What are single-event problems? A single event comes from one random experiment — tossing coins once, rolling dice once, drawing one card or one ball, or picking one number. Every such problem reduces to the same two-step recipe: list the sample space to find $n(S)$, then count the favourable outcomes to find $n(E)$, and finally $P(E)=\dfrac{n(E)}{n(S)}$.
The whole skill is in counting correctly. Below are the standard set-ups that cover almost every CBSE/ICSE question.
| Experiment | Sample space size $n(S)$ | Typical event |
|---|
| One coin | $2$ | Head |
| Two coins | $4$ | At least one head |
| Three coins | $8$ | Exactly two heads |
| One die | $6$ | Prime / even / $>4$ |
| Two dice | $36$ | A given sum |
| Deck of cards | $52$ | Spade / face card / ace |
| Bag of balls | total balls | A given colour |
| Numbers $1$ to $n$ | $n$ | Multiple / prime / perfect square |
Two dice — the sum table. When two dice are rolled there are $6\times6=36$ equally likely ordered pairs $(a,b)$. The number of ways to get each total is:
| Sum | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|
| Ways | 1 | 2 | 3 | 4 | 5 | 6 | 5 | 4 | 3 | 2 | 1 |
So, for example, $P(\text{sum}=7)=\dfrac{6}{36}=\dfrac{1}{6}$ (the most likely total) and $P(\text{sum}=2)=\dfrac{1}{36}$. The ways add to $36$, a useful check.
Choosing numbers. To count how many numbers from $a$ to $b$ are multiples of $d$, use $\left\lfloor\dfrac{b}{d}\right\rfloor-\left\lfloor\dfrac{a-1}{d}\right\rfloor$, or list them in an arithmetic pattern: count $=\dfrac{\text{last}-\text{first}}{d}+1$. Two-digit numbers run from $10$ to $99$, giving $99-10+1=90$ numbers.
Picking a letter. When a letter is chosen from a word, $n(S)$ is the total number of letters (counting repeats). For "MATHEMATICS" ($11$ letters) the vowels are A, A, E, I — that is $4$ favourable letters, so $P(\text{vowel})=\dfrac{4}{11}$.
Key counting traps:
- "Between" is exclusive. "A number between $2$ and $6$" on a die means $\{3,4,5\}$, not $\{2,3,4,5,6\}$. Read whether the endpoints are included.
- Order matters for coins and dice. Two coins give $4$ outcomes, not $3$; two dice give $36$ ordered pairs.
- "At most" and "at least". "At most two heads" on three coins means $0$, $1$ or $2$ heads; subtract the easy case from $1$ when faster.
- "Neither A nor B" equals the complement of "A or B".
- For mutually exclusive events, $P(A\text{ or }B)=P(A)+P(B)$; never add probabilities of events that can overlap.
Sometimes outcomes are not discrete. In a geometric (area) probability problem — like a dart landing on a board — the probability of landing in a region is $\dfrac{\text{area of favourable region}}{\text{total area}}$. The same favourable-over-total idea, measured by area instead of by counting.
Two coins are tossed simultaneously. Find the probability of getting exactly one head.
Solution- Step 1: $S=\{HH,HT,TH,TT\}$, so $n(S)=4$.
- Step 2: Exactly one head $=\{HT,TH\}$, so $n(E)=2$.
- Step 3: $P(\text{exactly one head})=\dfrac{2}{4}=\dfrac{1}{2}$.
Answer: $\dfrac{1}{2}$
A bag contains 5 red, 4 blue and 3 green balls. One ball is drawn at random. Find the probability that it is (i) red, (ii) not blue.
Solution- Step 1: Total balls $=5+4+3=12$, so $n(S)=12$.
- Step 2: (i) Red balls $=5$, so $P(\text{red})=\dfrac{5}{12}$.
- Step 3: (ii) Blue balls $=4$, so $P(\text{blue})=\dfrac{4}{12}=\dfrac{1}{3}$.
- Step 4: $P(\text{not blue})=1-\dfrac{1}{3}=\dfrac{2}{3}$.
Answer: (i) $\dfrac{5}{12}$; (ii) $\dfrac{2}{3}$
A number is chosen at random from the two-digit numbers 10 to 99. Find the probability that the number is a multiple of 5.
Solution- Step 1: Two-digit numbers from $10$ to $99$: total $=99-10+1=90$, so $n(S)=90$.
- Step 2: Multiples of $5$ are $10,15,20,\ldots,95$.
- Step 3: Count $=\dfrac{95-10}{5}+1=\dfrac{85}{5}+1=17+1=18$.
- Step 4: $P(\text{multiple of }5)=\dfrac{18}{90}=\dfrac{1}{5}$.
Answer: $\dfrac{1}{5}$
A die is thrown once. Find the probability of getting a number lying strictly between 2 and 6.
Solution- Step 1: $S=\{1,2,3,4,5,6\}$, so $n(S)=6$.
- Step 2: Numbers strictly between $2$ and $6$ are $\{3,4,5\}$ (endpoints excluded), so $n(E)=3$.
- Step 3: $P(E)=\dfrac{3}{6}=\dfrac{1}{2}$.
Answer: $\dfrac{1}{2}$
A letter is chosen at random from the word "MATHEMATICS". Find the probability that it is a vowel.
Solution- Step 1: "MATHEMATICS" has $11$ letters, so $n(S)=11$.
- Step 2: The vowels are A, A, E, I, so $n(E)=4$.
- Step 3: $P(\text{vowel})=\dfrac{4}{11}$.
Answer: $\dfrac{4}{11}$
Three coins are tossed simultaneously. Find the probability of getting exactly two heads.
Solution- Step 1: $S=\{HHH,HHT,HTH,THH,HTT,THT,TTH,TTT\}$, so $n(S)=8$.
- Step 2: Exactly two heads $=\{HHT,HTH,THH\}$, so $n(E)=3$.
- Step 3: $P(\text{exactly two heads})=\dfrac{3}{8}$.
Answer: $\dfrac{3}{8}$
Two dice are thrown together. Find the probability that the sum of the numbers appearing is 7.
Solution- Step 1: Two dice give $6\times6=36$ ordered outcomes, so $n(S)=36$.
- Step 2: Pairs with sum $7$: $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$, so $n(E)=6$.
- Step 3: $P(\text{sum}=7)=\dfrac{6}{36}=\dfrac{1}{6}$.
Answer: $\dfrac{1}{6}$
Two dice are thrown. Find the probability of getting a sum of at most 4 (i.e. 4 or less).
Solution- Step 1: $n(S)=36$.
- Step 2: Sum $2$: $(1,1)$ — $1$ way. Sum $3$: $(1,2),(2,1)$ — $2$ ways. Sum $4$: $(1,3),(2,2),(3,1)$ — $3$ ways.
- Step 3: Favourable $=1+2+3=6$.
- Step 4: $P(\text{sum}\le4)=\dfrac{6}{36}=\dfrac{1}{6}$.
Answer: $\dfrac{1}{6}$
A card is drawn from a well-shuffled deck of 52 cards. Find the probability that it is (i) a spade, (ii) a black face card.
Solution- Step 1: $n(S)=52$.
- Step 2: (i) Spades $=13$, so $P(\text{spade})=\dfrac{13}{52}=\dfrac{1}{4}$.
- Step 3: (ii) Black suits are spades and clubs; each has $3$ face cards, so black face cards $=2\times3=6$.
- Step 4: $P(\text{black face card})=\dfrac{6}{52}=\dfrac{3}{26}$.
Answer: (i) $\dfrac{1}{4}$; (ii) $\dfrac{3}{26}$
A box contains 90 discs numbered 1 to 90. One disc is drawn at random. Find the probability that the disc bears (i) a two-digit number, (ii) a perfect square.
Solution- Step 1: $n(S)=90$.
- Step 2: (i) Two-digit numbers are $10$ to $90$: count $=90-10+1=81$, so $P=\dfrac{81}{90}=\dfrac{9}{10}$.
- Step 3: (ii) Perfect squares from $1$ to $90$: $1,4,9,16,25,36,49,64,81$ — that is $9$ numbers.
- Step 4: $P(\text{perfect square})=\dfrac{9}{90}=\dfrac{1}{10}$.
Answer: (i) $\dfrac{9}{10}$; (ii) $\dfrac{1}{10}$
A number is selected at random from 1 to 25. Find the probability that it is a prime number.
Solution- Step 1: $n(S)=25$.
- Step 2: Primes up to $25$: $2,3,5,7,11,13,17,19,23$ — that is $9$ numbers.
- Step 3: $P(\text{prime})=\dfrac{9}{25}$.
Answer: $\dfrac{9}{25}$
In the figure, a circular dartboard of radius 1 unit has a smaller concentric circle of radius $\tfrac12$ unit shaded at its centre. A dart lands at random on the board. Find the probability that it lands in the shaded inner circle.
Solution- Step 1: This is a geometric (area) probability: $P=\dfrac{\text{favourable area}}{\text{total area}}$.
- Step 2: Total area $=\pi(1)^2=\pi$.
- Step 3: Shaded area $=\pi\left(\dfrac{1}{2}\right)^2=\dfrac{\pi}{4}$.
- Step 4: $P=\dfrac{\pi/4}{\pi}=\dfrac{1}{4}$.
Answer: $\dfrac{1}{4}$