In a bolt factory, machines $A$, $B$, $C$ manufacture $30\%$, $50\%$, $20\%$ of the total output respectively, and $4\%$, $5\%$, $3\%$ of their respective outputs are defective. A bolt is drawn at random from the total output and is found to be defective. Find the probability tha

Question

In a bolt factory, machines $A$, $B$, $C$ manufacture $30\%$, $50\%$, $20\%$ of the total output respectively, and $4\%$, $5\%$, $3\%$ of their respective outputs are defective. A bolt is drawn at random from the total output and is found to be defective. Find the probability that it was manufactured by machine $B$.

Accepted Answer

Concept understanding — Bayes' Theorem

Bayes' Theorem: Learning from Evidence

Imagine you have a bag with 3 red marbles and 7 blue marbles. If you pick one at random, the chance it's red is 3 out of 10 — that's straightforward. But now suppose someone picks a marble, doesn't show it to you, but tells you it's not blue. Suddenly, the only possibilities left are the red marbles. The probability that the hidden marble is red jumps to 1 (certainty). You just updated your belief based on new evidence.

That's the core idea of Bayes' Theorem: how to revise a probability when you get new information. It answers the question: Given that I now know B happened, how should I change my belief about A?

The Intuition in One Sentence

Bayes' Theorem says: The probability that A is true, given that B is true, equals the probability that B would happen if A were true, times the original probability of A, divided by the overall probability of B.

In other words: Your updated belief = (likelihood of the evidence under your hypothesis) × (your prior belief) / (total probability of the evidence).

The Precise Statement

Let $A$ and $B$ be two events. Then:

P(A \mid B) = \frac{P(B \mid A) \cdot P(A)}{P(B)}

Where:

$P(A \mid B)$ is the posterior probability — what you want: the probability of $A$ given that $B$ occurred.
$P(B \mid A)$ is the likelihood — how probable the evidence $B$ is if $A$ is true.
$P(A)$ is the prior probability — your initial belief about $A$ before seeing any evidence.
$P(B)$ is the marginal probability of $B$ — the total chance that $B$ happens, regardless of $A$ .

P(A \mid B) = \frac{P(B \mid A) \cdot P(A)}{P(B)}

Why It Works: A Simple Example

Suppose 1% of a population has a disease. A test for the disease is 99% accurate: it correctly identifies 99% of those who have it (true positive) and correctly says 99% of those who don't have it are negative (true negative). You take the test and get a positive result. What is the probability you actually have the disease?

Many people guess 99%. But that's wrong — and Bayes' Theorem shows why.

Let $D$ = "has the disease", $T^+$ = "tests positive". We know:

$P(D) = 0.01$ (prior)
$P(T^+ \mid D) = 0.99$ (likelihood)
$P(T^+ \mid \text{not } D) = 0.01$ (false positive rate)

First, find $P(T^+)$ , the total probability of a positive test:

P(T^+) = P(T^+ \mid D)P(D) + P(T^+ \mid \text{not } D)P(\text{not } D)

= (0.99)(0.01) + (0.01)(0.99) = 0.0099 + 0.0099 = 0.0198

Now apply Bayes:

P(D \mid T^+) = \frac{0.99 \times 0.01}{0.0198} = \frac{0.0099}{0.0198} = 0.5

So even with a positive test, there's only a 50% chance you have the disease. The test is good, but the disease is rare — most positive results come from the large number of healthy people who get false positives.

Watch out

A common mistake is to confuse $P(B \mid A)$ with $P(A \mid B)$ . In the disease example, $P(\text{positive} \mid \text{disease}) = 0.99$ , but $P(\text{disease} \mid \text{positive}) = 0.5$ . They are not the same.

The General Form (with Multiple Hypotheses)

Often you have several possible causes $A_1, A_2, \dots, A_n$ that partition the sample space. Then for any one cause $A_i$ :

P(A_i \mid B) = \frac{P(B \mid A_i) \cdot P(A_i)}{\sum_{j=1}^n P(B \mid A_j) \cdot P(A_j)}

The denominator is just the law of total probability applied to $B$ .

Why It Matters

Bayes' Theorem is the mathematical foundation of learning from data. It's used everywhere: spam filters update the probability an email is spam based on words it contains; doctors update the probability of a disease based on test results; machine learning algorithms update model parameters as new data arrives. Every time you change your mind because of new evidence, you're doing Bayesian reasoning — whether you know it or not.

The key takeaway: Bayes' Theorem is not a mysterious formula — it's just common sense made precise. It tells you how to weigh new evidence against your prior knowledge, and it prevents you from being fooled by rare events or misleading tests.

$P(D)=0.3(0.04)+0.5(0.05)+0.2(0.03)=0.012+0.025+0.006=0.043$ . $P(B\cap D)=0.5\times0.05=0.025$ . By Bayes' theorem, $P(B\mid D)=\dfrac{0.025}{0.043}=\dfrac{25}{43}\approx0.5814$ . [!ANSWER] $P(B\mid D)=\dfrac{25}{43}\approx0.581$ (about $58.1\%$ ).

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 mark
$Q. (Continuing the doctor case study of Q.36) Name the theorem of probability used in (ii).$
›Reveal solutionSolution

Finding the probability of a cause (mode of transport) given an observed effect (being late) is exactly what Bayes' theorem computes.

Computing $P(\text{he came by train}\mid\text{he is late})$ , i.e. reversing a conditional probability from $P(\text{late}\mid\text{cause})$ to $P(\text{cause}\mid\text{late})$ using the prior probabilities of each cause, is the defining application of Bayes' Theorem.

✓Final answer
$Bayes' Theorem$
CBSE 2026Set SEM31 markMCQ
$\dfrac{4}{5}$
(a) $\dfrac{3}{5}$
(b) $\dfrac{4}{5}$
(c) $\dfrac{4}{7}$
(d) $\dfrac{3}{7}$
›Reveal solutionSolution

Use Bayes' theorem with $P(D)=\tfrac15,\ P(S)=\tfrac45$ and the given sleeping probabilities.

This is a textbook Bayes'-theorem application from the NCERT/CBSE Class 12 probability chapter.

Let $D$ = watches Discovery, $S$ = watches Sports. Given $P(S)=\tfrac45$ , so $P(D)=\tfrac15$ . Also $P(\text{sleep}\mid D)=\tfrac34$ and $P(\text{sleep}\mid S)=\tfrac14$ .

Total probability of falling asleep:

$P(\text{sleep}) = \frac15\cdot\frac34 + \frac45\cdot\frac14 = \frac{3}{20} + \frac{4}{20} = \frac{7}{20}.$

By Bayes' theorem:

$P(D\mid\text{sleep}) = \frac{P(D)P(\text{sleep}\mid D)}{P(\text{sleep})} = \frac{3/20}{7/20} = \frac{3}{7}.$

✓Final answer
$P(\text{Discovery}\mid\text{asleep}) = \dfrac{3}{7}$
CBSE 2025Set ANNUAL1 mark
$Q. Given three identical Boxes-I, II and III, each containing two coins. In Box-I both coins are gold coins, in Box-II both coins are silver coins and in the Box-III, there is one gold coin and one silver coin. A person chooses a Box at random and takes out a coin. Based on the above information answer the following: If the coin drawn is gold, then what is the probability that it is drawn from Box-II?$
›Reveal solutionSolution

Since Box-II has no gold coins at all, it is impossible for a drawn gold coin to have come from Box-II: probability $=0$ , confirmed via Bayes' theorem.

Let $E_1, E_2, E_3$ be the events of selecting Box-I, Box-II, Box-III respectively, each equally likely:

$P(E_1) = P(E_2) = P(E_3) = \frac{1}{3}$

Let $G$ be the event of drawing a gold coin. The conditional probabilities are:

$P(G\mid E_1) = 1, \qquad P(G\mid E_2) = 0, \qquad P(G\mid E_3) = \frac{1}{2}$

By the law of total probability,

$P(G) = P(E_1)P(G\mid E_1) + P(E_2)P(G\mid E_2) + P(E_3)P(G\mid E_3)$

$P(G) = \frac{1}{3}(1) + \frac{1}{3}(0) + \frac{1}{3}\left(\frac{1}{2}\right) = \frac{1}{3} + 0 + \frac{1}{6} = \frac{1}{2}$

By Bayes' theorem,

$P(E_2 \mid G) = \frac{P(E_2)P(G\mid E_2)}{P(G)} = \frac{\tfrac13 \times 0}{\tfrac12} = 0$

✓Final answer
$P(\text{Box-II} \mid \text{gold}) = 0$
CBSE 2025Set ANNUAL1 mark
$Q. Given three identical Boxes-I, II and III, each containing two coins. In Box-I both coins are gold coins, in Box-II both coins are silver coins and in the Box-III, there is one gold coin and one silver coin. A person chooses a Box at random and takes out a coin. Based on the above information answer the following: If the coin drawn is gold coin, then what is the probability that the other coin in the Box is also a gold coin?$
›Reveal solutionSolution

The only box where the other coin is also gold is Box-I, so this probability equals $P(\text{Box-I}\mid \text{gold}) = \tfrac23$ by Bayes' theorem.

If the drawn coin is gold and the other coin in the same box is also gold, the box must be Box-I (the only box with two gold coins). So we need $P(E_1 \mid G)$ .

Using the same events and probabilities as before:

$P(E_1) = \frac{1}{3}, \quad P(G\mid E_1) = 1, \quad P(G) = \frac{1}{2} \ \text{(computed via total probability)}$

By Bayes' theorem,

$P(E_1 \mid G) = \frac{P(E_1)P(G\mid E_1)}{P(G)} = \frac{\tfrac13 \times 1}{\tfrac12} = \frac{1/3}{1/2} = \frac{2}{3}$

✓Final answer
$P(\text{other coin also gold} \mid \text{gold drawn}) = \dfrac{2}{3}$
CBSE 2020Set HE8231 mark
$A$
›Reveal solutionSolution

The statement is True — the computed probability is indeed $\dfrac{4}{5}$ .

Let $E_1$ : the coin actually shows head, $E_2$ : the coin actually shows tail, and $H$ : A reports that head appeared.

$P(E_1)=P(E_2)=\dfrac12$ (fair coin, each outcome equally likely).

Since A speaks the truth with probability $\dfrac45$ : $P(H/E_1)=\dfrac45$ (reports head truthfully when head occurs) and $P(H/E_2)=\dfrac15$ (lies, reporting head when actually tail occurred).

By Bayes' theorem:

$P(E_1/H)=\frac{P(E_1)P(H/E_1)}{P(E_1)P(H/E_1)+P(E_2)P(H/E_2)}=\frac{\tfrac12\cdot\tfrac45}{\tfrac12\cdot\tfrac45+\tfrac12\cdot\tfrac15}=\frac{\tfrac25}{\tfrac25+\tfrac1{10}}=\frac{4/10}{5/10}=\frac45.$

This matches the given value $\dfrac45$ .

✓Final answer
$True.$

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