Probability Class 10: Concepts, Formulas and Board Questions

What Probability is about

Weightage: This chapter is part of Unit VII, Statistics and Probability, which carries 11 of the 80 theory marks together with Statistics.Source: CBSE Mathematics syllabus 2026-27, Unit VII Statistics and Probability

In real board papers: Standard, asked in 21 of the 21 board papers we hold (2023 to 2026), on average 5.2 marks per paper; Basic, asked in 21 of the 21 board papers we hold (2023 to 2026), on average 4.9 marks per paper. See the questions

Key concepts

Random experiments, outcomes and equally likely outcomes

NCERT section 14.1 Probability — A Theoretical Approach

A random experiment is an action whose result you cannot predict in advance, such as tossing a coin or drawing a card. Each possible result is an outcome, and the list of all outcomes is the sample space. An event is a collection of outcomes you are interested in, like getting an even number on a die.

The classical definition only works when outcomes are equally likely, meaning none is favoured over another. A fair die gives six equally likely outcomes. But some lists are not equally likely. For example, the results of a cricket match (win, lose, draw) are not equally likely, so you cannot say the chance of winning is $\frac{1}{3}$.

Always write the sample space in a way that makes outcomes equally likely. With two coins, list HH, HT, TH, TT, not 'two heads, two tails, one of each'. Getting this step right prevents most wrong answers.

Worked example

Rahul tosses two fair coins together. He says: 'There are three outcomes: two heads, two tails, or one head and one tail. So the probability of getting one head and one tail is $\frac{1}{3}$.' Is Rahul correct? Give a reason and find the correct probability.

Try it first, then show the solutionSolution

1. The three outcomes Rahul lists are not equally likely, so the classical definition cannot be applied to them.
2. Write the equally likely outcomes: HH, HT, TH, TT. Total outcomes $=4$.
3. 'One head and one tail' happens in HT and TH, so favourable outcomes $=2$.
4. $P(\text{one head and one tail})=\frac{2}{4}=\frac{1}{2}$.
5. Check: $P(HH)+P(TT)+P(\text{one of each})=\frac{1}{4}+\frac{1}{4}+\frac{1}{2}=1$.

Answer: Rahul is wrong because his three outcomes are not equally likely. The correct probability is $\frac{1}{2}$.

Classical definition of probability

NCERT section 14.1 Probability — A Theoretical Approach

For an experiment with equally likely outcomes, the theoretical (classical) probability of an event $E$ is

$P(E)=\frac{\text{number of outcomes favourable to }E}{\text{number of all possible outcomes}}$.

Why does this work? If all outcomes are equally likely, each one carries the same share of chance. So the chance of $E$ is simply the fraction of outcomes that belong to $E$.

The method is always the same three steps:

• Count the total number of equally likely outcomes.
• Count the favourable outcomes, carefully, without missing or repeating any.
• Divide and write the fraction in its lowest terms.

When an event uses the word or, such as 'a multiple of 3 or a multiple of 5', count numbers that belong to both only once. Listing the favourable outcomes in your answer helps the examiner give you method marks even if you slip later.

Worked example

Cards numbered 1 to 50 are placed in a box and mixed well. One card is drawn at random. Find the probability that the number on the card is a perfect square or a multiple of 7.

Try it first, then show the solutionSolution

1. Total number of equally likely outcomes $=50$.
2. Perfect squares from 1 to 50: 1, 4, 9, 16, 25, 36, 49, which is 7 numbers.
3. Multiples of 7 from 1 to 50: 7, 14, 21, 28, 35, 42, 49, which is 7 numbers.
4. 49 appears in both lists, so count it once: favourable outcomes $=7+7-1=13$.
5. $P(E)=\frac{13}{50}$.

Answer: The required probability is $\frac{13}{50}$.

Range of probability, sure and impossible events, elementary events

NCERT section 14.1 Probability — A Theoretical Approach

Since the favourable outcomes can never be fewer than $0$ or more than the total, every probability satisfies $0\le P(E)\le 1$. A probability can be written as a fraction, decimal or percentage, but it can never be negative and never exceed $1$ (or $100\%$).

An event certain to happen, such as getting a number less than 7 on a die, is a sure event with probability $1$. An event that cannot happen, such as getting 8 on a die, is an impossible event with probability $0$.

An elementary event has exactly one outcome. The probabilities of all the elementary events of an experiment add up to $1$. More generally, if a set of events covers every outcome with no overlap, their probabilities add to $1$. This is very useful when some probabilities are given and you must find a missing one, and then a missing number of objects.

Worked example

A bag contains only red, blue and green balls. A ball is drawn at random. The probability of a red ball is $\frac{1}{4}$ and the probability of a blue ball is $\frac{1}{3}$. There are 10 green balls in the bag. Find the total number of balls and the number of red and blue balls.

Try it first, then show the solutionSolution

1. Every ball is red, blue or green, so $P(R)+P(B)+P(G)=1$.
2. $P(G)=1-\frac{1}{4}-\frac{1}{3}=\frac{12-3-4}{12}=\frac{5}{12}$.
3. Let the total be $n$. Then $\frac{10}{n}=\frac{5}{12}$, so $n=\frac{10\times 12}{5}=24$.
4. Red balls $=\frac{1}{4}\times 24=6$; blue balls $=\frac{1}{3}\times 24=8$.
5. Check: $6+8+10=24$.

Answer: There are 24 balls in total: 6 red, 8 blue and 10 green.

Complementary events

NCERT section 14.1 Probability — A Theoretical Approach

The event 'not $E$', written $\overline{E}$, happens exactly when $E$ does not happen. Every outcome belongs to either $E$ or $\overline{E}$, never both. So their favourable outcomes together make up the whole sample space, and

$P(E)+P(\overline{E})=1$, that is, $P(\overline{E})=1-P(E)$.

$E$ and $\overline{E}$ are called complementary events.

When to use it: when the event you want is awkward to count directly but its opposite is easy. Typical signals are the words 'not', 'at least one' and 'neither ... nor'. For 'at least one head', the opposite is 'no heads', which has only one outcome.

Using the complement also gives you a quick check: if you work out both $P(E)$ and $P(\overline{E})$ by counting, they must add to $1$.

Worked example

A number is chosen at random from the numbers 1 to 30. Find the probability that it is divisible by neither 3 nor 5.

Try it first, then show the solutionSolution

1. Total outcomes $=30$. Let $E$: the number is divisible by 3 or by 5.
2. Multiples of 3 up to 30: $10$. Multiples of 5 up to 30: $6$.
3. Multiples of both, that is of 15: 15 and 30, so $2$.
4. Favourable to $E$ $=10+6-2=14$, so $P(E)=\frac{14}{30}=\frac{7}{15}$.
5. 'Neither 3 nor 5' is $\overline{E}$, so $P(\overline{E})=1-\frac{7}{15}=\frac{8}{15}$.
6. Check: numbers not in $E$ $=30-14=16$, and $\frac{16}{30}=\frac{8}{15}$.

Answer: The required probability is $\frac{8}{15}$.

Problems on coins

NCERT section 14.1 Probability — A Theoretical Approach

A fair coin has two equally likely outcomes, H and T. When several coins are tossed together (or one coin is tossed several times), each coin doubles the number of outcomes: 2 for one coin, 4 for two coins, 8 for three coins.

Write the sample space in full. For three coins:

HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.

Then read the event carefully:

• exactly two heads means two heads and no more;
• at least two heads means two or three heads;
• at most one head means zero or one head.

Games are often set on coins: a player wins if some event happens and loses otherwise. Then $P(\text{lose})=1-P(\text{win})$, because winning and losing are complementary.

Worked example

In a game, three fair coins are tossed together. Priya wins if all three coins show the same face; otherwise she loses. Find (i) the probability that she loses, (ii) the probability of getting at least two heads.

Try it first, then show the solutionSolution

1. Sample space: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT, so total outcomes $=8$.
2. (i) All same face: HHH and TTT, so $P(\text{win})=\frac{2}{8}=\frac{1}{4}$.
3. Winning and losing are complementary, so $P(\text{lose})=1-\frac{1}{4}=\frac{3}{4}$.
4. (ii) At least two heads: HHH, HHT, HTH, THH, so 4 outcomes.
5. $P(\text{at least two heads})=\frac{4}{8}=\frac{1}{2}$.

Answer: (i) $\frac{3}{4}$ (ii) $\frac{1}{2}$

Problems on dice

NCERT section 14.1 Probability — A Theoretical Approach

A fair die has six equally likely faces, 1 to 6. When two dice are thrown, each face of the first can pair with each face of the second, giving $6\times 6=36$ outcomes, written as ordered pairs $(a,b)$.

Order matters: $(2,5)$ and $(5,2)$ are different outcomes, because the two dice are different objects. Only doublets like $(3,3)$ appear once.

The best way to count is to picture a $6\times 6$ grid, rows for the first die and columns for the second. Sums run from 2 to 12. A sum of 7 has the most outcomes (six), and the number falls by one as you move away from 7 in either direction. For products, differences or 'both even' type events, go row by row so nothing is missed.

Always state '36 outcomes' in your answer; it earns a step mark.

Worked example

Two different fair dice are thrown together. Find the probability that (i) the sum of the numbers is a prime number, (ii) the product of the numbers is a perfect square.

Try it first, then show the solutionSolution

1. Total outcomes $=6\times 6=36$.
2. (i) Prime sums possible: 2, 3, 5, 7, 11.
3. Sum 2: (1,1). Sum 3: (1,2), (2,1). Sum 5: (1,4), (4,1), (2,3), (3,2). Sum 7: (1,6), (6,1), (2,5), (5,2), (3,4), (4,3). Sum 11: (5,6), (6,5).
4. Favourable $=1+2+4+6+2=15$, so $P=\frac{15}{36}=\frac{5}{12}$.
5. (ii) Perfect square products: (1,1), (1,4), (4,1), (2,2), (3,3), (4,4), (5,5), (6,6).
6. Favourable $=8$, so $P=\frac{8}{36}=\frac{2}{9}$.

Answer: (i) $\frac{5}{12}$ (ii) $\frac{2}{9}$

Problems on playing cards

NCERT section 14.1 Probability — A Theoretical Approach

A standard deck has 52 cards in four suits: spades and clubs (black), hearts and diamonds (red). Each suit has 13 cards: ace, 2 to 10, jack, queen, king.

Key counts you must know:

• 26 red and 26 black cards;
• 4 aces, 4 kings, 4 queens, 4 jacks;
• face cards are jacks, queens and kings only, so there are $3\times 4=12$ face cards. An ace is not a face card.

Board questions often remove some cards first. Then two things change: the total becomes smaller, and the count for each event may change too. Work out exactly which cards are left before you count. For example, if red face cards are removed, the only face cards left are black ones. Write the new total clearly as a separate step.

Worked example

All the red face cards are removed from a well-shuffled deck of 52 playing cards. One card is then drawn at random from the remaining cards. Find the probability that it is (i) a black card, (ii) a face card, (iii) a red card that is not an ace.

Try it first, then show the solutionSolution

1. Red face cards: jack, queen, king of hearts and of diamonds, so 6 cards removed.
2. Remaining cards $=52-6=46$.
3. (i) No black card was removed, so black cards $=26$. $P=\frac{26}{46}=\frac{13}{23}$.
4. (ii) Only the 6 black face cards remain. $P=\frac{6}{46}=\frac{3}{23}$.
5. (iii) Red cards left $=26-6=20$; of these, 2 are red aces, so 18 are red non-aces.
6. $P=\frac{18}{46}=\frac{9}{23}$.

Answer: (i) $\frac{13}{23}$ (ii) $\frac{3}{23}$ (iii) $\frac{9}{23}$

Bags, balls and numbers with an unknown

NCERT section 14.1 Probability — A Theoretical Approach

Many longer questions turn the definition around. Instead of counting to get a probability, you are given a probability and must find a missing number.

The idea: if a bag has $n$ balls and $x$ of them are white, then $P(\text{white})=\frac{x}{n}$. Set this equal to the given value and solve the equation.

When balls are added or removed, both the favourable count and the total change. If 6 white balls are added to a bag of 12, the new white count is $x+6$ and the new total is $18$. Forgetting to change the total is the most common error here.

After solving, always check: the unknown must be a whole number, not more than the total, and substituting it back must give the stated probabilities.

Worked example

A bag contains 12 balls, of which $x$ are white and the rest are black. If 6 more white balls are put into the bag, the probability of drawing a white ball becomes double the probability before. Find $x$.

Try it first, then show the solutionSolution

1. Before: $P(\text{white})=\frac{x}{12}$.
2. After adding 6 white balls: white $=x+6$, total $=12+6=18$, so $P(\text{white})=\frac{x+6}{18}$.
3. Given: $\frac{x+6}{18}=2\times \frac{x}{12}=\frac{x}{6}$.
4. Cross-multiply: $6(x+6)=18x$, so $6x+36=18x$.
5. $12x=36$, so $x=3$.
6. Check: before $\frac{3}{12}=\frac{1}{4}$; after $\frac{9}{18}=\frac{1}{2}$, which is double.

Answer: $x=3$, so the bag first had 3 white balls.

Maths in real life: Probability

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Probability: Basic Terms and Classical Definition
• Properties of Probability and Complementary Events
• Probability Problems on Coins and Dice
• Probability Problems on Playing Cards
• Probability Problems on Bags, Balls and Numbers

Probability formula sheet

Classical probability

$P(E)=\frac{\text{number of outcomes favourable to }E}{\text{total number of equally likely outcomes}}$

The core formula of the chapter. Use only when all outcomes are equally likely.

Range of probability

$0\le P(E)\le 1$

Use to reject impossible values in MCQs and to check every answer you get.

Sure and impossible events

$P(\text{sure event})=1,P(\text{impossible event})=0$

A sure event contains every outcome; an impossible event contains none.

Complementary events

$P(E)+P(\overline{E})=1\Rightarrow P(\overline{E})=1-P(E)$

Use for 'not', 'at least one' and 'neither ... nor' questions, and to check answers.

Sum of elementary events

$P(E_1)+P(E_2)+\cdots +P(E_k)=1$

The probabilities of all elementary events (or of events covering all outcomes with no overlap) add to 1. Use to find a missing probability.

Number of outcomes for coins

$\text{1 coin: }2,\text{2 coins: }4,\text{3 coins: }8$

Each extra coin doubles the outcomes. Write the full list for 2 or 3 coins.

Number of outcomes for dice

$\text{1 die: }6,\text{2 dice: }6\times 6=36$

With two dice, (a, b) and (b, a) are different outcomes unless a = b.

Outcomes for a given sum with two dice

$\text{number of pairs with sum }s=6-|s-7|,2\le s\le 12$

Quick check: sum 7 has 6 pairs, sums 2 and 12 have 1 each.

Deck of playing cards

$52=4\text{ suits}\times 13;26\text{ red},26\text{ black};12\text{ face cards};4\text{ aces}$

Face cards are jack, queen and king only. Hearts and diamonds are red; spades and clubs are black.

Count of whole numbers in a range

$\text{numbers from }a\text{ to }b\text{ (inclusive)}=b-a+1$

For example, 10 to 99 has 90 numbers. Use when cards are numbered from a value other than 1.

Multiples of k from 1 to N

$\lfloor \frac{N}{k}\rfloor$

Take the whole-number part of N divided by k. For 'multiple of a or b', subtract the multiples of both so they are not counted twice.

Standard counts of primes and squares

$\text{primes up to }20:8,\text{up to }50:15,\text{up to }100:25;\text{squares from 1 to }N:\lfloor \sqrt{N}\rfloor$

1 is neither prime nor composite. Use these to check your counts quickly.

Finding an unknown number of objects

$\text{number of favourable objects}=P(E)\times \text{total}$

Use when a probability is given and you must find how many balls, cards or items there are.

How the board asks Probability

The question patterns to expect, and the method that works for each.

Tricks to remember Probability

Short memory aids. Say each one aloud once, then use it on a question the same day.

• F over T

  Probability is always Favourable over Total. Write 'Total = ...' on one line and 'Favourable = ...' on the next before you divide.

  Use it when: Keeps your working in the order the marking scheme rewards.

• The 7-peak for two dice

  Sums 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 have 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1 outcomes. It rises to 6 at 7 and falls back. These add to 36.

  Use it when: Instantly count outcomes for any sum, or check your listed pairs.

• 4-13-26-52 and JQK

  4 suits, 13 per suit, 26 per colour, 52 in all. Face cards are J, Q, K only, so 3 per suit and 12 in all. An ace is not a face.

  Use it when: Avoids the most common card-counting slip.

• At least one means one minus none

  For 'at least one' of something, find the probability of none and subtract from 1.

  Use it when: Much faster for coins and dice than counting all the favourable cases.

• The two-line final check

  Ask: is my answer between 0 and 1? Do my event and its opposite add to 1?

  Use it when: Catches counting errors in the last minute of a question.

• One is lonely

  1 is neither prime nor composite. Primes: 8 up to 20, 15 up to 50, 25 up to 100.

  Use it when: Quick check on numbered-card questions about primes.

Probability: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section E style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Probability

The slips examiners see most often, and what to write instead.

• ✗Mistake: Counting 1 as a prime number

  Students remember that a prime has no factors other than 1 and itself, and wrongly think 1 fits.

  Wrong working

  Primes from 1 to 20: 1, 2, 3, 5, 7, 11, 13, 17, 19, so $P=\frac{9}{20}$.

  Right working

  Primes from 1 to 20: 2, 3, 5, 7, 11, 13, 17, 19, so $P=\frac{8}{20}=\frac{2}{5}$.

  ✓Do this: A prime has exactly two different factors. 1 has only one factor, so it is not prime.

  Usually costs: Usually the answer mark, ½ to 1 mark.

• ✗Mistake: Treating (2, 5) and (5, 2) as the same outcome with two dice

  The two numbers look the same, so students count unordered pairs and get 21 outcomes.

  Wrong working

  Total outcomes $=21$; sum 7: {1,6}, {2,5}, {3,4}, so $P=\frac{3}{21}=\frac{1}{7}$.

  Right working

  Total outcomes $=36$; sum 7: (1,6), (6,1), (2,5), (5,2), (3,4), (4,3), so $P=\frac{6}{36}=\frac{1}{6}$.

  ✓Do this: The dice are different objects, so use ordered pairs and a 6 × 6 grid. Only doublets appear once.

  Usually costs: Often the whole question, 1 to 2 marks, because both total and count are wrong.

• ✗Mistake: Counting aces as face cards

  Aces have a letter, like J, Q and K, so they seem to belong with them.

  Wrong working

  Face cards $=16$, so $P(\text{face card})=\frac{16}{52}=\frac{4}{13}$.

  Right working

  Face cards (J, Q, K) $=12$, so $P(\text{face card})=\frac{12}{52}=\frac{3}{13}$.

  ✓Do this: Face cards show a face: jack, queen, king. Learn 12 face cards, 4 aces.

  Usually costs: 1 mark in an MCQ; ½ to 1 mark in a longer question.

• ✗Mistake: Not reducing the total after cards are removed

  Students fix '52' in their mind and forget the deck has changed.

  Wrong working

  King, queen and jack of clubs removed. $P(\text{heart})=\frac{13}{52}=\frac{1}{4}$.

  Right working

  Remaining cards $=52-3=49$, hearts still 13, so $P(\text{heart})=\frac{13}{49}$.

  ✓Do this: Write 'Remaining cards = ...' as your first line every time cards are removed.

  Usually costs: Every part of the question that uses the total, often 1 to 2 marks.

• ✗Mistake: Changing the favourable count but not the total when balls are added

  Students focus on the colour being added and forget the bag now has more balls.

  Wrong working

  12 balls, $x$ white, 6 white added: $P=\frac{x+6}{12}$.

  Right working

  New total $=12+6=18$, so $P=\frac{x+6}{18}$.

  ✓Do this: After any change, rewrite both the new count and the new total before forming the equation.

  Usually costs: The equation and everything after it, usually 2 to 3 marks in a long answer.

• ✗Mistake: Counting overlaps twice in 'or' events

  Students add the two lists without noticing numbers that appear in both.

  Wrong working

  Multiples of 2 or 3 from 1 to 20: $10+6=16$, so $P=\frac{16}{20}=\frac{4}{5}$.

  Right working

  Multiples of 6 (both): 6, 12, 18. Favourable $=10+6-3=13$, so $P=\frac{13}{20}$.

  ✓Do this: For 'a or b', subtract the numbers that are multiples of both, or simply list the favourable numbers.

  Usually costs: ½ to 1 mark for the wrong count and answer.

• ✗Mistake: Using outcomes that are not equally likely

  A short list like win, lose, draw looks complete, so students divide by 3.

  Wrong working

  A team can win, lose or draw, so $P(\text{win})=\frac{1}{3}$.

  Right working

  These outcomes are not equally likely, so the classical definition cannot give $\frac{1}{3}$.

  ✓Do this: Before dividing, ask: does each outcome have the same chance? If not, rewrite the sample space or explain why the formula does not apply.

  Usually costs: 1 mark in an assertion-reason or reasoning question.

NCERT exercise guide: Probability

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Probability: previous year questions

• Standard: Asked in 21 of the 21 board papers we hold (2023 to 2026), on average 5.2 marks per paper; 68 questions in all, counting OR alternatives. In 4 of 4 sample papers.
• Basic: Asked in 21 of the 21 board papers we hold (2023 to 2026), on average 4.9 marks per paper; 67 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Probability

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

Every Probability question, by year

Practice questions on Probability

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQBoard level

   Which of the following cannot be the probability of an event?

   (a) $0.003$

   (b) $\frac{17}{16}$

   (c) $0$

   (d) $33\%$

   Hint

   A probability must lie between 0 and 1, both included.

   Answer

   (b) $\frac{17}{16}$. It is greater than 1. The values $0.003$, $0$ and $33\%=0.33$ all lie between 0 and 1.

2. Q2 · 1 mark · Assertion-reasonBoard level

   A non-leap year is chosen at random.

   Assertion (A): The probability that it has 53 Sundays is $\frac{1}{7}$.

   Reason (R): A non-leap year has 52 complete weeks and 1 extra day.

   (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.

   Hint

   The 52 weeks already give 52 Sundays. Think about the extra day.

   Answer

   (a). $365=52\times 7+1$. The year has 53 Sundays only if the one extra day is a Sunday, and that day can be any of the 7 days, so $P=\frac{1}{7}$. R explains A.

3. Q3 · 2 marks · Very short answerBoard level

   Two different fair dice are thrown together. Find the probability that the sum of the numbers is at least 5.

   Hint

   Counting the sums less than 5 is quicker. Then use the complement.

   Answer

   Sum less than 5: (1,1), (1,2), (2,1), (1,3), (3,1), (2,2), which is 6 out of 36 outcomes.

   $P(\text{sum}\ge 5)=1-\frac{6}{36}=\frac{5}{6}$.

4. Q4 · 3 marks · Short answerHarder board level

   All the spades are removed from a well-shuffled deck of 52 playing cards. One card is drawn at random from the rest. Find the probability that it is (i) a red king, (ii) a face card, (iii) a black card.

   Hint

   Find the new total first. With the spades gone, the only black cards left are clubs.

   Answer

   Remaining cards $=52-13=39$.

   (i) Red kings: 2, so $P=\frac{2}{39}$. (ii) Face cards: $12-3=9$, so $P=\frac{9}{39}=\frac{3}{13}$. (iii) Black cards (only clubs): 13, so $P=\frac{13}{39}=\frac{1}{3}$.

5. Q5 · 3 marks · Short answerHarder board level

   Discs numbered 1 to 100 are put in a bag and mixed well. One disc is drawn at random. Find the probability that the number on it is (i) a two-digit number divisible by 4 but not by 6, (ii) a perfect cube.

   Hint

   A number divisible by both 4 and 6 is divisible by 12.

   Answer

   (i) Two-digit multiples of 4: $12$ to $96$, which is 22 numbers. Two-digit multiples of 12: $12$ to $96$, which is 8 numbers. Favourable $=22-8=14$, so $P=\frac{14}{100}=\frac{7}{50}$.

   (ii) Cubes: 1, 8, 27, 64, so $P=\frac{4}{100}=\frac{1}{25}$.

6. Q6 · 5 marks · Long answerHarder board level

   A bag contains white, black and red balls only. The number of black balls is twice the number of white balls, and the number of red balls is 5 more than the number of black balls. The probability of drawing a black ball at random is $\frac{1}{3}$. (i) Find the number of balls of each colour. (ii) Find the probability that a ball drawn at random is not red. (iii) How many white balls must be added to the bag so that the probability of drawing a white ball becomes $\frac{2}{7}$?

   Hint

   Let the number of white balls be $w$ and write the total in terms of $w$.

   Answer

   White $w$, black $2w$, red $2w+5$, total $5w+5$. Then $\frac{2w}{5w+5}=\frac{1}{3}$ gives $6w=5w+5$, so $w=5$. That is 5 white, 10 black, 15 red (30 balls).

   (ii) $P(\text{not red})=\frac{15}{30}=\frac{1}{2}$.

   (iii) $\frac{5+y}{30+y}=\frac{2}{7}$ gives $35+7y=60+2y$, so $y=5$. Check: $\frac{10}{35}=\frac{2}{7}$.

7. Q7 · 3 marks · Short answerHarder board level

   In a game, a fair coin is tossed and a fair die is thrown at the same time. Find the probability of getting (i) a head and an even number, (ii) a tail or a number greater than 4.

   Hint

   List the outcomes as H1, H2, ..., T6. Be careful not to count any outcome twice in (ii).

   Answer

   Total outcomes $=2\times 6=12$.

   (i) H2, H4, H6, so $P=\frac{3}{12}=\frac{1}{4}$.

   (ii) All 6 tails, plus H5 and H6, gives 8 outcomes, so $P=\frac{8}{12}=\frac{2}{3}$.

8. Q8 · 3 marks · Short answerHarder board level

   Two different fair dice are thrown together. Find the probability that (i) the number on the first die is a factor of the number on the second die, (ii) both numbers are prime.

   Hint

   For (i), go through each value of the first die and list the multiples of it up to 6.

   Answer

   (i) First die 1: 6 outcomes. 2: (2,2), (2,4), (2,6), so 3. 3: (3,3), (3,6), so 2. Then 4, 5 and 6 give 1 each. Total $=14$, so $P=\frac{14}{36}=\frac{7}{18}$.

   (ii) The primes on a die are 2, 3, 5. That gives $3\times 3=9$ outcomes, so $P=\frac{9}{36}=\frac{1}{4}$.

9. Q9 · 2 marks · Very short answerHOTS / competency

   A fair die is thrown once. Anu says: 'The events "a number less than 3" and "a number greater than 3" are complementary, so their probabilities add up to 1.' Check her claim, explain any error, and find the probability of the event that is actually complementary to 'a number less than 3'.

   Hint

   Complementary events must cover every outcome between them. Where does 3 go?

   Answer

   $P(\text{less than }3)=\frac{2}{6}$ and $P(\text{greater than }3)=\frac{3}{6}$. Their sum is $\frac{5}{6}$, not 1, because the outcome 3 is in neither event. So Anu is wrong.

   The true complement is 'a number 3 or more', with probability $1-\frac{2}{6}=\frac{2}{3}$.

10. Q10 · 3 marks · Short answerHOTS / competency

    A leap year is chosen at random. Find the probability that it has (i) 53 Sundays, (ii) 53 Sundays or 53 Saturdays.

    Hint

    366 days is 52 weeks and 2 extra days. List the 7 possible pairs of consecutive days.

    Answer

    The two extra days can be (Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat) or (Sat, Sun), which is 7 equally likely pairs.

    (i) A Sunday appears in (Sat, Sun) and (Sun, Mon), so $P=\frac{2}{7}$.

    (ii) A Sunday or a Saturday appears in (Fri, Sat), (Sat, Sun) and (Sun, Mon), so $P=\frac{3}{7}$.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.State the classical definition of probability.Show answer

   $P(E)=\frac{\text{favourable outcomes}}{\text{total equally likely outcomes}}$.

2. 2.Between which values does every probability lie?Show answer

   $0\le P(E)\le 1$; it can never be negative or greater than 1.

3. 3.What is the probability of a sure event and of an impossible event?Show answer

   Sure event: $1$. Impossible event: $0$.

4. 4.If $P(E)=0.35$, what is $P(\overline{E})$?Show answer

   $1-0.35=0.65$.

5. 5.How many outcomes when three coins are tossed together?Show answer

   $2\times 2\times 2=8$.

6. 6.How many outcomes when two dice are thrown, and how many are doublets?Show answer

   36 outcomes; 6 doublets, so $P(\text{doublet})=\frac{1}{6}$.

7. 7.How many face cards and how many aces are in a deck of 52?Show answer

   12 face cards (J, Q, K of each suit) and 4 aces.

8. 8.What do the probabilities of all elementary events of an experiment add up to?Show answer

   They add up to $1$.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you write the classical definition and say when it can be used?
• ☐Can you list all 8 outcomes for three coins without missing one?
• ☐Can you say how many pairs give each sum from 2 to 12 with two dice?
• ☐Can you state the counts in a deck: red, black, each suit, face cards, aces?
• ☐Can you find a probability after some cards are removed, starting with the new total?
• ☐Can you use $P(\overline{E})=1-P(E)$ for 'not' and 'at least one' questions?
• ☐Can you count multiples of a or b in a range without double counting?
• ☐Can you set up and solve an equation when balls are added and a probability changes?
• ☐Can you explain why win, lose, draw does not give each a probability of one-third?
• ☐Can you spot instantly which MCQ option cannot be a probability?

Probability: frequently asked questions