Probability: Miscellaneous Exercise - Class 12 Maths NCERT

Unpacking the Miscellaneous Exercise

The "Miscellaneous Exercise" in any NCERT chapter is strategically placed to challenge your holistic understanding, and in Probability, this is particularly true. It acts as a comprehensive review, weaving together different topics like a tapestry. You might encounter problems that require you to first calculate conditional probabilities, then apply the Theorem of Total Probability, and finally use Bayes' Theorem to find a reverse probability. Other questions might involve defining a random variable, determining its probability distribution, and then calculating its mean and variance, possibly in a context that also includes Bernoulli trials.

Success in this exercise hinges on your ability to break down complex word problems into smaller, manageable parts. It requires careful identification of events, understanding dependencies, and choosing the correct formula or theorem for each step. This section serves as an excellent proving ground for your analytical skills and problem-solving aptitude, essential for both your board exams and future academic pursuits in STEM fields. Don't be intimidated; approach each problem systematically, and you'll find the solutions.

Essential Probability Concepts at a Glance

Conditional Probability

The probability of an event A occurring, given that another event B has already occurred. Denoted as P(A|B) = P(A ∩ B) / P(B), where P(B) > 0.

Theorem of Total Probability

If E1, E2, ..., En are a partition of the sample space S, and A is any event, then P(A) = P(E1)P(A|E1) + P(E2)P(A|E2) + ... + P(En)P(A|En).

Bayes' Theorem

Used to find the probability of a cause, given the effect. P(Ei|A) = [P(Ei)P(A|Ei)] / Σ[P(Ej)P(A|Ej)], where E1, E2, ..., En are mutually exclusive and exhaustive events.

Bernoulli Trials

A sequence of independent trials where each trial has only two possible outcomes (success or failure), and the probability of success remains constant for each trial.

Binomial Distribution

Describes the probability of getting exactly 'x' successes in 'n' Bernoulli trials, where P(X=x) = nCx p^x q^(n-x), with p as probability of success and q as probability of failure (1-p).

Probability Distribution of a Random Variable

A table or function that lists all possible values a random variable can take and their corresponding probabilities. The sum of all probabilities must be 1.

Mean (Expectation) of a Random Variable

E(X) = Σ (xi * P(X=xi)), for a discrete random variable X, representing the average value of the random variable.

Variance of a Random Variable

Var(X) = E(X^2) - [E(X)]^2 = Σ (xi^2 * P(X=xi)) - [E(X)]^2. It measures the spread of the distribution around its mean.

Step-by-Step Solved Problems

• Example 1: Combining Conditional Probability and Bayes' Theorem A factory has two machines A and B. Machine A produces 60% of the items and Machine B produces 40% of the items. 2% of items produced by Machine A are defective, and 1% of items produced by Machine B are defective. An item is chosen at random from the total production and is found to be defective. What is the probability that it was produced by Machine B? Solution: Step 1: Define events. Let E1 be the event that the item is produced by Machine A. Let E2 be the event that the item is produced by Machine B. Let D be the event that the item is defective. Step 2: Write down the given probabilities. P(E1) = 0.60 (60% of items from Machine A) P(E2) = 0.40 (40% of items from Machine B) P(D|E1) = 0.02 (2% defective from Machine A) P(D|E2) = 0.01 (1% defective from Machine B) Step 3: Identify what needs to be found. We need to find the probability that the item was produced by Machine B, given that it is defective, i.e., P(E2|D). Step 4: Apply Bayes' Theorem. P(E2|D) = [P(E2) P(D|E2)] / [P(E1) P(D|E1) + P(E2) P(D|E2)] P(E2|D) = [0.40 0.01] / [0.60 0.02 + 0.40 0.01] P(E2|D) = 0.0040 / [0.0120 + 0.0040] P(E2|D) = 0.0040 / 0.0160 P(E2|D) = 4/16 = 1/4 Final answer: The probability that the defective item was produced by Machine B is 1/4 or 0.25. Example 2: Probability Distribution and Expectation Two cards are drawn simultaneously (or successively without replacement) from a well-shuffled pack of 52 cards. Find the probability distribution of the number of aces. Also, find the mean of this distribution. Solution: Step 1: Define the random variable. Let X be the random variable representing the number of aces drawn when two cards are drawn from a pack of 52 cards. The possible values for X are 0, 1, or 2 aces. Step 2: Calculate the total number of ways to draw 2 cards. Total ways = 52C2 = (52 51) / (2 1) = 26 51 = 1326. Step 3: Calculate probabilities for each value of X. There are 4 aces and 48 non-aces in a deck. Case 1: X = 0 (No aces) P(X=0) = (Number of ways to choose 0 aces and 2 non-aces) / Total ways P(X=0) = (4C0 48C2) / 52C2 P(X=0) = (1 (48 47) / (2 1)) / 1326 P(X=0) = (1 1128) / 1326 = 1128 / 1326 = 188 / 221 Case 2: X = 1 (One ace) P(X=1) = (Number of ways to choose 1 ace and 1 non-ace) / Total ways P(X=1) = (4C1 48C1) / 52C2 P(X=1) = (4 48) / 1326 P(X=1) = 192 / 1326 = 32 / 221 Case 3: X = 2 (Two aces) P(X=2) = (Number of ways to choose 2 aces and 0 non-aces) / Total ways P(X=2) = (4C2 48C0) / 52C2 P(X=2) = ((4 3) / (2 1) 1) / 1326 P(X=2) = (6 1) / 1326 = 6 / 1326 = 1 / 221 Step 4: Form the probability distribution. | X | 0 | 1 | 2 | |--------|-----------|-----------|-----------| | P(X) | 188/221 | 32/221 | 1/221 | Step 5: Verify the sum of probabilities. 188/221 + 32/221 + 1/221 = (188 + 32 + 1) / 221 = 221 / 221 = 1. (Correct) Step 6: Calculate the mean (Expectation) E(X). E(X) = Σ [xi P(X=xi)] E(X) = (0 188/221) + (1 32/221) + (2 * 1/221) E(X) = 0 + 32/221 + 2/221 E(X) = 34/221 = 2/13 Final answer: The probability distribution is as shown in the table, and the mean of the distribution is 2/13.

Exam Strategies and Common Pitfalls

Navigating the miscellaneous exercise effectively requires more than just knowing formulas; it demands a strategic approach. Here are some key tips and common pitfalls to avoid:

1. Read Carefully, Identify Events: The biggest mistake is misinterpreting the problem statement. Clearly define all events (E1, E2, A, B, etc.) and write down all given probabilities. A tree diagram can be incredibly helpful for visualizing complex scenarios, especially those involving multiple stages or conditional probabilities.
2. Choose the Right Tool: Don't rush to apply Bayes' Theorem or Binomial Distribution. First, determine what kind of probability is being asked: conditional, total probability, or reverse probability. Is it a sequence of independent trials? Is there a random variable defined?
3. Check for Independence/Dependence: This is crucial. If events are independent, P(A∩B) = P(A)P(B). If dependent, you'll need conditional probability P(A∩B) = P(A)P(B|A).
4. Mutually Exclusive and Exhaustive Events: For Total Probability Theorem and Bayes' Theorem, ensure your 'causes' (E1, E2, ...) are mutually exclusive (cannot happen simultaneously) and exhaustive (cover all possibilities of the sample space).
5. Probability Distribution Validation: When forming a probability distribution, always verify that the sum of all probabilities is exactly 1. If it's not, you've made a calculation error.
6. Calculations and Simplification: Be meticulous with your calculations. Probability values are often fractions or decimals, and errors can easily creep in. Simplify fractions only at the end if required, or keep them for intermediate steps if that makes the calculation clearer.
7. Practice Diverse Problems: The miscellaneous exercise covers a broad range. Don't stick to one type of problem. Practice various combinations of concepts to build versatility.

Practice Questions with Solutions

• Q: A die is thrown again and again until three sixes are obtained. Find the probability of obtaining the third six in the 6th throw of the die. A: Step 1: Identify the type of problem. This is a problem involving Bernoulli trials and the concept of negative binomial distribution implicitly. We need exactly two sixes in the first 5 throws and then a six on the 6th throw. Step 2: Define success and failure. Success (S): Getting a six (P(S) = 1/6) Failure (F): Not getting a six (P(F) = 5/6) Step 3: Apply the binomial probability for the first 5 throws. We need 2 successes (sixes) in the first 5 trials. Let Y be the number of sixes in 5 throws. P(Y=2) = 5C2 (1/6)^2 (5/6)^(5-2) = 10 (1/36) (125/216) = 1250 / 7776. Step 4: Consider the 6th throw. The 6th throw must be a six, and its probability is 1/6. Since the throws are independent, we multiply the probabilities. Step 5: Calculate the final probability. Required Probability = P(Y=2 in first 5 throws) P(six on 6th throw) Required Probability = (1250 / 7776) (1/6) = 1250 / 46656 = 625 / 23328. Final answer: The probability of obtaining the third six in the 6th throw is 625/23328.
• Q: An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted, and it is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn, and then a ball is drawn at random. What is the probability that the second ball is red? A: Step 1: Define events and their initial probabilities. Let R1 be the event that the first ball drawn is red. P(R1) = 5/10 = 1/2. Let B1 be the event that the first ball drawn is black. P(B1) = 5/10 = 1/2. Let R2 be the event that the second ball drawn is red. Step 2: Calculate conditional probabilities for the second draw. If R1 occurred: The urn now has (5+2) = 7 red balls and 5 black balls (total 12 balls). P(R2|R1) = 7/12. If B1 occurred: The urn now has 5 red balls and (5+2) = 7 black balls (total 12 balls). P(R2|B1) = 5/12. Step 3: Apply the Theorem of Total Probability. P(R2) = P(R1)P(R2|R1) + P(B1)P(R2|B1) P(R2) = (1/2) (7/12) + (1/2) (5/12) P(R2) = 7/24 + 5/24 = 12/24 = 1/2. Final answer: The probability that the second ball drawn is red is 1/2.
• Q: In a group of 100 students, 20 are engineers and 80 are non-engineers. 10% of engineers are girls and 40% of non-engineers are girls. If a student is chosen at random and is found to be a girl, what is the probability that she is an engineer? A: Step 1: Define events and initial probabilities. Let E be the event that a student is an engineer. P(E) = 20/100 = 0.2. Let NE be the event that a student is a non-engineer. P(NE) = 80/100 = 0.8. Let G be the event that a student is a girl. Step 2: Write down conditional probabilities of being a girl. P(G|E) = 10% = 0.1 (10% of engineers are girls) P(G|NE) = 40% = 0.4 (40% of non-engineers are girls) Step 3: Identify what needs to be found. We need to find P(E|G), the probability that the chosen student is an engineer given that she is a girl. Step 4: Apply Bayes' Theorem. P(E|G) = [P(E) P(G|E)] / [P(E) P(G|E) + P(NE) P(G|NE)] P(E|G) = [0.2 0.1] / [0.2 0.1 + 0.8 0.4] P(E|G) = 0.02 / [0.02 + 0.32] P(E|G) = 0.02 / 0.34 = 2 / 34 = 1 / 17. Final answer: The probability that the girl chosen is an engineer is 1/17.
• Q: A random variable X has the following probability distribution: | X | 0 | 1 | 2 | 3 | 4 | |--------|-----|-------|-------|-------|-------| | P(X) | 0.1 | 0.2 | k | 2k | 0.2 | Find the value of k, and then calculate E(X) and Var(X). A: Step 1: Find the value of k. For any probability distribution, the sum of all probabilities must be 1. ΣP(X) = P(0) + P(1) + P(2) + P(3) + P(4) = 1 0.1 + 0.2 + k + 2k + 0.2 = 1 0.5 + 3k = 1 3k = 1 - 0.5 3k = 0.5 k = 0.5 / 3 = 1/6. Step 2: Write the complete probability distribution. | X | 0 | 1 | 2 | 3 | 4 | |--------|-----|-------|--------|--------|-------| | P(X) | 0.1 | 0.2 | 1/6 | 2/6 = 1/3 | 0.2 | Step 3: Calculate the Expectation E(X). E(X) = Σ [xi P(X=xi)] E(X) = (0 0.1) + (1 0.2) + (2 1/6) + (3 1/3) + (4 0.2) E(X) = 0 + 0.2 + 1/3 + 1 + 0.8 E(X) = 2 + 1/3 = 7/3. Step 4: Calculate E(X^2). E(X^2) = Σ [xi^2 P(X=xi)] E(X^2) = (0^2 0.1) + (1^2 0.2) + (2^2 1/6) + (3^2 1/3) + (4^2 0.2) E(X^2) = 0 + (1 0.2) + (4 1/6) + (9 1/3) + (16 0.2) E(X^2) = 0.2 + 4/6 + 3 + 3.2 E(X^2) = 0.2 + 2/3 + 3 + 3.2 = 6.4 + 2/3 = 19.2/3 + 2/3 = 21.2/3. Step 5: Calculate Variance Var(X). Var(X) = E(X^2) - [E(X)]^2 Var(X) = 21.2/3 - (7/3)^2 Var(X) = 21.2/3 - 49/9 Var(X) = (3 * 21.2 - 49) / 9 = (63.6 - 49) / 9 = 14.6 / 9. Final answer: k = 1/6, E(X) = 7/3, Var(X) = 14.6/9.

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