NCERT Solutions & Concepts for Class 12 Maths Exercise 13.3: Bayes' Theorem

Understanding Partitions and the Theorem of Total Probability

Before diving into Bayes' Theorem, we must understand the structure of our sample space. A set of events $E_1, E_2, \\dots, E_n$ is said to represent a partition of the sample space $S$ if they are pairwise mutually exclusive ($E_i \\cap E_j = \\phi$ for $i \\neq j$), collectively exhaustive ($E_1 \\cup E_2 \\cup \\dots \\cup E_n = S$), and have non-zero probabilities ($P(E_i) > 0$). When these conditions are met, any arbitrary event $A$ associated with $S$ can happen in conjunction with any of these partitions. The Theorem of Total Probability states that the total probability of event $A$ is the sum of the conditional probabilities across all partitions: $P(A) = \\sum_{j=1}^{n} P(E_j) P(A|E_j)$. Think of this as tracing all possible paths that lead to event $A$ and summing their individual likelihoods.

The Step-by-Step Process to Apply Bayes' Theorem

1. Identify the Partitions (Prior Events) — Determine the mutually exclusive and exhaustive events $E_1, E_2, \\dots, E_n$ that can cause the final outcome. Write down their individual probabilities $P(E_i)$.
2. Identify the Common Observed Event — Define the event $A$ that has already occurred or been observed (e.g., 'a red ball is drawn' or 'the medical test returns positive').
3. Determine the Conditional Probabilities — Find the probability of the observed event occurring under each partition, denoted as $P(A|E_i)$.
4. Apply the Bayes' Formula — To find the reverse probability $P(E_k|A)$ (the probability that partition $E_k$ caused event $A$), calculate: $P(E_k|A) = \\frac{P(E_k)P(A|E_k)}{\\sum_{i=1}^{n} P(E_i)P(A|E_i)}$.

Detailed Worked Examples for Exercise 13.3

• Example 1: Bag Selection and Ball Drawing\nProblem: Bag I contains 3 red and 4 black balls, while Bag II contains 5 red and 6 black balls. One bag is chosen at random and a ball is drawn from it. If the ball drawn is red, find the probability that it was drawn from Bag II.\nStep 1: Define the events.\nLet $E_1$ = Bag I is selected, $E_2$ = Bag II is selected. Let $A$ = Drawn ball is red.\nStep 2: Calculate prior probabilities.\nSince both bags are equally likely to be chosen, $P(E_1) = 1/2$ and $P(E_2) = 1/2$.\nStep 3: Calculate conditional probabilities.\n$P(A|E_1) = 3/7$ and $P(A|E_2) = 5/11$.\nStep 4: Apply Bayes' Theorem to find $P(E_2|A)$.\n$P(E_2|A) = \\frac{P(E_2)P(A|E_2)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)} = \\frac{\\frac{1}{2} \\times \\frac{5}{11}}{\\left(\\frac{1}{2} \\times \\frac{3}{7}\\right) + \\left(\\frac{1}{2} \\times \\frac{5}{11}\\right)} = \\frac{\\frac{5}{22}}{\\frac{3}{14} + \\frac{5}{22}} = \\frac{\\frac{5}{22}}{\\frac{66 + 70}{308}} = \\frac{35}{68}$.\nFinal Answer: The probability that the red ball was drawn from Bag II is $\\mathbf{\\frac{35}{68}}$.
• Example 2: Disease Diagnostic Testing\nProblem: A laboratory blood test is 99% effective in detecting a disease when it is present. However, the test yields a false positive result for 0.5% of healthy patients. If 0.1% of the population has the disease, what is the probability that a person actually has the disease given their test result is positive?\nStep 1: Define the events.\nLet $E_1$ = Person has the disease, $E_2$ = Person is healthy. Let $A$ = Test result is positive.\nStep 2: Write prior probabilities.\n$P(E_1) = 0.1\\% = 0.001$, so $P(E_2) = 1 - 0.001 = 0.999$.\nStep 3: Write conditional probabilities.\n$P(A|E_1) = 99\\% = 0.99$ and $P(A|E_2) = 0.5\\% = 0.005$.\nStep 4: Apply Bayes' Theorem to calculate $P(E_1|A)$.\n$P(E_1|A) = \\frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)} = \\frac{0.001 \\times 0.99}{(0.001 \\times 0.99) + (0.999 \\times 0.005)} = \\frac{0.00099}{0.00099 + 0.004995} = \\frac{0.00099}{0.005985} = \\frac{22}{133}$.\nFinal Answer: The probability that the person actually has the disease given a positive test is $\\mathbf{\\frac{22}{133}}$ (approx 16.5%).

Exam Traps & Board Tips for Exercise 13.3

• Confusing $P(A|E_i)$ and $P(E_i|A)$: This is the absolute biggest trap in CBSE exams. $P(A|E_i)$ is the forward probability (likelihood of outcome $A$ given condition $E_i$). $P(E_i|A)$ is the backward Bayesian probability (likelihood that condition $E_i$ was the cause given outcome $A$).
• Sum of Priors: Always verify that $P(E_1) + P(E_2) + \\dots + P(E_n) = 1$. If your partition probabilities do not sum to 1, you have misidentified the partitions.
• Simplify Calculations at the End: Do not convert fractions to decimals mid-way. Keep them as fractions; the common denominators almost always cancel out, saving you valuable time during your board exam.

Practice Questions with Solutions

• Q: An insurance company insured 2000 scooter drivers, 4000 car drivers, and 6000 truck drivers. The probability of an accident is 0.01, 0.03, and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?\nA: Step 1: Identify the partitions and prior probabilities. Total drivers = 2000 + 4000 + 6000 = 12000. Let $E_1$ = Scooter driver, $E_2$ = Car driver, $E_3$ = Truck driver. $P(E_1) = 2000/12000 = 1/6$ $P(E_2) = 4000/12000 = 1/3$ $P(E_3) = 6000/12000 = 1/2$ Step 2: Define the observed event $A$ (driver meets with an accident) and conditional probabilities. $P(A|E_1) = 0.01 = 1/100$ $P(A|E_2) = 0.03 = 3/100$ $P(A|E_3) = 0.15 = 15/100$ Step 3: Apply Bayes' Theorem to find $P(E_1|A)$. $P(E_1|A) = \\frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2) + P(E_3)P(A|E_3)}$ Numerator = $(1/6) \\times (1/100) = 1/600$ Denominator = $(1/600) + (1/3 \\times 3/100) + (1/2 \\times 15/100) = 1/600 + 1/100 + 15/200 = \\frac{1 + 6 + 45}{600} = 52/600$ $P(E_1|A) = \\frac{1/600}{52/600} = 1/52$ Final answer: The probability that the accident-prone driver is a scooter driver is $\\mathbf{\\frac{1}{52}}$.
• Q: In a factory which manufactures bolts, machines A, B, and C manufacture respectively 25%, 35%, and 40% of the bolts. Of their outputs, 5, 4, and 2 percent are respectively defective bolts. A bolt is drawn at random from the product and is found to be defective. What is the probability that it was manufactured by machine B?\nA: Step 1: Define events for machine selection. Let $E_1$ = manufactured by machine A, $E_2$ = manufactured by machine B, $E_3$ = manufactured by machine C. $P(E_1) = 25/100$, $P(E_2) = 35/100$, $P(E_3) = 40/100$. Step 2: Let $D$ = drawn bolt is defective. $P(D|E_1) = 5/100$, $P(D|E_2) = 4/100$, $P(D|E_3) = 2/100$. Step 3: Apply Bayes' Theorem to calculate $P(E_2|D)$. $P(E_2|D) = \\frac{P(E_2)P(D|E_2)}{P(E_1)P(D|E_1) + P(E_2)P(D|E_2) + P(E_3)P(D|E_3)}$ Numerator = $\\frac{35}{100} \\times \\frac{4}{100} = \\frac{140}{10000}$ Denominator = $\\frac{(25 \\times 5) + (35 \\times 4) + (40 \\times 2)}{10000} = \\frac{125 + 140 + 80}{10000} = \\frac{345}{10000}$ $P(E_2|D) = \\frac{140}{345} = \\frac{28}{69}$ Final answer: The probability that the defective bolt was manufactured by machine B is $\\mathbf{\\frac{28}{69}}$.
• Q: A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.\nA: Step 1: Define partition events. Let $E_1$ = lost card is a diamond, $E_2$ = lost card is not a diamond. $P(E_1) = 13/52 = 1/4$ $P(E_2) = 39/52 = 3/4$ Step 2: Let $A$ = drawing two diamond cards from the remaining 51 cards. If $E_1$ occurred (diamond lost), 12 diamonds remain: $P(A|E_1) = \\frac{^{12}C_2}{^{51}C_2} = \\frac{12 \\times 11}{51 \\times 50}$ If $E_2$ occurred (non-diamond lost), 13 diamonds remain: $P(A|E_2) = \\frac{^{13}C_2}{^{51}C_2} = \\frac{13 \\times 12}{51 \\times 50}$ Step 3: Apply Bayes' Theorem to find $P(E_1|A)$. $P(E_1|A) = \\frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}$ Numerator = $\\frac{1}{4} \\times \\frac{12 \\times 11}{51 \\times 50}$ Denominator = $\\frac{1}{4} \\times \\frac{12 \\times 11}{51 \\times 50} + \\frac{3}{4} \\times \\frac{13 \\times 12}{51 \\times 50}$ Canceling the common term $\\frac{12}{4 \\times 51 \\times 50}$: $P(E_1|A) = \\frac{11}{11 + (3 \\times 13)} = \\frac{11}{11 + 39} = \\frac{11}{50}$ Final answer: The probability that the lost card was a diamond is $\\mathbf{\\frac{11}{50}}$.
• Q: A doctor is to visit a patient. The probabilities that he will come by train, bus, scooter, or by other means of transport are 3/10, 1/5, 1/10, and 2/5 respectively. The probabilities that he will be late are 1/4, 1/3, and 1/12, if he comes by train, bus, and scooter respectively, but if he comes by other means of transport, he will not be late. When he arrives, he is late. What is the probability that he came by train?\nA: Step 1: Define the partition events. Let $E_1, E_2, E_3, E_4$ be the events that the doctor comes by train, bus, scooter, and other means of transport respectively. $P(E_1) = 3/10$, $P(E_2) = 1/5 = 2/10$, $P(E_3) = 1/10$, $P(E_4) = 2/5 = 4/10$. Step 2: Let $L$ = doctor is late. $P(L|E_1) = 1/4$, $P(L|E_2) = 1/3$, $P(L|E_3) = 1/12$, $P(L|E_4) = 0$. Step 3: Apply Bayes' Theorem to compute $P(E_1|L)$. $P(E_1|L) = \\frac{P(E_1)P(L|E_1)}{\\sum P(E_i)P(L|E_i)}$ Numerator = $\\frac{3}{10} \\times \\frac{1}{4} = \\frac{3}{40}$ Denominator = $\\left(\\frac{3}{10} \\times \\frac{1}{4}\\right) + \\left(\\frac{1}{5} \\times \\frac{1}{3}\\right) + \\left(\\frac{1}{10} \\times \\frac{1}{12}\\right) + 0 = \\frac{3}{40} + \\frac{1}{15} + \\frac{1}{120} = \\frac{9 + 8 + 1}{120} = \\frac{18}{120} = \\frac{3}{20}$ $P(E_1|L) = \\frac{3/40}{3/20} = \\frac{3}{40} \\times \\frac{20}{3} = \\frac{1}{2}$ Final answer: The probability that the doctor came by train is $\\mathbf{\\frac{1}{2}}$.

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