Conditional Probability, Total Probability and Bayes' Theorem

Every probability question hands you a sample space. Some questions then hand you something extra: a piece of information that has already come true. "Given that the ball drawn is white." "It is known that the candidate reached late." "Three balls are drawn and all are found to be black." The moment those words appear, the sample space you started with is no longer the sample space you must count in.

Conditional probability is the machinery for that shrinking. Total probability and Bayes' theorem are the two directions you can run it in. Between them they carried six of the nineteen JEE Main 2026 shifts, and they are the single most reliable place in the syllabus to pick up four marks in ninety seconds.

Fundamental concept of conditioning

Row 5 is where marks are lost. Students write $P(A\mid B)=\dfrac{P(A)}{P(B)}$. The numerator is the overlap, not the whole of $A$. If you can only remember one line from this post, remember that the numerator and the denominator must both live inside $B$.

Venn diagram of two overlapping circles A and B inside a rectangle S, with the lens-shaped intersection shaded gold and labelled as the favourable part, and B labelled as the new sample space

The sample space really does shrink

Here is the same pair of events drawn twice. On the left nothing is known, so the rectangle $S$ is the universe and $P(A)$ is the share of the rectangle that $A$ covers. On the right you have been told $B$ happened, so everything outside $B$ is dead. The circle $B$ is now the universe, and the only part of $A$ that survives is $A\cap B$.

Two panels side by side. Left panel shows sample space S with circles A and B overlapping. Right panel greys out everything except B, which is outlined in gold, with A shown dashed and only the overlap shaded

Two consequences follow immediately, and both get asked.

Conditional probability is itself a probability. Fix $B$ and let $A$ vary. Then $P(\cdot\mid B)$ satisfies every axiom that $P(\cdot)$ does. In particular

$$P(A\mid B)+P(A'\mid B)=1$$

Proof in one line: $A\cap B$ and $A'\cap B$ are disjoint and their union is $B$, so $P(A\cap B)+P(A'\cap B)=P(B)$. Divide by $P(B)$.

The addition rule survives conditioning. For any events $A$, $B$ and any $E$ with $P(E)\neq 0$,

$$P\!\left(\frac{A\cup B}{E}\right)=P\!\left(\frac{A}{E}\right)+P\!\left(\frac{B}{E}\right)-P\!\left(\frac{A\cap B}{E}\right)$$

Proof: $(A\cup B)\cap E=(A\cap E)\cup(B\cap E)$, apply the ordinary addition rule to those two sets, note that $(A\cap E)\cap(B\cap E)=A\cap B\cap E$, then divide throughout by $P(E)$.

Two panels. Left shows B outlined in gold split into the parts A cap B and A-complement cap B. Right shows a large gold circle E containing two smaller overlapping circles A and B, illustrating the conditional addition rule

The multiplication theorem: reading the formula backwards

The definition $P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}$ is usually taught as a way to compute a conditional probability. In JEE it is used far more often in the other direction, cleared of its denominator:

$$P(A\cap B)=P(B)\,P(A\mid B)=P(A)\,P(B\mid A)$$

That is the multiplication theorem. It says: to make two things happen, make the first happen, then make the second happen given the first. Drawn as a tree, it says multiply along a path.

Two-stage probability tree from a start node splitting into A and A-complement, with A splitting further into B and B-complement, and the product of the two edge probabilities labelled as P of A intersect B

For three events, condition on more and more history:

$$P(A\cap B\cap C)=P(A)\,P(B\mid A)\,P(C\mid A\cap B)$$

and in general

$$P(A_1\cap A_2\cap\cdots\cap A_n)=P(A_1)\,P(A_2\mid A_1)\,P(A_3\mid A_1\cap A_2)\cdots P(A_n\mid A_1\cap\cdots\cap A_{n-1})$$

This is the whole theory of "drawing without replacement". Second draw depends on the first, third depends on the first two, and the chain rule is exactly the bookkeeping for that dependence.

Example 1. (JEE Main 2026, 6 Apr Shift 2) A bag contains $6$ blue and $6$ green balls. Pairs of balls are drawn without replacement until the bag is empty. Find the probability that every drawn pair has one blue and one green ball.

▸ SHOW SOLUTION

Think of the twelve balls being pulled out in order, positions $1$ to $12$, all $12!$ orders equally likely. The pairs are positions $(1,2)$, $(3,4)$, ..., $(11,12)$.

Do it as a chain, pair by pair. Before the first pair there are $6$ blue and $6$ green. Pick any ball; whatever colour it is, $6$ of the remaining $11$ balls have the other colour, so

$$P(\text{first pair is mixed})=\frac{6}{11}$$

Now $5$ blue and $5$ green remain. By the same argument $P(\text{second pair mixed}\mid\text{first mixed})=\dfrac{5}{9}$, then $\dfrac{4}{7}$, then $\dfrac{3}{5}$, then $\dfrac{2}{3}$, and the last pair is forced, probability $1$.

Multiply along the path:

$$P=\frac{6}{11}\cdot\frac{5}{9}\cdot\frac{4}{7}\cdot\frac{3}{5}\cdot\frac{2}{3}\cdot 1=\frac{720}{10395}=\frac{16}{231}$$

Answer: $\dfrac{16}{231}$, option (3).

Notice what the chain rule bought you. The counting solution needs $\dfrac{6!\,6!\,2^6}{12!}$ and a careful argument about unordered pairs. The conditional chain needs no counting at all.

Example 2. (JEE Main 2026, 24 Jan Shift 1) From a lot of $10$ defective and $90$ non-defective bulbs, $8$ bulbs are selected one by one with replacement. Find the probability of getting at least $7$ defective bulbs.

▸ SHOW SOLUTION

With replacement means the eight draws are independent, so every conditional probability in the chain collapses to the unconditional one: $p=\dfrac{10}{100}=\dfrac1{10}$ each time.

$$P(X=7)={}^8C_7\left(\frac1{10}\right)^{7}\left(\frac9{10}\right)=8\cdot\frac{9}{10^8}=\frac{72}{10^8}$$$$P(X=8)=\left(\frac1{10}\right)^{8}=\frac{1}{10^8}$$$$P(X\ge 7)=\frac{72}{10^8}+\frac{1}{10^8}=\frac{73}{10^8}$$

Answer: $\dfrac{73}{10^{8}}$, option (4).

The words "with replacement" are the whole question. Change them to "without replacement" and every factor becomes a genuine conditional probability and the binomial formula dies.

Independence, and the one statement that is false

$A$ and $B$ are independent when knowing one changes nothing about the other:

$$P(A\mid B)=P(A)\quad\Longleftrightarrow\quad P(A\cap B)=P(A)\,P(B)$$

The product form is the definition worth carrying, because it is symmetric in $A$ and $B$ and it stays legal when $P(B)=0$.

Three facts that follow, each worth proving once:

(i) If $A$ and $B$ are independent, so are $A$ and $B'$.

$$P(A\cap B')=P(A)-P(A\cap B)=P(A)-P(A)P(B)=P(A)\bigl(1-P(B)\bigr)=P(A)P(B')$$

Apply it twice and $A'$ and $B'$ are independent too.

(ii) For independent $A$, $B$: $P(A\cup B)=1-P(A')P(B')$.

(iii) $A$ and $S$ are independent, and $A$ and $\varphi$ are independent, for every $A$.

Now the trap.

Two panels. Left shows two disjoint circles A and B labelled mutually exclusive. Right shows two overlapping circles with the lens shaded, labelled independent

Example 3. (JEE Main 2026, 24 Jan Shift 2) $S$ is a set of $5$ elements and $P(S)$ is its power set. An ordered pair $(A,B)$ is chosen at random from $P(S)\times P(S)$. If $P(A\cap B=\varphi)=\dfrac{3^p}{2^q}$, find $p+q$.

▸ SHOW SOLUTION

Handle the five elements independently. That is the whole idea: the pair $(A,B)$ is built by making five separate, unrelated decisions.

For one element $x$ there are four equally likely states: in $A$ only, in $B$ only, in both, in neither. So $4^5$ ordered pairs in total, which matches $|P(S)|^2=(2^5)^2=4^5$.

$A\cap B=\varphi$ fails exactly when some element sits in both. So for each element only $3$ of the $4$ states are allowed, and the elements are independent:

$$P(A\cap B=\varphi)=\left(\frac34\right)^{5}=\frac{3^5}{4^5}=\frac{3^5}{2^{10}}$$

So $p=5$, $q=10$ and $p+q=15$.

Answer: $15$.

Example 4. (JEE Main 2026, 4 Apr Shift 1) A coin is tossed $8$ times. If $p$ is the probability that exactly $4$ heads appear in the first six tosses and exactly $3$ heads appear in the last five tosses, find $96p$.

▸ SHOW SOLUTION

The two events are not independent, because tosses $4$, $5$, $6$ belong to both windows. Split on the overlap, which is the standard move.

Let $k$ be the number of heads among tosses $4,5,6$. Then tosses $1,2,3$ must supply $4-k$ heads and tosses $7,8$ must supply $3-k$ heads. Feasibility needs $0\le 4-k\le 3$ and $0\le 3-k\le 2$, so $k\in\{1,2,3\}$.

Count the favourable strings of length $8$:

$$k=1:\ {}^3C_3\cdot{}^3C_1\cdot{}^2C_2=1\cdot 3\cdot 1=3$$$$k=2:\ {}^3C_2\cdot{}^3C_2\cdot{}^2C_1=3\cdot 3\cdot 2=18$$$$k=3:\ {}^3C_1\cdot{}^3C_3\cdot{}^2C_0=3\cdot 1\cdot 1=3$$

Total favourable $=3+18+3=24$, and all $2^8=256$ strings are equally likely, so $p=\dfrac{24}{256}$ and

$$96p=96\cdot\frac{24}{256}=9$$

Answer: $9$.

Partition of a sample space

Everything that follows rests on one picture: the sample space chopped into non-overlapping pieces that between them use up all of $S$.

$E_1,E_2,\ldots,E_n$ form a partition of $S$ when all three hold:

$$E_i\cap E_j=\varphi\ \text{ for } i\neq j,\qquad E_1\cup E_2\cup\cdots\cup E_n=S,\qquad P(E_i)>0\ \text{ for every } i$$

In words: mutually exclusive, exhaustive, and none of them impossible. In a question the partition is almost never called a partition. It is "the three bags", "bus or scooter or car", "the coin is fair or two-headed", "the letter came from KANPUR or from ANANTPUR". Spotting it is the entire skill.

The smallest partition is $\{E,E'\}$ for any event $E$ with $0<P(E)<1$. That is why so many two-case questions can be done without ever naming a partition.

A rectangle S divided into five vertical bands labelled E1 to E5, with a gold ellipse A lying across all five bands and the band boundaries dashed inside A

Now look at the ellipse $A$ in that picture. The five dashed cuts break $A$ into five disjoint pieces:

$$A=(E_1\cap A)\cup(E_2\cap A)\cup\cdots\cup(E_n\cap A)$$

and the pieces are pairwise disjoint because the $E_i$ already are. So probabilities simply add:

$$P(A)=P(E_1\cap A)+P(E_2\cap A)+\cdots+P(E_n\cap A)$$

Law of total probability

Apply the multiplication theorem to each term above, $P(E_i\cap A)=P(E_i)\,P(A\mid E_i)$, and you have it.

Theorem. If $E_1,\ldots,E_n$ partition $S$ and $A$ is any event, then

$$P(A)=P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)+\cdots+P(E_n)P(A\mid E_n)=\sum_{i=1}^{n}P(E_i)\,P(A\mid E_i)$$

Proof. $A=A\cap S=A\cap(E_1\cup\cdots\cup E_n)=(A\cap E_1)\cup\cdots\cup(A\cap E_n)$. The sets $A\cap E_i$ are pairwise disjoint, since $(A\cap E_i)\cap(A\cap E_j)\subseteq E_i\cap E_j=\varphi$. By the addition rule for disjoint events, $P(A)=\sum_i P(A\cap E_i)$. Each $P(E_i)>0$, so the multiplication theorem gives $P(A\cap E_i)=P(E_i)P(A\mid E_i)$. Substitute. $\blacksquare$

Probability tree with a start node branching to E1, E2, E3 with probabilities P of Ei, each branching to A with probability P of A given Ei, and the three path products braced together into the total probability formula

Bayes' theorem

Total probability runs forward: causes are known, and you want the chance of the effect. Bayes runs backward: the effect has happened, and you want the chance that a particular cause was responsible.

Two panels separated by a dashed line. Left panel has an arrow from cause Ei to effect A labelled P of A given Ei, marked FORWARD. Right panel has the arrow reversed, from effect A back to cause Ei, labelled P of Ei given A, marked REVERSE

Theorem. Let $E_1,\ldots,E_n$ partition $S$ and let $A$ be an event with $P(A)>0$. Then for each $i$,

$$P(E_i\mid A)=\frac{P(E_i)\,P(A\mid E_i)}{\displaystyle\sum_{j=1}^{n}P(E_j)\,P(A\mid E_j)}$$

Proof. Start from the definition and use the multiplication theorem on the numerator, in the form that conditions on $E_i$:

$$P(E_i\mid A)=\frac{P(E_i\cap A)}{P(A)}=\frac{P(E_i)\,P(A\mid E_i)}{P(A)}$$

Now replace $P(A)$ in the denominator by the law of total probability. $\blacksquare$

That is all Bayes is: the definition of conditional probability, with the denominator expanded. Nothing new is being assumed.

The one-sentence version. $P(E_i\mid A)$ is your path divided by all the paths that reach $A$.

A horizontal stacked bar split into three segments labelled E1, E2, E3, each segment labelled P of Ei times P of A given Ei, with the whole bar labelled P of A and the first segment marked your slice

Vocabulary the paper uses.

Two shortcuts that save real time in the exam.

(a) Equal priors. If $P(E_1)=P(E_2)=\cdots=P(E_n)$, that common value cancels from numerator and denominator, and

$$P(E_k\mid A)=\frac{P(A\mid E_k)}{P(A\mid E_1)+P(A\mid E_2)+\cdots+P(A\mid E_n)}$$

(b) Never simplify the fractions. Put every path product over one common denominator and then throw the denominator away. In a Bayes ratio it cancels, so $\dfrac{12/150}{37/150}=\dfrac{12}{37}$ with no arithmetic at all.

Type 1 — the counting layer underneath

Before a conditional probability can be computed, the plain probabilities in it have to be computed. Every 2026 shift tested that layer, so it is worth clearing first. The three tools are: favourable over total, the complement rule $P(A)=1-P(A')$, and the addition rule.

Example 5. (JEE Main 2026, 22 Jan Shift 1) Two distinct numbers $a$ and $b$ are chosen at random from $1,2,\ldots,50$. Find the probability that $ab$ is divisible by $3$.

▸ SHOW SOLUTION

Direct counting would need three cases. The complement needs one.

$ab$ is not divisible by $3$ exactly when neither $a$ nor $b$ is a multiple of $3$.

Multiples of $3$ in the list: $\left\lfloor \dfrac{50}{3}\right\rfloor =16$. Non-multiples: $50-16=34$.

$$P(\text{not divisible})=\frac{{}^{34}C_2}{{}^{50}C_2}=\frac{34\cdot 33}{50\cdot 49}=\frac{1122}{2450}=\frac{561}{1225}$$$$P(ab\text{ divisible by }3)=1-\frac{561}{1225}=\frac{664}{1225}$$

Answer: $\dfrac{664}{1225}$, option (2).

Example 6. (JEE Main 2026, 21 Jan Shift 1) The mean and variance of the seven observations $2,4,10,x,12,14,y$ (with $x>y$) are $8$ and $16$. Two numbers are then chosen one after another without replacement from $\{1,2,3,x-4,y,5\}$. Find the probability that the smaller of the two is less than $4$.

▸ SHOW SOLUTION

First pin down $x$ and $y$. Sum of the five known values is $2+4+10+12+14=42$, so

$$\frac{42+x+y}{7}=8\ \Rightarrow\ x+y=14$$

Sum of their squares is $4+16+100+144+196=460$, and variance $=\overline{x^2}-\bar x^2$ gives

$$\frac{460+x^2+y^2}{7}-64=16\ \Rightarrow\ x^2+y^2=100$$

Then $2xy=(x+y)^2-(x^2+y^2)=196-100=96$, so $xy=48$. Hence $x,y$ are roots of $t^2-14t+48=0$, that is $t=6$ or $t=8$. Since $x>y$, $x=8$ and $y=6$.

So the set is $\{1,2,3,4,6,5\}$, six distinct numbers. Two are drawn without replacement, giving $6\times 5=30$ equally likely ordered outcomes.

Complement again: "smaller $\ge 4$" means both chosen numbers are $\ge 4$, that is both come from $\{4,5,6\}$, which happens in $3\times 2=6$ ways.

$$P(\text{smaller}<4)=1-\frac{6}{30}=1-\frac15=\frac45$$

Answer: $\dfrac45$, option (2).

Example 7. (JEE Main 2026, 23 Jan Shift 1) From the first $100$ natural numbers, $a$ is chosen and then $b$ is chosen, without replacement. If $P(a-b\ge 10)=\dfrac{m}{n}$ in lowest terms, find $m+n$.

▸ SHOW SOLUTION

Order matters here, so the sample space has $100\times 99=9900$ ordered pairs.

Fix $b$. Then $a$ must satisfy $a\ge b+10$, so $a$ ranges over $b+10,\,b+11,\,\ldots,\,100$, which is $100-(b+10)+1=91-b$ values. This is positive only for $b\le 90$.

$$\text{favourable}=\sum_{b=1}^{90}(91-b)=\sum_{k=1}^{90}k=\frac{90\cdot 91}{2}=4095$$$$P=\frac{4095}{9900}=\frac{91}{220}$$

Since $91=7\cdot 13$ and $220=2^2\cdot 5\cdot 11$ share no factor, $m=91$, $n=220$ and $m+n=311$.

Answer: $311$.

Type 2 — probability wearing another chapter's clothes

A large share of 2026 probability questions were really determinant, quadratic or sequence questions with a probability wrapper. The wrapper is always the same: work out which cases succeed, count them, divide by the total.

Example 8. (JEE Main 2026, 22 Jan Shift 2) $n$ is the number obtained on rolling a fair die. If the probability that the system

$$x-ny+z=6,\qquad x+(n-2)y+(n+1)z=8,\qquad (n-1)y+z=1$$

has a unique solution is $\dfrac{k}{6}$, find the sum of $k$ and all possible values of $n$.

▸ SHOW SOLUTION

A linear system has a unique solution exactly when $\Delta\neq 0$.

$$\Delta=\begin{vmatrix} 1 & -n & 1\\ 1 & n-2 & n+1\\ 0 & n-1 & 1\end{vmatrix}$$

Expand along the first column:

$$\Delta=1\cdot\bigl[(n-2)(1)-(n+1)(n-1)\bigr]-1\cdot\bigl[(-n)(1)-(1)(n-1)\bigr]$$$$\Delta=(n-2-n^2+1)-(-2n+1)=-n^2+3n-2$$

So $\Delta\neq 0\iff n^2-3n+2\neq 0\iff (n-1)(n-2)\neq 0\iff n\neq 1,2$.

A die gives $n\in\{1,2,3,4,5,6\}$, so $4$ of the $6$ outcomes work and $\dfrac{k}{6}=\dfrac46$ gives $k=4$.

$$\text{Sum}=k+(3+4+5+6)=4+18=22$$

Answer: $22$, option (4).

Example 9. (JEE Main 2026, 2 Apr Shift 1) Let $a,b,c\in\{1,2,3,4\}$. If $P\bigl(ax^2+2\sqrt2\,bx+c>0 \text{ for all } x\in\mathbb{R}\bigr)=\dfrac{m}{n}$ in lowest terms, find $m+n$.

▸ SHOW SOLUTION

Since $a\ge 1>0$ always, the quadratic is positive for every real $x$ exactly when its discriminant is negative:

$$(2\sqrt2\,b)^2-4ac<0\ \Rightarrow\ 8b^2-4ac<0\ \Rightarrow\ 2b^2<ac$$

Count by $b$.

$b=1$, need $ac>2$: $a=1$ gives $c\in\{3,4\}$; $a=2$ gives $c\in\{2,3,4\}$; $a=3$ and $a=4$ give all four values of $c$. Count $=2+3+4+4=13$.

$b=2$, need $ac>8$: $a=3$ gives $c\in\{3,4\}$; $a=4$ gives $c\in\{3,4\}$. Count $=4$.

$b=3$ needs $ac>18$ and $b=4$ needs $ac>32$, but $ac\le 16$. Count $=0$.

Favourable $=17$ out of $4^3=64$, and $\gcd(17,64)=1$.

$$m+n=17+64=81$$

Answer: $81$.

Example 10. (JEE Main 2026, 4 Apr Shift 2) From a month of $31$ days, $3$ different dates are chosen at random. If the probability that they are in increasing A.P. is $\dfrac{a}{b}$ in lowest terms, find $a+b$.

▸ SHOW SOLUTION

$$\text{Total}={}^{31}C_3=\frac{31\cdot 30\cdot 29}{6}=4495$$

For $a<b<c$ in A.P., $b=\dfrac{a+c}{2}$, so $b$ is determined by $a$ and $c$, and the only requirement is that $a$ and $c$ have the same parity.

Among $1,\ldots,31$ there are $16$ odd and $15$ even numbers, so

$$\text{favourable}={}^{16}C_2+{}^{15}C_2=120+105=225$$$$P=\frac{225}{4495}=\frac{45}{899}$$

$899=29\cdot 31$ and $45=3^2\cdot 5$, so the fraction is in lowest terms and $a+b=45+899=944$.

Answer: $944$.

Example 11. (JEE Main 2026, 2 Apr Shift 2) A man throws a fair coin repeatedly, scoring $10$ for each head and $5$ for each tail. If $P(\text{exactly }30\text{ points})=\dfrac{m}{n}$ in lowest terms, find $m+n$.

▸ SHOW SOLUTION

"Exactly $30$ points" means the game is described by the pair $(h,t)$ with $10h+5t=30$, that is $2h+t=6$, with $h,t\ge 0$ integers:

$$(h,t)=(0,6),\ (1,4),\ (2,2),\ (3,0)$$

Each such outcome is a specific string of $h+t$ tosses, and the number of arrangements is ${}^{h+t}C_h$, each of probability $\left(\tfrac12\right)^{h+t}$.

$$P={}^6C_0\left(\frac12\right)^{6}+{}^5C_1\left(\frac12\right)^{5}+{}^4C_2\left(\frac12\right)^{4}+{}^3C_3\left(\frac12\right)^{3}$$$$P=\frac{1}{64}+\frac{5}{32}+\frac{6}{16}+\frac{1}{8}=\frac{1+10+24+8}{64}=\frac{43}{64}$$

$\gcd(43,64)=1$, so $m+n=43+64=107$.

Answer: $107$, option (C).

Type 3 — total probability

Recognition test: the question asks for the plain probability of an event, but that event can arrive by more than one route, and the routes have different chances. No "given that" appears in the question. Draw the tree, multiply along paths, add.

Example 12. (JEE Main 2026, 23 Jan Shift 2) Bag A contains $9$ white and $8$ black balls; bag B contains $6$ white and $4$ black balls. One ball is picked at random from bag B and dropped into bag A. A ball is then drawn at random from bag A. If $P(\text{white})=\dfrac{p}{q}$ in lowest terms, find $p+q$.

▸ SHOW SOLUTION

The transferred ball is the partition. Two cases, and they are mutually exclusive and exhaustive.

$$P(\text{white transferred})=\frac{6}{10}=\frac35,\qquad P(\text{black transferred})=\frac{4}{10}=\frac25$$

After the transfer bag A holds $18$ balls in both cases, but the composition differs:

white transferred $\Rightarrow$ $10$ white out of $18$; black transferred $\Rightarrow$ $9$ white out of $18$.

$$P(\text{white drawn})=\frac35\cdot\frac{10}{18}+\frac25\cdot\frac{9}{18}=\frac{30}{90}+\frac{18}{90}=\frac{48}{90}=\frac{8}{15}$$

So $p=8$, $q=15$ and $p+q=23$.

Answer: $23$, option (2).

Two-branch tree. From pick from bag B, the white branch with probability 6 over 10 leads to bag A with 10 white and 8 black, then to white with probability 10 over 18. The black branch with probability 4 over 10 leads to bag A with 9 white and 9 black, then to white with probability 9 over 18

Example 13. (JEE Main 2026, 5 Apr Shift 2) The probabilities that players $A$ and $B$ are selected for the captaincy are $0.6$ and $0.4$. If $A$ is selected the team wins with probability $0.8$; if $B$ is selected it wins with probability $0.7$. Find the probability that the team wins.

▸ SHOW SOLUTION

$\{A \text{ captain},\ B\text{ captain}\}$ is a partition: $0.6+0.4=1$.

$$P(W)=P(A)P(W\mid A)+P(B)P(W\mid B)=(0.6)(0.8)+(0.4)(0.7)$$$$P(W)=0.48+0.28=0.76$$

Answer: $0.76$, option (2).

Sanity check worth doing in three seconds: the answer must lie between the two conditional values $0.7$ and $0.8$, because it is a weighted average of them. It does, and it is closer to $0.8$ because $A$ is the more likely captain.

Example 14. (JEE Main 2026, 6 Apr Shift 1) A bag contains $(N+1)$ coins: $N$ fair coins and one coin with heads on both sides. A coin is chosen at random and tossed. If $P(\text{Head})=\dfrac{9}{16}$, find $N$.

▸ SHOW SOLUTION

The partition is which coin got picked.

$$P(\text{two-headed})=\frac{1}{N+1},\qquad P(\text{fair})=\frac{N}{N+1}$$$$P(H\mid\text{two-headed})=1,\qquad P(H\mid\text{fair})=\frac12$$$$P(H)=\frac{1}{N+1}\cdot 1+\frac{N}{N+1}\cdot\frac12=\frac{2+N}{2(N+1)}$$

Set this equal to $\dfrac{9}{16}$:

$$16(N+2)=18(N+1)\ \Rightarrow\ 16N+32=18N+18\ \Rightarrow\ 2N=14\ \Rightarrow\ N=7$$

Answer: $N=7$, option (2).

Check: with $N=7$, $P(H)=\dfrac{9}{16}=0.5625$, a little above $\tfrac12$, which is right because one coin is biased towards heads.

Two-branch tree from pick a coin. The two-headed branch has probability 1 over N plus 1 and then Head with probability 1. The fair branch has probability N over N plus 1 and then Head with probability one half

Type 4 — Bayes' theorem

Recognition test: the question tells you the effect already happened, and asks which cause was behind it. The give-away phrases are "given that", "it is known that", "is found to be", "was observed to be". The tree is exactly the total-probability tree; only the final step changes, from adding all paths to dividing one path by all of them.

Example 15. (JEE Main 2026, 8 Apr Shift 2) A candidate travels to the exam centre by exactly one of bus, scooter or car, with probabilities $\dfrac25,\dfrac15,\dfrac25$. The probabilities of reaching late are $\dfrac15,\dfrac13,\dfrac14$ respectively. Given that the candidate reached late, find the probability that the candidate travelled by bus.

▸ SHOW SOLUTION

Priors first, and check they add to $1$: $\dfrac25+\dfrac15+\dfrac25=1$. Good.

Path products, all over the common denominator $150$:

$$\text{bus and late}=\frac25\cdot\frac15=\frac{2}{25}=\frac{12}{150}$$$$\text{scooter and late}=\frac15\cdot\frac13=\frac{1}{15}=\frac{10}{150}$$$$\text{car and late}=\frac25\cdot\frac14=\frac{1}{10}=\frac{15}{150}$$$$P(\text{late})=\frac{12+10+15}{150}=\frac{37}{150}$$$$P(\text{Bus}\mid\text{late})=\frac{12/150}{37/150}=\frac{12}{37}$$

Answer: $\dfrac{12}{37}$, option (2).

Once every product sits over $150$, the division is just $12$ against $37$. That is why you never simplify the individual fractions.

Three-branch tree from candidate. Bus with probability 2 over 5 then late with 1 over 5, scooter with 1 over 5 then late with 1 over 3, car with 2 over 5 then late with 1 over 4, and the three path products shown as 12, 10 and 15 over 150

Example 16. (JEE Main 2026, 5 Apr Shift 1) A letter is known to have arrived either from KANPUR or from ANANTPUR. On the envelope just two consecutive letters "AN" are visible. Find the probability that the letter came from ANANTPUR.

▸ SHOW SOLUTION

A word of length $L$ contains $L-1$ consecutive letter pairs, and the visible pair is one of them chosen at random.

KANPUR has $6$ letters, so $5$ pairs: KA, AN, NP, PU, UR. "AN" occurs once.

$$P(\text{AN}\mid\text{KANPUR})=\frac15$$

ANANTPUR has $8$ letters, so $7$ pairs: AN, NA, AN, NT, TP, PU, UR. "AN" occurs twice.

$$P(\text{AN}\mid\text{ANANTPUR})=\frac27$$

The two cities are equally likely, so the priors are $\tfrac12$ each and they cancel:

$$P(\text{ANANTPUR}\mid\text{AN})=\frac{\frac12\cdot\frac27}{\frac12\cdot\frac15+\frac12\cdot\frac27}=\frac{\frac27}{\frac15+\frac27}$$$$\frac15+\frac27=\frac{7+10}{35}=\frac{17}{35}\ \Rightarrow\ P=\frac{2}{7}\cdot\frac{35}{17}=\frac{10}{17}$$

Answer: $\dfrac{10}{17}$, option (2).

Two-branch tree from envelope. The KANPUR branch lists its five letter pairs and carries probability 1 over 5 for AN. The ANANTPUR branch lists its seven pairs and carries probability 2 over 7 for AN

Type 5 — Bayes with many hypotheses

Sometimes the partition has more than three pieces, and occasionally it has infinitely many in principle. The formula does not change. What changes is that you must sum a series in the denominator, which is where the identity ${}^{r}C_r+{}^{r+1}C_r+\cdots+{}^{m}C_r={}^{m+1}C_{r+1}$ earns its place.

Example 17. (JEE Main 2026, 28 Jan Shift 1) A bag contains $10$ balls, of which $k$ are red and $(10-k)$ are black, where $0\le k\le 10$. Three balls are drawn at random without replacement and all are found to be black. Find the probability that the bag contains $1$ red and $9$ black balls.

▸ SHOW SOLUTION

The hypotheses are the eleven possible compositions. Since nothing is said to favour any of them, they are equally likely:

$$P(E_k)=\frac{1}{11},\qquad k=0,1,2,\ldots,10$$

The evidence is $A$: all three drawn balls are black. Under $E_k$ the bag has $10-k$ black balls, so

$$P(A\mid E_k)=\frac{{}^{10-k}C_3}{{}^{10}C_3}\quad\text{for } 10-k\ge 3,\qquad P(A\mid E_k)=0\ \text{ for } k=8,9,10$$

Apply Bayes. Both $\dfrac{1}{11}$ and $\dfrac{1}{{}^{10}C_3}$ appear in every term, so both cancel:

$$P(E_1\mid A)=\frac{{}^{9}C_3}{\displaystyle\sum_{k=0}^{7}{}^{10-k}C_3}=\frac{{}^{9}C_3}{{}^{3}C_3+{}^{4}C_3+\cdots+{}^{10}C_3}$$

The denominator is the hockey-stick sum:

$$\sum_{m=3}^{10}{}^{m}C_3={}^{11}C_4=\frac{11\cdot 10\cdot 9\cdot 8}{24}=330$$

and the numerator is ${}^{9}C_3=\dfrac{9\cdot 8\cdot 7}{6}=84$. Hence

$$P(E_1\mid A)=\frac{84}{330}=\frac{14}{55}$$

Answer: $\dfrac{14}{55}$, option (4).

Bar chart of the eleven weights, 10 minus k choose 3, against k from 0 to 10. The bar at k equals 1 has height 84 and is highlighted in gold, the bars at k equals 8, 9 and 10 are zero

Two things are worth noticing in that solution, because they generalise.

Equal priors let you throw the priors away. All eleven $P(E_k)$ were $\dfrac{1}{11}$, so the answer became a ratio of pure likelihoods. Whenever a question says "chosen at random" about the hypotheses and gives you no other information, expect this.

The impossible hypotheses drop out on their own. For $k=8,9,10$ the bag has fewer than three black balls, so $P(A\mid E_k)=0$ and those branches contribute nothing. You do not need a separate argument to exclude them; the likelihood does it for you. Notice also that the posterior for $k=0$ is the largest of all, $\dfrac{120}{330}$, which matches intuition: three black balls in a row is most easily explained by a bag with no red balls at all.

Type 6 — straight from the definition

Not every conditional question needs a tree. When the sample space is small enough to list, or when you are handed $P(A)$, $P(B)$ and $P(A\cup B)$, the definition alone does the job. These five are IITianForum originals, built to drill the four situations that keep recurring.

Example 18. A fair die is rolled twice. Given that the sum of the two numbers is $8$, find the probability that at least one roll showed a $5$.

▸ SHOW SOLUTION

Condition first, count second. The given event is $B$: sum $=8$. List it:

$$B=\{(2,6),(3,5),(4,4),(5,3),(6,2)\},\qquad n(B)=5$$

Those five ordered pairs are now the entire sample space. Among them, the ones containing a $5$ are $(3,5)$ and $(5,3)$.

$$P(\text{at least one }5\mid \text{sum}=8)=\frac{2}{5}$$

Answer: $\dfrac25$.

The unconditional answer would have been $\dfrac{11}{36}$. The condition changed it, which is exactly what a condition is supposed to do.

Example 19. For two events, $P(A)=0.4$, $P(B)=0.5$ and $P(A\cup B)=0.7$. Find $P(A\mid B)$ and $P(B\mid A')$, and decide whether $A$ and $B$ are independent.

▸ SHOW SOLUTION

From the addition rule,

$$P(A\cap B)=P(A)+P(B)-P(A\cup B)=0.4+0.5-0.7=0.2$$$$P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{0.2}{0.5}=0.4$$

For the second one, use $P(B\cap A')=P(B)-P(A\cap B)=0.5-0.2=0.3$ and $P(A')=0.6$:

$$P(B\mid A')=\frac{0.3}{0.6}=0.5$$

Independence check: $P(A)P(B)=0.4\times 0.5=0.2=P(A\cap B)$, so yes, independent. Consistently, $P(A\mid B)=0.4=P(A)$ and $P(B\mid A')=0.5=P(B)$.

Answer: $P(A\mid B)=0.4$, $P(B\mid A')=0.5$, and $A$, $B$ are independent.

Example 20. A box holds $4$ red and $6$ green pens. Three pens are drawn one by one without replacement. Find (i) $P(\text{all three green})$ and (ii) $P(\text{third is red}\mid\text{first two are green})$.

▸ SHOW SOLUTION

(i) Chain rule along the path:

$$P(GGG)=\frac{6}{10}\cdot\frac{5}{9}\cdot\frac{4}{8}=\frac{120}{720}=\frac16$$

(ii) Do not compute anything. Once two greens are gone the box literally holds $4$ red and $4$ green, so

$$P(\text{third red}\mid GG)=\frac{4}{8}=\frac12$$

Answer: (i) $\dfrac16$, (ii) $\dfrac12$.

Part (ii) is the point of conditional probability in one line: after the condition, you are looking at a genuinely different box.

Example 21. Urn I has $3$ white and $2$ black balls; Urn II has $2$ white and $4$ black balls. A fair die is thrown. If it shows $1$ or $2$, a ball is drawn from Urn I; otherwise from Urn II. The ball drawn turns out to be white. Find the probability that it came from Urn I.

▸ SHOW SOLUTION

Priors from the die:

$$P(\text{I})=\frac{2}{6}=\frac13,\qquad P(\text{II})=\frac{4}{6}=\frac23$$

Likelihoods from the urns:

$$P(W\mid \text{I})=\frac35,\qquad P(W\mid \text{II})=\frac{2}{6}=\frac13$$

Path products, over a common denominator $45$:

$$P(\text{I and }W)=\frac13\cdot\frac35=\frac15=\frac{9}{45},\qquad P(\text{II and }W)=\frac23\cdot\frac13=\frac29=\frac{10}{45}$$$$P(\text{I}\mid W)=\frac{9/45}{9/45+10/45}=\frac{9}{19}$$

Answer: $\dfrac{9}{19}$.

The prior favoured Urn II two to one, but the evidence favours Urn I, and the posterior lands just under a half. Bayes weighs the two effects against each other rather than letting either win outright.

Example 22. $A$ speaks the truth $\dfrac56$ of the time and $B$ speaks the truth $\dfrac35$ of the time. They independently describe the same incident. Find the probability that they contradict each other.

▸ SHOW SOLUTION

They contradict exactly when one tells the truth and the other does not. The two ways are mutually exclusive, and within each way the two speakers are independent:

$$P(\text{contradict})=P(A\text{ true})P(B\text{ false})+P(A\text{ false})P(B\text{ true})$$$$=\frac56\cdot\frac25+\frac16\cdot\frac35=\frac{10}{30}+\frac{3}{30}=\frac{13}{30}$$

Answer: $\dfrac{13}{30}$.

Watch the wording. "Contradict" is this. "Agree" is $\dfrac56\cdot\dfrac35+\dfrac16\cdot\dfrac25=\dfrac{17}{30}$, and the two add to $1$ as they must. "Both are believed" and "the statement is true" are different questions again, and each needs its own sentence read carefully.

Which tool? A thirty-second decision

Almost every mistake in this chapter is a tool-selection mistake made in the first ten seconds. Run the question through this chart before you write anything.

Flow chart. Read the question, then ask whether a condition is given. If NO, break the event over the causes and add, which is total probability. If YES, ask whether the given event is the cause or the effect. Cause given leads to the multiplication theorem, effect given leads to Bayes theorem

In words, and this is worth writing on the inside cover of your notebook.

Quick recall

Conditional probability rewards a habit rather than a memory. Draw the tree, even when the numbers look easy enough to do in your head, because the tree is what stops you from losing a branch. Six questions across JEE Main 2026 came down to exactly that discipline.

Want the full lecture-notes PDF for this chapter, with every theorem proved, all the Venn and tree diagrams drawn out, and a graded problem bank with complete solutions? It is linked with these notes on IITIANFORUM. Real Study, Genuine Results.