A bag contains 8 red and 5 white balls. Three balls are drawn at random. Find the probability that (a) All the

NCERT Exemplar · Q13

Q. A bag contains 8 red and 5 white balls. Three balls are drawn at random. Find the probability that

(a) All the three balls are white.

(b) All the three balls are red.

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Concept understanding — Permutations Without Repetition

Permutations Without Repetition – The Idea of Arranging Things

Imagine you have three different books on a shelf: a Physics book, a Chemistry book, and a Maths book. How many different ways can you arrange them in a row?

You could try listing them out:

Physics, Chemistry, Maths
Physics, Maths, Chemistry
Chemistry, Physics, Maths
Chemistry, Maths, Physics
Maths, Physics, Chemistry
Maths, Chemistry, Physics

That's 6 arrangements. Notice that each arrangement uses all three books exactly once — no book is repeated, and no book is left out. This is the core idea: permutations without repetition count the number of ways to arrange a set of distinct objects in order, using each object exactly once.

Why "Without Repetition"?

The phrase "without repetition" means that once you place an object in a position, you cannot use it again. In our book example, once you put the Physics book in the first slot, you cannot put it in the second or third slot. Each object appears exactly once in the arrangement.

This is different from "permutations with repetition" (like creating 3-letter codes from the letters A, B, C where you can reuse letters — e.g., AAA, AAB, etc.). Here, no repeats allowed.

The Counting Logic – Why Multiply?

Let's build the arrangement step by step for 3 distinct books:

First position: You have 3 choices (any of the 3 books).
Second position: After placing the first book, only 2 books remain — so 2 choices.
Third position: Only 1 book is left — so 1 choice.

Total arrangements = $3 \times 2 \times 1 = 6$ .

This product $3 \times 2 \times 1$ is called 3 factorial, written as $3!$ .

P(n) = n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1

For $n$ distinct objects, the number of permutations (arrangements in order) is $n!$ .

What If You Only Arrange Some of Them?

Suppose you have 5 different books, but you only want to arrange 3 of them on a shelf. How many ways?

First position: 5 choices
Second position: 4 choices
Third position: 3 choices

Total = $5 \times 4 \times 3 = 60$ .

This is a permutation of 5 objects taken 3 at a time, written as $P(5,3)$ or ${}^5P_3$ .

P(n, r) = \frac{n!}{(n-r)!} = n \times (n-1) \times \cdots \times (n-r+1)

Here $n$ is the total number of distinct objects, and $r$ is how many you are arranging. The formula works because:

Numerator $n!$ counts all arrangements of all $n$ objects.
Denominator $(n-r)!$ removes the arrangements of the $n-r$ objects you are not using.

Key Points to Remember

Important

Order matters — swapping two objects gives a different permutation.
No repetition — each object is used at most once.
For arranging all $n$ objects: $n!$
For arranging r out of $n$ objects: $\frac{n!}{(n-r)!}$

Common Mistake to Avoid …

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We treat the draw as a combination (order irrelevant) and count favourable outcomes over total outcomes using $^{n}C_{r}$ . The probabilities are: (a) $\frac{10}{286}$ ,

(b) $\frac{56}{286}$ ,

The core idea here is classical probability:

P(\text{event}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of equally likely outcomes}}

Since we are drawing three balls at random from a bag, and we don't care about the order in which they appear, each outcome is a combination — a selection of 3 balls from the total. This is the Permutations Without Repetition idea applied to counting selections: order doesn't matter, so we use $^{n}C_{r}$ .

^{n}C_{r} = \frac{n!}{r!(n-r)!}

Let’s first find the total number of ways to draw any 3 balls from the bag.

Total balls = $8$ red + $5$ white = $13$ balls. Total outcomes = number of ways to choose any 3 from 13:

^{13}C_{3} = \frac{13 \times 12 \times 11}{3 \times 2 \times 1} = 286

So there are 286 equally likely sets of three balls.

Now we handle each part separately.

(a) All three balls are white

We need to choose 3 white balls from the 5 available.

Favourable outcomes = $^{5}C_{3}$ :

^{5}C_{3} = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10

So the probability is:

P(\text{all white}) = \frac{10}{286}

Tip

Notice that $^{5}C_{3} = ^{5}C_{2}$ — choosing which 3 to take is the same as choosing which 2 to leave out. This symmetry often simplifies calculations.

(b) All three balls are red

We need to choose 3 red balls from the 8 available.

Favourable outcomes = $^{8}C_{3}$ :

^{8}C_{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56

So:

$P(\text{all red}) = \frac{56}{286}$ …

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Q.A bag contains 8 red and 5 white balls. Three balls are drawn at random. Find the probability that

(a) All three balls are white

(b) All three balls are red

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