Chapter 13 · Probability

13.1 Conditional Probability

Once you know that an event $F$ has happened, only the outcomes inside $F$ are still possible. Conditional probability recalculates every probability inside that smaller sample space: $P(E\mid F)=\dfrac{P(E\cap F)}{P(F)}$.

Two dice are thrown behind a screen. What is the probability that you have thrown a double? There are $36$ equally likely outcomes and $6$ of them are doubles, so the answer is $\tfrac{6}{36}=\tfrac16$. Now a friend peeps behind the screen and tells you one fact: "the sum is at least $10$". Is the chance of a double still $\tfrac16$? The fact rules out $30$ of the $36$ outcomes. Only $(4,6)$, $(5,5)$, $(6,4)$, $(5,6)$, $(6,5)$, $(6,6)$ are still possible, they are still equally likely, and two of them — $(5,5)$ and $(6,6)$ — are doubles. Given the new information, the chance of a double has doubled to $\tfrac26=\tfrac13$.

This is how probabilities are updated whenever information arrives: a doctor revises the chance of a disease after a test result, a weather service after a satellite image, a commentator after every over. The information "the event $F$ has happened" shrinks the sample space from $S$ to $F$, and every probability is recalculated inside $F$. In this section you will turn that idea into a formula, learn its rules, and practise the most common mistake to avoid — confusing "$E$ given $F$" with "$F$ given $E$".

Probability Given That an Event Has Happened

Look again at the dice. Write $E$ for "a double" and $F$ for "the sum is at least $10$". The answer $\tfrac13$ came from counting outcomes in $E\cap F$ and dividing by the number of outcomes in $F$:

$$\frac{n(E\cap F)}{n(F)}=\frac{2}{6}=\frac13 .$$

Dividing the top and the bottom by $n(S)=36$ turns the counts into probabilities: $\dfrac{n(E\cap F)/36}{n(F)/36}=\dfrac{P(E\cap F)}{P(F)}$. This second form makes sense even when the outcomes are not equally likely, so it is taken as the definition.

📘 Definition 13.1 — Conditional probability

Let $E$ and $F$ be events of the same sample space $S$ of a random experiment, with $P(F)\ne0$. The conditional probability of $E$ given $F$ — the probability that $E$ happens, given that $F$ has happened — is $$P(E\mid F)=\frac{P(E\cap F)}{P(F)} .$$ The event $F$ is called the conditioning event. When all the outcomes of $S$ are equally likely, this becomes $$P(E\mid F)=\frac{n(E\cap F)}{n(F)} .$$

Read $P(E\mid F)$ as "the probability of $E$ given $F$". The vertical bar is not a division sign. Notice that only the part of $E$ that lies inside $F$ counts: once we know $F$ has happened, the outcomes of $E$ outside $F$ are impossible. That is why the formula has $E\cap F$ on top and not $E$. In effect $F$ becomes a reduced sample space, and $P(E\mid F)$ is the share of $F$ that is taken up by $E$.

✏️ Example 1 — Reducing the sample space

A fair coin is tossed three times. Let $E$: "exactly two heads" and $F$: "the first toss is a head". Find $P(E\mid F)$ and $P(F\mid E)$.

💡 Show solution▶

Sample space $S=\{HHH,HHT,HTH,HTT,$ $THH,THT,TTH,TTT\}$, eight equally likely outcomes.

Events $E=\{HHT,HTH,THH\}$, $F=\{HHH,HHT,HTH,HTT\}$, so $E\cap F=\{HHT,HTH\}$.

Given $F$ $P(E\mid F)=\dfrac{n(E\cap F)}{n(F)}=\dfrac24=\dfrac12$.

Given $E$ Now $E$ is the reduced sample space: $P(F\mid E)=\dfrac{n(E\cap F)}{n(E)}=\dfrac23$.

Check With the formula: $P(E\cap F)=\tfrac28$, $P(F)=\tfrac48$, $P(E)=\tfrac38$, so $P(E\mid F)=\tfrac{2/8}{4/8}=\tfrac12$ and $P(F\mid E)=\tfrac{2/8}{3/8}=\tfrac23$ ✓. The two answers differ: $P(E\mid F)$ and $P(F\mid E)$ are different questions.

✏️ Example 2 — Using the formula

For events $E$ and $F$, $P(E)=0.6$, $P(F)=0.4$ and $P(E\cap F)=0.25$. Find (a) $P(E\mid F)$, (b) $P(F\mid E)$, (c) $P(E\cup F)$, (d) $P(E\mid F')$.

💡 Show solution▶

(a) $P(E\mid F)=\dfrac{0.25}{0.4}=0.625=\dfrac58$.

(b) $P(F\mid E)=\dfrac{0.25}{0.6}=\dfrac{25}{60}=\dfrac5{12}$.

(d) $E\cap F'$ is "$E$ but not $F$", so $P(E\cap F')=P(E)-P(E\cap F)=0.35$, and $P(F')=0.6$. Hence $P(E\mid F')=\dfrac{0.35}{0.6}=\dfrac7{12}$.

Check $P(E)$ should be the weighted average of $P(E\mid F)$ and $P(E\mid F')$ with weights $P(F)$, $P(F')$: $0.4\cdot\tfrac58+0.6\cdot\tfrac7{12}=0.25+0.35=0.6$ ✓. (Section 13.3 turns this check into a theorem.)

The $6\times6$ grid is the sample space of two dice (first die across, second die down). Outcomes in the event $A$ carry a red dot. Choose the information $B$ you are given: the outcomes outside $B$ fade away, and the blue cells form the reduced sample space. Count the dots that survive: $P(A\mid B)=\tfrac{n(A\cap B)}{n(B)}$. Press Play to step through the choices of $B$.

event $A$:given $B$:

$n(B)$—

$n(A\cap B)$—

$P(A)$—

$P(A\mid B)$—

Properties of Conditional Probability

Fix an event $F$ with $P(F)\ne0$. Giving each event $E$ the number $P(E\mid F)$ is a new way of assigning probabilities — the one that is right once $F$ is known. It obeys the same rules as an ordinary probability.

📐 Theorem 13.1 — Properties of conditional probability

Let $E$, $A$, $B$ and $F$ be events of a sample space $S$, with $P(F)\ne0$. Then

(i) $P(S\mid F)=P(F\mid F)=1$;
(ii) $P\big((A\cup B)\mid F\big)$ $=P(A\mid F)+P(B\mid F)$ $-P\big((A\cap B)\mid F\big)$; in particular, if $A$ and $B$ are mutually exclusive, $P\big((A\cup B)\mid F\big)=P(A\mid F)+P(B\mid F)$;
(iii) $P(E'\mid F)=1-P(E\mid F)$;
(iv) $0\le P(E\mid F)\le1$.

Proof. (i) $S\cap F=F$ and $F\cap F=F$, so both are $\dfrac{P(F)}{P(F)}=1$. (ii) By the distributive law, $(A\cup B)\cap F=(A\cap F)\cup(B\cap F)$, and the intersection of these two events is $(A\cap B)\cap F$. The addition rule of Class 11 gives

$$P\big((A\cup B)\cap F\big)=P(A\cap F)+P(B\cap F)-P(A\cap B\cap F).$$

Divide every term by $P(F)$. If $A\cap B=\phi$ the last term is $0$. (iii) $E$ and $E'$ are mutually exclusive and $E\cup E'=S$, so by (ii) and (i), $P(E\mid F)+P(E'\mid F)=P(S\mid F)=1$. (iv) $E\cap F\subset F$, so $0\le P(E\cap F)\le P(F)$; divide by $P(F)\gt0$. $\square$

✏️ Example 3 — Two children

A family has two children. Assume the four outcomes $BB$, $BG$, $GB$, $GG$ (older child first; $B$ = boy, $G$ = girl) are equally likely. Find the probability that both children are boys, given that (a) the older child is a boy; (b) at least one child is a boy.

💡 Show solution▶

Event $E=\{BB\}$, "both are boys".

(a) $F_1=\{BB,BG\}$, so $P(E\mid F_1)=\dfrac{n(E\cap F_1)}{n(F_1)}=\dfrac12$.

(b) $F_2=\{BB,BG,GB\}$, so $P(E\mid F_2)=\dfrac13$.

Why different? "At least one boy" rules out only $GG$, leaving three cases; "the older is a boy" rules out two cases. The precise wording of the information matters — many people wrongly answer $\tfrac12$ to (b).

✏️ Example 4 — Cards

One card is drawn from a well-shuffled pack of $52$ cards. Find (a) $P(\text{king}\mid\text{face card})$; (b) $P(\text{face card}\mid\text{king})$; (c) $P(\text{red}\mid\text{king or queen})$; (d) $P(\text{heart}\mid\text{red})$. (The face cards are the $12$ kings, queens and jacks.)

💡 Show solution▶

(a) Reduced sample space: $12$ face cards, of which $4$ are kings: $\dfrac4{12}=\dfrac13$.

(b) Every king is a face card: $\dfrac44=1$. (Whenever $F\subset E$, $P(E\mid F)=1$.)

(d) $26$ red cards, $13$ hearts: $\dfrac{13}{26}=\dfrac12$. Compare $P(\text{heart})=\tfrac14$: knowing the card is red doubles the chance.

✏️ Example 5 — Using the properties

Two fair dice are thrown and the sum is found to be $6$. Find the probability that (a) at least one die shows a $2$; (b) neither die shows a $2$.

💡 Show solution▶

Given $F$: "sum is $6$" $=\{(1,5),(2,4),(3,3),(4,2),(5,1)\}$, $n(F)=5$.

(a) Let $A$: "first die shows $2$", $B$: "second die shows $2$". $A\cap F=\{(2,4)\}$, $B\cap F=\{(4,2)\}$, and $A\cap B\cap F=\phi$ because $(2,2)$ has sum $4$. By Theorem 13.1(ii): $P\big((A\cup B)\mid F\big)=\tfrac15+\tfrac15-0=\tfrac25$.

(b) "Neither shows a $2$" is $(A\cup B)'$. By Theorem 13.1(iii): $1-\tfrac25=\tfrac35$.

Check The outcomes of $F$ with no $2$ are $(1,5)$, $(3,3)$, $(5,1)$: $\tfrac35$ ✓.

✏️ Example 6 — Reading a two-way table

A coaching institute surveyed $400$ Class 12 students about a mock test. (The numbers are illustrative.)

	passed	failed	total
attends coaching	120	40	160
no coaching	150	90	240
total	270	130	400

A student is chosen at random. Find (a) $P(\text{passed}\mid\text{coaching})$; (b) $P(\text{passed}\mid\text{no coaching})$; (c) $P(\text{coaching}\mid\text{passed})$.

💡 Show solution▶

(a) Reduce to the row "attends coaching": $\dfrac{120}{160}=\dfrac34$.

(b) Row "no coaching": $\dfrac{150}{240}=\dfrac58$.

Interpret Conditioning on a row and conditioning on a column answer different questions: $\tfrac34$ is the pass rate of coached students, $\tfrac49$ is the share of passers who were coached. (A survey like this shows an association, not that coaching causes passing.)

⚠️ Common mistakes when studying alone

1. Swapping the condition. $P(E\mid F)$ and $P(F\mid E)$ have the same numerator $P(E\cap F)$ but different denominators. They are equal only when $P(E)=P(F)$.

2. Using $n(E)$ instead of $n(E\cap F)$. Only the part of $E$ inside $F$ survives the information.

3. Complementing the wrong event. $P(E'\mid F)=1-P(E\mid F)$ is true; $P(E\mid F')=1-P(E\mid F)$ is false in general (Exercise III.5).

4. Conditioning on an impossible event. $P(E\mid F)$ is defined only when $P(F)\ne0$.

5. "Given" in disguise. Words like "if it is known that", "it is found that", "among those who…" all signal a conditional probability.

Conditional probability as a long-run frequency. The lab throws two dice again and again (a seeded random sequence, so every run is the same). Throws that are not in $B$ are thrown away (grey). Among the throws that are kept, the green curve tracks the fraction that are also in $A$; it settles on $P(A\mid B)$ (green dashed line). The grey curve tracks the fraction of all throws in $A$, which settles on $P(A)$.

pair:

throws—

kept (in $B$)—

also in $A$—

exact $P(A\mid B)$—

Exercises

Using the definition.

1 $P(E)=0.5$, $P(F)=0.4$ and $P(E\cap F)=0.2$. Find $P(E\mid F)$ and $P(F\mid E)$.
2 $P(A)=\tfrac38$, $P(B)=\tfrac12$ and $P(A\cap B)=\tfrac14$. Find $P(A\mid B)$ and $P(B\mid A)$.
3 $2P(A)=P(B)=\tfrac5{13}$ and $P(A\mid B)=\tfrac25$. Find $P(A\cap B)$ and $P(A\cup B)$.
4 $P(A)=0.6$, $P(B\mid A)=0.5$ and $P(B)=0.45$. Find $P(A\cap B)$ and $P(A\mid B)$.
5 $P(E)=0.35$, $P(F)=0.45$ and $P(E\cup F)=0.6$. Find $P(E\mid F)$ and $P(F\mid E)$.
6 If $P(B)\gt0$ and $P(A\cap B)=P(B)$, find $P(A\mid B)$.
7 $A\subset B$ with $P(A)=0.2$ and $P(B)=0.5$. Find $P(A\mid B)$ and $P(B\mid A)$.
8 $A$ and $B$ are mutually exclusive and $P(B)\gt0$. Find $P(A\mid B)$.

Check: conditional probabilities from $P(E)$, $P(F)$ and $P(E\cap F)$

P(E):P(F):P(E ∩ F):

—

Enter decimals or fractions such as 5/13. The lab also tells you if the three numbers cannot belong to real events.

💡 Show solutions▶

1 $P(E\mid F)=\tfrac{0.2}{0.4}=0.5$; $P(F\mid E)=\tfrac{0.2}{0.5}=0.4$.

2 $P(A\mid B)=\tfrac{1/4}{1/2}=\tfrac12$; $P(B\mid A)=\tfrac{1/4}{3/8}=\tfrac23$.

3 $P(A)=\tfrac5{26}$. $P(A\cap B)=P(B)\,P(A\mid B)=\tfrac5{13}\cdot\tfrac25=\tfrac2{13}$. $P(A\cup B)=\tfrac5{26}+\tfrac{10}{26}-\tfrac4{26}=\tfrac{11}{26}$.

4 $P(A\cap B)=0.6\times0.5=0.3$; $P(A\mid B)=\tfrac{0.3}{0.45}=\tfrac23$.

5 $P(E\cap F)=0.35+0.45-0.6=0.2$. $P(E\mid F)=\tfrac{0.2}{0.45}=\tfrac49$; $P(F\mid E)=\tfrac{0.2}{0.35}=\tfrac47$.

6 $P(A\mid B)=\tfrac{P(B)}{P(B)}=1$: whenever $B$ happens, $A$ happens too.

7 $A\cap B=A$: $P(A\mid B)=\tfrac{0.2}{0.5}=0.4$ and $P(B\mid A)=\tfrac{0.2}{0.2}=1$.

8 $A\cap B=\phi$, so $P(A\mid B)=\tfrac0{P(B)}=0$: if $B$ has happened, $A$ cannot have.

Counting in the reduced sample space. Coins and dice are fair; cards come from a well-shuffled pack of $52$.

1 A coin is tossed twice. Find the probability of two heads given that at least one head appears.
2 A coin is tossed three times. Find the probability of exactly two heads given that at least one tail appears.
3 A die is thrown. Find the probability that the number is prime, given that it is odd.
4 A die is thrown. Find the probability that the number is greater than $3$, given that it is even.
5 Two dice are thrown. Find the probability that the sum is $8$, given that the first die shows $5$.
6 Two dice are thrown. Find the probability of a double, given that the sum is $8$.
7 Two dice are thrown. Find the probability that the sum is at least $10$, given that at least one die shows a $5$.
8 Two dice are thrown. Find the probability that the product of the numbers is $12$, given that their sum is $7$.
9 One card is drawn. Find $P(\text{ace}\mid\text{black card})$ and $P(\text{spade}\mid\text{ace})$.
10 A number is chosen at random from $1,2,\dots,20$. Find the probability that it is a multiple of $3$, given that it is even.
11 A family has three children, and all $8$ sequences of boys and girls are equally likely. Find the probability that all three are girls, given that at least two are girls.

💡 Show solutions▶

1 $F=\{HH,HT,TH\}$, $E\cap F=\{HH\}$: $\tfrac13$.

2 At least one tail: all outcomes except $HHH$, so $n(F)=7$. Exactly two heads: $HHT,HTH,THH$, all in $F$. Answer $\tfrac37$.

3 $F=\{1,3,5\}$; primes in $F$: $3,5$. Answer $\tfrac23$.

4 $F=\{2,4,6\}$; greater than $3$: $4,6$. Answer $\tfrac23$.

5 $F=\{(5,1),\dots,(5,6)\}$, $n(F)=6$; only $(5,3)$ has sum $8$. Answer $\tfrac16$.

6 Sum $8$: $(2,6),(3,5),(4,4),(5,3),(6,2)$; one double. Answer $\tfrac15$.

7 At least one $5$: $11$ outcomes. Of these, sum $\ge10$: $(5,5),(5,6),(6,5)$. Answer $\tfrac3{11}$.

8 Sum $7$: six outcomes; product $12$: $(3,4),(4,3)$. Answer $\tfrac26=\tfrac13$.

9 $26$ black cards, $2$ black aces: $\tfrac2{26}=\tfrac1{13}$. Of the $4$ aces, $1$ is a spade: $\tfrac14$.

10 $10$ even numbers; multiples of $3$ among them: $6,12,18$. Answer $\tfrac3{10}$.

11 At least two girls: $GGB,GBG,BGG,GGG$ — four outcomes; one is $GGG$. Answer $\tfrac14$.

III

Properties and reasoning. In each part assume the conditioning events have non-zero probability.

1 If $P(E\mid F)=0.35$, find $P(E'\mid F)$, and state which property you used.
2 $P(A\mid C)=0.5$, $P(B\mid C)=0.3$ and $P\big((A\cap B)\mid C\big)=0.1$. Find $P\big((A\cup B)\mid C\big)$ and $P\big((A\cup B)'\mid C\big)$.
3 $P(A)=0.8$, $P(B)=0.5$ and $P(B\mid A)=0.4$. Find $P(A\cap B)$, $P(A\mid B)$ and $P(A\cup B)$.
4 Show that $P(E\mid F)\ge P(E\cap F)$, with equality when $P(F)=1$.
5 A die is thrown; $E=\{1,2\}$ and $F=\{1,2,3\}$. Find $P(E\mid F)$ and $P(E\mid F')$, and show that $P(E\mid F')\ne1-P(E\mid F)$.
6 Show that if $P(E\cap F)\gt0$, then $P(E\mid F)=P(F\mid E)$ exactly when $P(E)=P(F)$.
7 Show that if $P(A\mid B)\gt P(A)$, then $P(B\mid A)\gt P(B)$. (Assume $P(A)\gt0$, $P(B)\gt0$.) Put this into words.

💡 Show solutions▶

1 Theorem 13.1(iii): $P(E'\mid F)=1-0.35=0.65$.

2 Theorem 13.1(ii): $0.5+0.3-0.1=0.7$; then (iii): $1-0.7=0.3$.

3 $P(A\cap B)=P(A)P(B\mid A)=0.32$; $P(A\mid B)=\tfrac{0.32}{0.5}=0.64$; $P(A\cup B)=0.8+0.5-0.32=0.98$.

4 $0\lt P(F)\le1$, so dividing $P(E\cap F)\ge0$ by $P(F)$ cannot make it smaller: $\dfrac{P(E\cap F)}{P(F)}\ge P(E\cap F)$. If $P(F)=1$, the division changes nothing.

5 $P(E\mid F)=\tfrac23$. $F'=\{4,5,6\}$ and $E\cap F'=\phi$, so $P(E\mid F')=0$, while $1-P(E\mid F)=\tfrac13$.

6 $\dfrac{P(E\cap F)}{P(F)}=\dfrac{P(E\cap F)}{P(E)}$; since the common numerator is not zero, cancel it: $P(F)=P(E)$. Conversely equal denominators give equal fractions.

7 $\dfrac{P(A\cap B)}{P(B)}\gt P(A)$ gives $P(A\cap B)\gt P(A)P(B)$; divide by $P(A)$: $P(B\mid A)\gt P(B)$. In words: if knowing $B$ makes $A$ more likely, then knowing $A$ makes $B$ more likely.

Applications. All data are illustrative.

1 A repair shop logged $500$ phones: brand X had $60$ screen faults and $140$ battery faults; brand Y had $120$ screen faults and $180$ battery faults. For a phone chosen at random, find (a) $P(\text{screen fault}\mid X)$; (b) $P(X\mid\text{screen fault})$; (c) $P(\text{battery fault}\mid Y)$.
2 In a city, the probability of rain on a given day is $0.3$, of a major traffic jam is $0.4$, and of both is $0.24$. Find $P(\text{jam}\mid\text{rain})$ and $P(\text{rain}\mid\text{jam})$.
3 A cricket team wins $60\%$ of its matches and wins the toss in $50\%$ of them; it wins both the toss and the match in $35\%$. Find the probability that it wins a match given that it has won the toss, and the probability that it won the toss given that it won the match.
4 A 3-digit PIN is chosen at random from $000$ to $999$. Given that its three digits are all different, find the probability that (a) it begins with $0$; (b) its digits are in increasing order.
5 In a screening programme, $1000$ people were tested: $20$ had a certain condition and $18$ of them tested positive; of the $980$ without the condition, $49$ tested positive. Find $P(\text{positive}\mid\text{condition})$ and $P(\text{condition}\mid\text{positive})$, and comment on the difference.

💡 Show solutions▶

1 Brand X: $200$ phones; screen faults in all: $180$. (a) $\tfrac{60}{200}=0.3$; (b) $\tfrac{60}{180}=\tfrac13$; (c) $\tfrac{180}{300}=0.6$.

2 $P(\text{jam}\mid\text{rain})=\tfrac{0.24}{0.3}=0.8$; $P(\text{rain}\mid\text{jam})=\tfrac{0.24}{0.4}=0.6$.

3 $P(\text{win}\mid\text{toss})=\tfrac{0.35}{0.5}=0.7$; $P(\text{toss}\mid\text{win})=\tfrac{0.35}{0.6}=\tfrac7{12}$.

4 PINs with all digits different: $10\cdot9\cdot8=720$. (a) First digit $0$: $9\cdot8=72$, so $\tfrac{72}{720}=\tfrac1{10}$. (b) Each choice of $3$ different digits can be written in increasing order in exactly one way: $^{10}C_3=120$, so $\tfrac{120}{720}=\tfrac16$.

5 $P(\text{positive}\mid\text{condition})=\tfrac{18}{20}=0.9$. Positives in all: $18+49=67$, so $P(\text{condition}\mid\text{positive})=\tfrac{18}{67}\approx0.27$. The test catches $90\%$ of those with the condition, yet most positives are false alarms, because the healthy group is so much larger. Section 13.3 studies this with Bayes' theorem.

🎯 Exam corner Board · 1 markBoard · 2 marksBoard · Case studyJEE MainJEE Advanced

How this is examined: Board papers ask for $P(E\mid F)$ from given probabilities in 1-mark MCQs and Assertion–Reason items, set 2-mark questions that combine the definition with the addition rule or complements, and build case studies on two-way tables. JEE problems hide the conditioning event in the wording and test whether you count the reduced sample space correctly.

E1. Board · 1 mark If $P(A)=0.4$, $P(B)=0.8$ and $P(B\mid A)=0.6$, then $P(A\cup B)$ is (A) $0.96$ (B) $0.24$ (C) $0.56$ (D) $0.60$

💡 Show solution▶

Idea $P(A\cap B)=P(A)\,P(B\mid A)$ $=0.4\times0.6=0.24$, so $P(A\cup B)=0.4+0.8-0.24=0.96$.

Answer (A). Trap (B) is $P(A\cap B)$, where the calculation stops halfway; (C) is $P(B)-P(A\cap B)$; (D) subtracts the conditional probability $0.6$ instead of the intersection.

E2. Board · 1 mark Assertion–Reason. $E$ and $F$ are events with $0\lt P(F)\lt1$. Assertion (A): $P(E\mid F)+P(E'\mid F)=1$. Reason (R): $P(E\mid F)+P(E\mid F')=1$. (A) Both A and R are true and R is the correct explanation of A (B) Both A and R are true but R is not the correct explanation of A (C) A is true but R is false (D) A is false but R is true

💡 Show solution▶

Idea A is Theorem 13.1(iii). R is false: with a die, $E=\{1,2\}$, $F=\{1,2,3\}$ give $P(E\mid F)+P(E\mid F')=\tfrac23+0\ne1$.

Answer (C). The complement that adds up to $1$ is the complement of the event before the bar, not of the condition.

E3. Board · 2 marks For events $A$ and $B$, $P(A)=0.3$, $P(B)=0.4$ and $P(A\cup B)=0.5$. Find $P(A\mid B)$ and $P(A'\mid B')$.

💡 Show solution▶

Step 1 $P(A\cap B)=0.3+0.4-0.5=0.2$, so $P(A\mid B)=\tfrac{0.2}{0.4}=\tfrac12$. (1 mark)

Step 2 By De Morgan's law $A'\cap B'=(A\cup B)'$, so $P(A'\cap B')=1-0.5=0.5$, and $P(B')=0.6$. Hence $P(A'\mid B')=\tfrac{0.5}{0.6}=\tfrac56$. (1 mark)

Answer $P(A\mid B)=\tfrac12$, $P(A'\mid B')=\tfrac56$.

E4. Board · Case study A city survey of $600$ office workers (illustrative data) recorded how they commute and whether they reached work late on the survey day: metro $250$ workers, of whom $25$ were late; bus $200$, of whom $50$ were late; car $150$, of whom $45$ were late. One worker is chosen at random. (i) Find $P(\text{late}\mid\text{bus})$ and $P(\text{on time}\mid\text{metro})$. (2 marks) (ii) Find the probability that the worker came by bus, given that the worker was late. (1 mark) (iii) Find $P(\text{late}\mid\text{did not use the metro})$. (1 mark)

💡 Show solution▶

(i) $P(\text{late}\mid\text{bus})=\tfrac{50}{200}=\tfrac14$; $P(\text{on time}\mid\text{metro})=\tfrac{250-25}{250}=\tfrac9{10}$.

(ii) Late workers: $25+50+45=120$. $P(\text{bus}\mid\text{late})=\tfrac{50}{120}=\tfrac5{12}$.

(iii) Not metro: $200+150=350$ workers, of whom $50+45=95$ were late: $\tfrac{95}{350}=\tfrac{19}{70}$.

E5. JEE Main A number is chosen at random from $1,2,\dots,100$. Given that it is divisible by $4$, the probability that it is divisible by $6$ is (A) $\tfrac8{25}$ (B) $\tfrac4{25}$ (C) $\tfrac16$ (D) $\tfrac2{25}$

💡 Show solution▶

Method Reduced sample space: the $25$ multiples of $4$. A multiple of $4$ is divisible by $6$ exactly when it is divisible by $\operatorname{lcm}(4,6)=12$; there are $8$ multiples of $12$ up to $100$ ($12,24,\dots,96$). So the probability is $\tfrac8{25}$.

Answer (A). Trap (B): $\tfrac{16}{100}$ is the unconditional $P(\text{divisible by }6)$. (D) $=\tfrac8{100}$ divides by $100$ instead of $25$. (C) treats divisibility by $6$ as if it ignored the information.

⏱ Fast route Among multiples of $4$, i.e. $4k$ with $k=1,\dots,25$, you need $3\mid k$: $k=3,6,\dots,24$, eight values.

E6. JEE Advanced (One or more options correct.) A fair die is thrown three times. Let $F$ be the event "the three numbers are all different". Then (A) $P(F)=\tfrac59$ (B) $P(\text{at least one six}\mid F)=\tfrac12$ (C) $P(\text{the sum is }10\mid F)=\tfrac3{20}$ (D) $P(F\mid\text{at least one six})=\tfrac12$

💡 Show solution▶

(A) $n(S)=216$, $n(F)=6\cdot5\cdot4=120$: $P(F)=\tfrac{120}{216}=\tfrac59$ ✓.

(B) Outcomes of $F$ with no six: $5\cdot4\cdot3=60$, so $120-60=60$ have a six: $\tfrac{60}{120}=\tfrac12$ ✓.

(C) Sets of three different numbers with sum $10$: $\{1,3,6\}$, $\{1,4,5\}$, $\{2,3,5\}$, each in $3!=6$ orders: $\tfrac{18}{120}=\tfrac3{20}$ ✓.

(D) At least one six: $216-125=91$ outcomes, of which $60$ are in $F$: $\tfrac{60}{91}\ne\tfrac12$ ✗ — the trap is to swap the condition from (B).

Answer (A), (B), (C).

🔬 Research–Teaching NexusMini project · about 60 minutes

Research question. In your class, does the chance of getting at least $7$ hours of sleep depend on whether a phone is kept in the bedroom at night — and does the answer change if you turn the question round?

Why it matters. Health agencies around the world advise teenagers to sleep $8$–$10$ hours, and many studies report that screens at bedtime are linked with shorter sleep. Every such report is a statement about conditional probabilities: "among teenagers with a phone in the bedroom, the proportion who sleep enough is …". Learning to read and compute these figures — and to notice which event is the condition — protects you from misleading headlines.

What to do.

With your teacher's permission, run an anonymous survey of your class (or several classes): two yes/no questions — "Is your phone in your bedroom at night?" and "Did you sleep at least 7 hours last night?"
Organise the answers in a $2\times2$ two-way table with row and column totals.
Compute $P(\text{enough sleep}\mid\text{phone})$, $P(\text{enough sleep}\mid\text{no phone})$, $P(\text{phone}\mid\text{enough sleep})$ and $P(\text{enough sleep})$.
Check that $P(\text{enough sleep})$ lies between the two conditional probabilities of step 3, and explain why it must.
Write two headlines, one using a row-conditional and one using a column-conditional probability, and say which is fairer.

Tools: Paper questionnaire or an online form; a spreadsheet for the table and the fractions.

Hand in: The table, the four probabilities as fractions and decimals, the two headlines, and a paragraph on why a survey like this shows an association but not a cause.

Going further. Collect the same data a week later. Do the conditional probabilities stay roughly the same? How large a sample would you want before trusting a difference of $0.1$?

Instructor note: with counts $a$ (phone, enough), $b$ (phone, not enough), $c$ (no phone, enough), $d$ (no phone, not enough), the four answers are $\tfrac{a}{a+b}$, $\tfrac{c}{c+d}$, $\tfrac{a}{a+c}$ and $\tfrac{a+c}{a+b+c+d}$. Step 4 holds because the overall proportion is a weighted average of the two row proportions — a first look at the theorem of total probability (Section 13.3). Keep the survey anonymous and voluntary.

🧪 Self-check quiz

0/5 correct

Q1. If $P(F)=0.5$ and $P(E\cap F)=0.2$, then $P(E\mid F)$ is

Q2. Two fair coins are tossed. Given that at least one shows a head, the probability that both show heads is

Q3. If $P(E\mid F)=0.3$, then $P(E'\mid F)$

Q4. One card is drawn from a pack of $52$. Given that it is a face card (king, queen or jack), the probability that it is a queen is

Q5. $A$ and $B$ are mutually exclusive events with $P(A)\gt0$ and $P(B)\gt0$. Then $P(A\mid B)$ equals

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