skillfully apply your knowledge of sample spaces, events, mutually exclusive events, independent events, conditional probability, and the multiplication theorem. Some problems might even indirectly touch upon the total probability concept by requiring careful partitioning of the sample space. The true value of this exercise lies in its ability to enhance your problem-solving skills, teaching you to break down complex questions into simpler, manageable parts. It's an excellent opportunity to identify areas where your understanding might be weak and to solidify your grasp of foundational probability theory before moving on to more advanced topics in higher classes.

Key Probability Concepts Revisited

Sample Space (S)

The set of all possible outcomes of a random experiment.

Event (E)

A subset of the sample space. An event occurs if any outcome in that subset occurs.

Mutually Exclusive Events

Two events A and B are mutually exclusive if they cannot occur at the same time, i.e., their intersection is an empty set (A ∩ B = ∅).

Independent Events

Two events A and B are independent if the occurrence of one does not affect the probability of the occurrence of the other. Mathematically, P(A ∩ B) = P(A)P(B).

Conditional Probability (P(A|B))

The probability of event A occurring, given that event B has already occurred. It is calculated as P(A ∩ B) / P(B), provided P(B) > 0.

Multiplication Theorem on Probability

For any two events A and B, P(A ∩ B) = P(A)P(B|A) = P(B)P(A|B), provided P(A) > 0 and P(B) > 0. For independent events, this simplifies to P(A ∩ B) = P(A)P(B).

Strategizing Your Solutions

1. Understand the Experiment & Define Sample Space — Carefully read the problem to identify the random experiment. List or describe the entire sample space (S) and determine n(S), the total number of possible outcomes. For complex scenarios, use combinations (nCr) or permutations (nPr) if order matters or not.
2. Identify Events of Interest — Clearly define the events mentioned in the question (e.g., Event A: 'getting a head on the first toss', Event B: 'drawing a red ball'). Write them down precisely to avoid confusion.
3. Determine Relationships Between Events — Ask yourself: Are the events mutually exclusive? Are they independent? Is one event conditional on another? This step is crucial for selecting the correct formula. If events are dependent, conditional probability will likely be involved.
4. Choose the Correct Probability Formula — Based on the relationships identified, apply the appropriate formula: P(A∪B) = P(A) + P(B) - P(A∩B) for general events; P(A∪B) = P(A) + P(B) for mutually exclusive events; P(A∩B) = P(A)P(B) for independent events; P(A|B) = P(A∩B)/P(B) for conditional probability, etc.
5. Calculate Probabilities Systematically — Calculate the individual probabilities needed for your chosen formula. Show your steps clearly. For problems involving 'at least', consider using the complementary event (1 - P(none)).
6. Review and Verify — After arriving at an answer, check if it makes sense. Probabilities must always be between 0 and 1, inclusive. Re-read the question to ensure you have answered exactly what was asked.

Worked Examples to Master Miscellaneous Problems

• Example 1: Conditional Probability with Card Draws Question: A bag contains 5 red and 3 black balls. Two balls are drawn without replacement. What is the probability that the second ball drawn is red, given that the first ball drawn was black? Solution: Step 1: Define events. Let R1 be the event that the first ball drawn is red. Let B1 be the event that the first ball drawn is black. Let R2 be the event that the second ball drawn is red. Let B2 be the event that the second ball drawn is black. Step 2: Identify the given information and what needs to be found. We are given that the first ball drawn was black (event B1 occurred). We need to find the probability that the second ball drawn is red (event R2 occurs) given B1. This is a conditional probability problem: P(R2 | B1). Step 3: Calculate relevant probabilities. Initially, there are 5 red and 3 black balls, total 8 balls. P(B1) = Probability that the first ball is black = 3/8. Now, if the first ball drawn was black, the bag now contains: 5 red balls (since a black ball was removed) 2 black balls (since one black ball was removed) Total balls remaining = 7. Step 4: Calculate the conditional probability. The probability of drawing a red ball as the second ball, given that the first was black, is the number of red balls remaining divided by the total remaining balls. P(R2 | B1) = Number of red balls left / Total balls left = 5 / 7. Final Answer: The probability that the second ball drawn is red, given the first was black, is 5/7.
• Example 2: Probability with Combinations and 'At Least' Question: From a group of 5 men and 4 women, a committee of 3 is to be formed. What is the probability that the committee consists of at least 2 men? Solution: Step 1: Determine the total sample space. Total number of people = 5 men + 4 women = 9. A committee of 3 is to be formed. The order of selection does not matter, so we use combinations. Total number of ways to form a committee of 3 from 9 people, n(S) = ⁹C₃ = (9 × 8 × 7) / (3 × 2 × 1) = 3 × 4 × 7 = 84. Step 2: Define the event of interest. Let E be the event that the committee consists of at least 2 men. 'At least 2 men' means the committee can have: (a) 2 men and 1 woman OR (b) 3 men and 0 women. Step 3: Calculate the number of outcomes for each case of event E. Case (a): 2 men and 1 woman. Number of ways to choose 2 men from 5 = ⁵C₂ = (5 × 4) / (2 × 1) = 10. Number of ways to choose 1 woman from 4 = ⁴C₁ = 4. Number of ways for Case (a) = ⁵C₂ × ⁴C₁ = 10 × 4 = 40. Case (b): 3 men and 0 women. Number of ways to choose 3 men from 5 = ⁵C₃ = (5 × 4 × 3) / (3 × 2 × 1) = 10. Number of ways to choose 0 women from 4 = ⁴C₀ = 1. Number of ways for Case (b) = ⁵C₃ × ⁴C₀ = 10 × 1 = 10. Step 4: Calculate the total number of outcomes for event E. Total number of ways for event E, n(E) = 40 (for 2 men, 1 woman) + 10 (for 3 men, 0 women) = 50. Step 5: Calculate the probability. P(E) = n(E) / n(S) = 50 / 84. Step 6: Simplify the fraction. P(E) = 25 / 42. Final Answer: The probability that the committee consists of at least 2 men is 25/42.

Exam Strategies and Common Pitfalls

When tackling the miscellaneous exercises in probability for your CBSE Class 11 Maths exam, it's easy to fall into common traps. One frequent mistake is confusing mutually exclusive events with independent events. Remember, mutually exclusive events cannot happen together (P(A∩B) = 0), while independent events mean the occurrence of one doesn't influence the other (P(A∩B) = P(A)P(B)). They are distinct concepts! Another pitfall is incorrectly determining the sample space, especially in problems involving combinations or permutations; always decide if order matters and if replacement is allowed. For 'at least' or 'at most' scenarios, consider using the complementary event to simplify calculations (P(A) = 1 - P(A')). Lastly, always write down your assumptions and clearly define your events. This structured approach not only helps in solving complex problems but also earns you partial marks even if your final answer is slightly off.

Practice Questions with Solutions

• Q: A die is thrown twice. What is the probability that the sum of the numbers appearing is 7, given that the first number was 4? A: Step 1: Define events. Let A be the event that the sum of the numbers is 7. Let B be the event that the first number was 4. Step 2: Determine the sample space for B. If the first number was 4, the possible outcomes are (4,1), (4,2), (4,3), (4,4), (4,5), (4,6). So, n(B) = 6. Step 3: Determine A ∩ B. For the sum to be 7, given the first number is 4, the second number must be 3. So, A ∩ B = {(4,3)}. Step 4: Calculate P(A|B). P(A|B) = n(A ∩ B) / n(B) = 1 / 6. Final answer: The probability is 1/6.
• Q: In a class of 60 students, 30 like maths, 20 like science, and 10 like both. A student is selected at random. What is the probability that the student likes maths or science? A: Step 1: Define events and given probabilities. Let M be the event that a student likes Maths. P(M) = 30/60 = 1/2. Let S be the event that a student likes Science. P(S) = 20/60 = 1/3. Let M ∩ S be the event that a student likes both Maths and Science. P(M ∩ S) = 10/60 = 1/6. Step 2: Apply the addition theorem for probability. P(M ∪ S) = P(M) + P(S) - P(M ∩ S). Step 3: Substitute the values. P(M ∪ S) = (1/2) + (1/3) - (1/6) = (3/6) + (2/6) - (1/6) = 4/6. Step 4: Simplify the result. P(M ∪ S) = 2/3. Final answer: The probability that the student likes maths or science is 2/3.
• Q: A box contains 10 electric bulbs, out of which 4 are defective. If two bulbs are chosen at random without replacement, what is the probability that both are non-defective? A: Step 1: Determine total number of bulbs and non-defective bulbs. Total bulbs = 10. Defective bulbs = 4. Non-defective bulbs = 10 - 4 = 6. Step 2: Calculate probability of the first bulb being non-defective. P(1st non-defective) = 6/10. Step 3: Calculate probability of the second bulb being non-defective, given the first was non-defective (without replacement). After drawing one non-defective bulb, 5 non-defective bulbs remain and 9 total bulbs remain. P(2nd non-defective | 1st non-defective) = 5/9. Step 4: Apply the multiplication theorem. P(both non-defective) = P(1st non-defective) × P(2nd non-defective | 1st non-defective) = (6/10) × (5/9). Step 5: Calculate and simplify. (6/10) × (5/9) = (3/5) × (5/9) = 3/9 = 1/3. Final answer: The probability that both bulbs are non-defective is 1/3.
• Q: Three coins are tossed simultaneously. Consider the event E 'at least two heads' and event F 'no head'. Are E and F mutually exclusive? A: Step 1: List the sample space. When three coins are tossed, the sample space S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}. Step 2: Define event E 'at least two heads'. E = {HHH, HHT, HTH, THH}. Step 3: Define event F 'no head'. F = {TTT}. Step 4: Check for intersection of E and F. E ∩ F = { } (empty set). Step 5: Conclude based on intersection. Since the intersection of E and F is an empty set, E and F are mutually exclusive events. Final answer: Yes, E and F are mutually exclusive events.

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