MathsCBSE Class 10 · Probability

Probability Problems on Bags, Balls and Numbers for Class 10

Updated 13 October 2026 · Fact-checked

These are probability questions where you draw a ball from a bag, a numbered card or ticket from a box, or spin a wheel with equal sectors. You count the total and favourable outcomes and divide. The board also gives a probability and asks you to find an unknown number of balls, often after some are added or removed.

Count every ball as one equally likely outcome: red balls over all balls gives 5/8, even though there are only two colours.

Illustration drawn for QuestionBuddy.

Understand Probability Problems on Bags, Balls and Numbers

Picture a bag with 5 red and 3 green balls, all the same size and well mixed. When you pull one out without looking, each of the 8 balls has the same chance. So the balls themselves are the equally likely outcomes. The colour is not the outcome. There are only two colours, but red is more likely because there are more red balls.

That is why the classical definition works here. Favourable outcomes are the balls of the colour you want, and the total is every ball in the bag. So $P (red) = \frac{5}{8}$ . The same idea works for numbered cards, tickets and discs, where each card is one outcome. It also works for a spinner, but only if its sectors are equal in size. Then each sector is one equally likely outcome.

Numbered-card questions test your counting. You need the total in a range, which is $b - a + 1$ for numbers from $a$ to $b$ . You also need to count primes, squares and multiples. For 'a multiple of 4 or 6', numbers that are multiples of both are counted once. For 'neither', use the complement from the earlier topic: $P (\overline{E}) = 1 - P (E)$ .

The harder board questions turn the definition around. You are told a probability and must find how many balls there are. If a bag has $n$ balls and $x$ are white, then $\frac{x}{n}$ equals the given value. Solve this linear equation. When balls are added or removed, both the count of that colour and the total change. So $k$ white balls added gives $\frac{x + k}{n + k}$ .

Two checks protect your marks. First, if a bag has only certain colours, their probabilities add up to $1$ . Second, every unknown you find must be a whole number no bigger than the total. If you get $6.5$ balls, look for an error in your working.

Probability Problems on Bags, Balls and Numbers: formulas to remember

Probability of drawing a colour: $P (colour) = \frac{number of balls of that colour}{total number of balls}$; Each ball is one equally likely outcome, not each colour.
Total numbers in a range: $a to b inclusive: b - a + 1$; Cards numbered 11 to 60 give 50 cards, not 49.
Multiples of k from 1 to N: $⌊ \frac{N}{k} ⌋$; For 'multiple of a or b', add the two counts and subtract the multiples of the LCM of a and b.
Sum of colour probabilities: $P (red) + P (blue) + P (green) = 1$; Holds when every ball in the bag is one of these colours. Use it to find a missing probability.
Number of objects from a probability: $favourable count = P (E) \times total$; Answer must be a whole number.
After adding k balls of one colour: $P_{new} = \frac{x + k}{n + k}$; For removing k balls, use \frac{x - k}{n - k}. Both numerator and denominator change.
Spinner with equal sectors: $P (outcome) = \frac{sectors showing that outcome}{total number of sectors}$; Valid only when all sectors are equal in size.

How to solve Probability Problems on Bags, Balls and Numbers questions

Use this routine on any bag, card, ticket or spinner question.

1Write the total number of equally likely outcomes: all the balls, all the cards (use b − a + 1), or all the equal sectors.
2If anything is added or removed, write the new total as a separate line before you count anything.
3List or count the favourable outcomes. For numbers, write them out. For 'or', remove any double counting.
4For 'not' or 'neither', find the opposite event and use P(not E) = 1 − P(E).
5If a probability is given and a count is unknown, call the unknown x and write each probability as count over total.
6Form the linear equation, cross-multiply and solve.
7Check: the answer is a whole number, no bigger than the total, and substituting it back gives every stated probability.
8Write each final probability in lowest terms.

Probability Problems on Bags, Balls and Numbers in real life

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.

Finding an unknown after adding balls

Suppose your school stationery cupboard has a box of 36 pens. They are all the same shape, and some are blue and the rest are black. Your class teacher says that if you pick a pen without looking, the probability that it is blue is $\frac{2}{3}$ .

Find the number of blue and black pens. How many more black pens must the teacher add so that a pen picked at random is equally likely to be blue or black?

Try it first, then show the solutionSolution

Blue pens $= \frac{2}{3} \times 36 = 24$ .
Black pens $= 36 - 24 = 12$ .
Let $k$ black pens be added. New total $= 36 + k$ .
For equal chances, black pens must equal blue pens: $12 + k = 24$ .
So $k = 12$ .
Check: total $48$ , $P (black) = \frac{24}{48} = \frac{1}{2}$ .

Answer: 24 blue pens and 12 black pens; 12 more black pens must be added.

Memory hook: Add pens and the box grows: change the top and the bottom.

Numbered tickets with 'but not'

Imagine a sweet shop near your home runs a festival lucky draw with 200 coupons numbered 1 to 200, all mixed in a drum. One coupon is drawn. A customer wins a box of sweets if the number is a multiple of 8 but not a multiple of 12.

Find the probability that the coupon drawn wins the box of sweets.

Try it first, then show the solutionSolution

Total coupons $= 200$ .
Multiples of 8 from 1 to 200: $\frac{200}{8} = 25$ .
A number that is a multiple of both 8 and 12 is a multiple of their LCM, 24.
Multiples of 24 up to 200: $24, 48, \dots, 192$ , which is $8$ numbers.
Winning numbers $= 25 - 8 = 17$ .
$P (win) = \frac{17}{200}$ .

Answer: $\frac{17}{200}$

Memory hook: 'But not' means count the group, then take out the overlap through the LCM.

Tricks to remember Probability Problems on Bags, Balls and Numbers

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

Balls, not colours
Count the balls, never the colours. Two colours does not mean a half.
Use it when: Stops the error of saying P(red) = 1/2 just because there are two colours.
Add one, add both
Whatever goes into or out of the bag changes the top AND the bottom of the fraction.
Use it when: Use in every question where balls are added or removed.
Plus one for ranges
From a to b, it is b minus a, plus one.
Use it when: Getting the total right for cards numbered from any start value.
OR needs the LCM
Multiples of a or b: add both counts, take away the multiples of the LCM.
Use it when: Avoids double counting numbers like 12 in 'multiple of 4 or 6'.

Probability Problems on Bags, Balls and Numbers: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Numbered cards from 11 to 60 · 2 marks · Very short answerBoard level

Cards numbered 11 to 60 are placed in a box and mixed well. One card is drawn at random. Find the probability that the number on the card is (i) a perfect square, (ii) a multiple of 7.

Try it first, then show the solutionSolution

Total cards $= 60 - 11 + 1 = 50$ . (½ mark)
Perfect squares: $16, 25, 36, 49$ , so 4 cards. (½ mark)
$P (square) = \frac{4}{50} = \frac{2}{25}$ .
Multiples of 7: $14, 21, 28, 35, 42, 49, 56$ , so 7 cards. (½ mark)
$P (multiple of 7) = \frac{7}{50}$ . (½ mark)

Answer: (i) $\frac{2}{25}$ (ii) $\frac{7}{50}$

Example 2: Unknown red balls when the probability changes · 3 marks · Short answerHarder board level

A bag contains 18 balls, of which $x$ are red. If 2 more red balls are put in the bag, the probability of drawing a red ball becomes $\frac{9}{8}$ times the earlier probability. Find $x$ .

Try it first, then show the solutionSolution

Earlier, $P (red) = \frac{x}{18}$ . (½ mark)
After adding 2 red balls, red $= x + 2$ and total $= 20$ , so $P = \frac{x + 2}{20}$ . (½ mark)
Given: $\frac{x + 2}{20} = \frac{9}{8} \times \frac{x}{18} = \frac{x}{16}$ . (½ mark)
Cross-multiply: $16 x + 32 = 20 x$ . (½ mark)
$4 x = 32$ , so $x = 8$ . (½ mark)
Check: $\frac{8}{18} = \frac{4}{9}$ and $\frac{10}{20} = \frac{1}{2} = \frac{9}{8} \times \frac{4}{9}$ . (½ mark)

Answer: $x = 8$ red balls

Example 3: Tickets 1 to 100 with 'or' and 'neither' · 3 marks · Short answerHarder board level

Tickets numbered 1 to 100 are mixed and one is drawn at random. Find the probability that the number is (i) a multiple of 4 or 6, (ii) a multiple of neither 4 nor 6, (iii) a two-digit number whose digits add up to 9.

Try it first, then show the solutionSolution

Total tickets $= 100$ .
Multiples of 4: $25$ . Multiples of 6: $⌊ \frac{100}{6} ⌋ = 16$ . (½ mark)
Multiples of both are multiples of 12: $⌊ \frac{100}{12} ⌋ = 8$ . (½ mark)
Multiples of 4 or 6 $= 25 + 16 - 8 = 33$ , so $P = \frac{33}{100}$ . (½ mark)
Neither: $P = 1 - \frac{33}{100} = \frac{67}{100}$ . (½ mark)
Digit sum 9: $18, 27, 36, 45, 54, 63, 72, 81, 90$ , so 9 tickets. (½ mark)
$P = \frac{9}{100}$ . (½ mark)

Answer: (i) $\frac{33}{100}$ (ii) $\frac{67}{100}$ (iii) $\frac{9}{100}$

Example 4: Three colours with two conditions · 5 marks · Long answerHarder board level

A bag contains red, blue and green balls only. The number of blue balls is twice the number of red balls. The probability of drawing a green ball is $\frac{2}{5}$ . If 6 more red balls are put into the bag, the probability of drawing a red ball becomes $\frac{1}{4}$ . (i) Find the number of balls of each colour in the bag at the start. (ii) After the 6 red balls are added, find the probability that a ball drawn is not blue.

Try it first, then show the solutionSolution

Let red $= x$ , blue $= 2 x$ , green $= y$ . Total $= 3 x + y$ . (½ mark)
$\frac{y}{3 x + y} = \frac{2}{5}$ gives $5 y = 6 x + 2 y$ . (½ mark)
So $3 y = 6 x$ , that is, $y = 2 x$ , and total $= 5 x$ . (1 mark)
After adding 6 red: red $= x + 6$ , total $= 5 x + 6$ , so $\frac{x + 6}{5 x + 6} = \frac{1}{4}$ . (½ mark)
$4 x + 24 = 5 x + 6$ , so $x = 18$ . (1 mark)
Red $= 18$ , blue $= 36$ , green $= 36$ , total $= 90$ . Check: $\frac{36}{90} = \frac{2}{5}$ and $\frac{24}{96} = \frac{1}{4}$ . (½ mark)
After adding, not blue $= 24 + 36 = 60$ out of $96$ . (½ mark)
$P (not blue) = \frac{60}{96} = \frac{5}{8}$ . (½ mark)

Answer: (i) 18 red, 36 blue, 36 green (ii) $\frac{5}{8}$

Example 5: Case study: the school fair spinner · 4 marks · Case-basedHOTS / competency

At a school fair, a wheel is divided into 20 equal sectors: 8 red, 6 blue, 4 green and 2 golden. The wheel is spun once and stops on one sector at random. A player wins a prize if it stops on golden.

(i) Find the probability that the wheel stops on blue. (1 mark)

(ii) Find the probability that the wheel does not stop on red. (1 mark)

(iii) The organisers want the chance of winning to be $\frac{1}{4}$ . They repaint some red sectors golden. How many red sectors must be repainted, and what is the new probability of stopping on red? (2 marks)

Try it first, then show the solutionSolution

The sectors are equal, so each of the 20 sectors is equally likely.
(i) $P (blue) = \frac{6}{20} = \frac{3}{10}$ . (1 mark)
(ii) Not red $= 20 - 8 = 12$ sectors, so $P = \frac{12}{20} = \frac{3}{5}$ . (1 mark)
(iii) Let $k$ red sectors be repainted. Golden $= 2 + k$ and the total stays 20: $\frac{2 + k}{20} = \frac{1}{4}$ . (½ mark)
$2 + k = 5$ , so $k = 3$ . (½ mark)
Red sectors left $= 8 - 3 = 5$ . (½ mark)
$P (red) = \frac{5}{20} = \frac{1}{4}$ . (½ mark)

Answer: (i) $\frac{3}{10}$ (ii) $\frac{3}{5}$ (iii) 3 sectors; new $P (red) = \frac{1}{4}$

Where students lose marks in Probability Problems on Bags, Balls and Numbers

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

Mistake: Not changing the total when balls are added
Students update the favourable count but forget the bag has grown.
Wrong working
10 balls, 4 white; 5 white added: $P = \frac{9}{10}$
Right working
$P = \frac{4 + 5}{10 + 5} = \frac{9}{15} = \frac{3}{5}$
Do this: Write 'new total = ...' as its own line before any fraction.
Usually costs: Usually all marks after the equation step.
Mistake: Wrong total for a range of cards
Subtracting the end numbers misses one card.
Wrong working
Cards 11 to 60: total $= 60 - 11 = 49$
Right working
Total $= 60 - 11 + 1 = 50$
Do this: Always add 1 for an inclusive range.
Usually costs: Every probability in the question becomes wrong.
Mistake: Double counting in 'or' events
Numbers that are multiples of both values get counted twice.
Wrong working
1 to 100, multiple of 4 or 6: $25 + 16 = 41$
Right working
$25 + 16 - 8 = 33$ , removing multiples of 12
Do this: Subtract the multiples of the LCM.
Usually costs: 1 mark or more in a 3-mark question.
Mistake: Treating 1 as a prime
1 has only one factor, so it is not prime.
Wrong working
Primes from 1 to 10: 1, 2, 3, 5, 7, so $P = \frac{1}{2}$
Right working
Primes: 2, 3, 5, 7, so $P = \frac{4}{10} = \frac{2}{5}$
Do this: Start prime lists at 2.
Usually costs: The answer mark, and the whole MCQ.
Mistake: Accepting a non-whole number of balls
An arithmetic slip gives a fraction of a ball, and the student does not notice.
Wrong working
Getting $g = 6.5$ green balls and moving on
Right working
Recheck the equation. A count of balls must be a whole number.
Do this: Substitute your answer back into every condition.
Usually costs: Final answer marks.

Exam tips

Write the total number of outcomes as a separate line. It is a step mark on its own.
List favourable numbers in full for card questions, so the examiner can give method marks even if you slip later.
In 'find the unknown' questions, write both probabilities as fractions, show the equation and then solve. Finish with a check line.
For spinners, say that the sectors are equal. That is what makes the outcomes equally likely.
Give every final probability in lowest terms and check that it lies between 0 and 1.

Practice questions on Probability Problems on Bags, Balls and Numbers

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

Q1 · 1 mark · MCQBoard level
Cards numbered 1 to 25 are mixed well and one card is drawn at random. The probability that the number is a multiple of 3 or 5 is
(a) $\frac{13}{25}$
(b) $\frac{12}{25}$
(c) $\frac{11}{25}$
(d) $\frac{2}{5}$
Hint
Count multiples of 3 and of 5, then remove multiples of 15.
Answer
(b) $\frac{12}{25}$ . Multiples of 3: 8, of 5: 5, of 15: 1. So $8 + 5 - 1 = 12$ .
Q2 · 3 marks · Short answerHarder board level
A jar has 24 balls, some red and the rest white. If 4 red balls are taken out and 10 white balls are put in, the probability of drawing a red ball becomes $\frac{1}{3}$ . Find the number of red balls at the start, the original probability of drawing a red ball, and the new probability of drawing a white ball.
Hint
Let red = x. The new total is 24 − 4 + 10 and the new red count is x − 4.
Answer
New total $= 24 - 4 + 10 = 30$ . $\frac{x - 4}{30} = \frac{1}{3}$ gives $x - 4 = 10$ , so $x = 14$ red balls (and 10 white). Original $P (red) = \frac{14}{24} = \frac{7}{12}$ . New white $= 10 + 10 = 20$ , so new $P (white) = \frac{20}{30} = \frac{2}{3}$ . Check: $\frac{10}{30} + \frac{20}{30} = 1$ .
Q3 · 3 marks · Short answerHarder board level
A box has cards numbered with all the two-digit numbers, 10 to 99. One card is drawn at random. Find the probability that the number (i) has both digits the same, (ii) is divisible by 5, (iii) is a perfect square.
Hint
Total is 99 − 10 + 1. List squares from 16 to 81.
Answer
Total $= 90$ . (i) $11, 22, \dots, 99$ : 9, so $\frac{1}{10}$ . (ii) $10, 15, \dots, 95$ : 18, so $\frac{1}{5}$ . (iii) $16, 25, 36, 49, 64, 81$ : 6, so $\frac{1}{15}$ .
Q4 · 5 marks · Long answerHarder board level
A bag contains 6 red balls, 8 white balls and some green balls. The probability of drawing a green ball is $\frac{1}{3}$ . (i) Find the number of green balls. (ii) How many white balls must be added so that the probability of drawing a white ball becomes $\frac{1}{2}$ ? (iii) After this, find the probability of drawing a green ball.
Hint
Let green = g; the total is 14 + g. For (ii), only white and the total change.
Answer
(i) $\frac{g}{14 + g} = \frac{1}{3}$ gives $3 g = 14 + g$ , so $g = 7$ and the total is 21. (ii) $\frac{8 + w}{21 + w} = \frac{1}{2}$ gives $16 + 2 w = 21 + w$ , so $w = 5$ . Check: $\frac{13}{26} = \frac{1}{2}$ . (iii) $P (green) = \frac{7}{26}$ .
Q5 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a game stall at a school mela has a bag of red, white and black balls of the same size. A ball is drawn at random. The stall board says $P (red) = \frac{1}{3}$ and $P (white) = \frac{1}{4}$ . To make the game easier, the organisers add 6 black balls, and then the probability of drawing a black ball becomes $\frac{1}{2}$ .
(i) Find the probability of drawing a black ball before the 6 balls are added. (1 mark)
(ii) A student says the bag could have held 30 balls at the start. Explain why this is impossible. (1 mark)
(iii) Find the number of balls of each colour at the start. (2 marks)
Hint
Let the starting total be n. Red is n/3, white is n/4, so black is 5n/12. After adding, black is 5n/12 + 6 and the total is n + 6.
Answer
(i) $P (black) = 1 - \frac{1}{3} - \frac{1}{4} = \frac{5}{12}$ . (ii) Red $= \frac{n}{3}$ and white $= \frac{n}{4}$ must be whole numbers, so $n$ must be a multiple of 12. With 30 balls, white would be $7.5$ , which is impossible. (iii) $\frac{\frac{5 n}{12} + 6}{n + 6} = \frac{1}{2}$ gives $\frac{5 n}{6} + 12 = n + 6$ , so $\frac{n}{6} = 6$ and $n = 36$ . Red $= 12$ , white $= 9$ , black $= 15$ . Check: after adding, black $= 21$ out of $42$ , which is $\frac{1}{2}$ .

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.Bag: 4 red, 6 green. P(green)?Show answer
$\frac{6}{10} = \frac{3}{5}$
2.How many cards are numbered 5 to 54?Show answer
$54 - 5 + 1 = 50$
3.P(red) = 2/5 in a bag of 30 balls. How many red?Show answer
$\frac{2}{5} \times 30 = 12$
4.Bag has n balls, x white; k white added. New P(white)?Show answer
$\frac{x + k}{n + k}$
5.P(red) = 0.3, P(blue) = 0.5, and the rest are green. P(green)?Show answer
$1 - 0.8 = 0.2$

Probability Problems on Bags, Balls and Numbers: frequently asked questions

Why is the probability not one half when a bag has two colours?

The colours are not the equally likely outcomes; the individual balls are. If there are more red balls than blue, red is more likely. Always divide by the number of balls.

How do I find the number of balls when a probability is given?

Multiply the probability by the total if the total is known. If the total is unknown, let the count be x, write the probability as a fraction and solve the equation. Check that the answer is a whole number.

Can a spinner question use the classical definition?

Yes, if all the sectors are equal in size. Then each sector is one equally likely outcome, and you count sectors just as you would count balls.

Are these questions asked as case studies?

They can be. Section E cases often describe a lucky draw, a game or a box of items. The last sub-part usually asks you to find an unknown number, using the same method as on this page.