There are three bags, each containing 100 marbles. Bag 1 has 75 red and 25 blue marbles. Bag 2 has 60 red and

9.4 · Q47

Q. There are three bags, each containing 100 marbles. Bag 1 has 75 red and 25 blue marbles. Bag 2 has 60 red and 40 blue marbles and Bag 3 has 45 red and 55 blue marbles. One of the bags is chosen at random and a marble is picked from the chosen bag. What is the probability that the chosen marble is red?

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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.

Each bag is chosen with prior probability $1/3$ . $P(\text{red}/\text{Bag1})=0.75$ , $P(\text{red}/\text{Bag2})=0.60$ , $P(\text{red}/\text{Bag3})=0.45$ . By the law of total probability, $P(\text{red})=\frac{1}{3}(0.75)+\frac{1}{3}(0.60)+\frac{1}{3}(0.45)=\frac{0.75+0.60+0.45}{3}=\frac{1.8}{3}=0.6$ .

✓Final answer

0.6=\frac{3}{5}

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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