AFCAT Numerical Ability · Probability

Choosing Several Balls at Once: Combinations

When several balls are taken together, count selections with nCr: the ways to get the mix you want, divided by the ways to choose that many balls from the whole bag.

All Probability notes

Why this matters

This is the hardest page of the chapter: well over a third of its questions are HARD. Each one is two nCr counts and a division, and the marks go on the cases: exactly against at least, same colour summed over every colour, and different colours, where the complement is often quicker.

Concept 1 of 4: A given mix of colours: choose from each colour and multiply

Choosing 3 balls from 5 red and 4 blue, exactly 1 red means 1 of the 5 reds and 2 of the 4 blues:

5 \times 6 = 30

ways out of

^9C_3 = 84

, so

\tfrac{30}{84} = \tfrac{5}{14}

Definition

The total number of ways to choose r balls from n is $^nC_r$ . The order of the chosen balls does not matter.
The ways to get a mix are $^aC_x \times {}^bC_y \times \dots$ , one factor for each colour, with $x + y + \dots = r$ .
'Drawn together', 'selected' or 'chosen' means combinations. Drawing one by one without replacement gives the same probability, worked step by step as on the last page.
Useful values: $^nC_2 = \tfrac{n(n - 1)}{2}$ and $^nC_3 = \tfrac{n(n - 1)(n - 2)}{6}$ .

A mix of colours

P = \dfrac{{}^aC_x \times {}^bC_y}{{}^nC_r}

Worked example

A bag has 4 red and 6 green balls. Three balls are drawn together. Find the probability that exactly 2 are green.

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionModerate

Example 1 · Probability · Selecting Balls with Combinations

A box has

14

balls:

6

red,

4

blue, and

4

yellow. If

3

balls are selected, what is the probability that exactly

2

are red?

Every ball is in the total

Choosing 2 balls from 3 red, 4 blue and 5 white, the total is

^{12}C_2 = 66

, even when the question asks only about red.

Exactly two red needs the rest non-red

Exactly 2 red among 3 chosen means the third ball is not red. Multiply by the ways to choose that ball;

^rC_2

over the total alone misses the factor.

Drill 11 more on a given mix of colours: choose from each colour and multiply

Concept 2 of 4: Same colour: add over every colour

Two balls of the same colour can be both red, or both blue, or both white. Count each case and add.

Definition

Same colour for r balls: add $^{(\text{that colour})}C_r$ over every colour with at least r balls.
Divide by $^nC_r$ .
'Exactly two of the same colour' among three: pick the colour, pick two of it, then pick the third ball from the other colours.

Same colour

P(\text{same}) = \dfrac{{}^aC_r + {}^bC_r + {}^cC_r}{{}^nC_r}

Worked example

A bag has 2 red, 4 blue and 7 white balls. Two are drawn together. Find the probability that they are the same colour.

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionModerate

Example 2 · Probability · Selecting Balls with Combinations

From a bag with

4

red,

5

blue, and

3

yellow balls, what is the probability of drawing

2

balls of the same color?

Do not stop at one colour

Same colour from 2 red, 4 blue and 7 white is

1 + 6 + 21 = 28

pairs. Counting only the white pairs gives

\tfrac{21}{78}

Drill 5 more on same colour: add over every colour

Concept 3 of 4: Different colours: one minus same, or one of each

Two balls of different colours are every pair that is not the same colour. For three balls from three colours, 'all different' means one of each colour.

Definition

Two balls of different colours: $1 - P(\text{same colour})$ .
r balls, all different, from exactly r colours: multiply the colour counts. From more colours than r, add over each choice of r colours.
With fewer colours than balls, 'all different' is impossible: the probability is 0.

Different colours

P(\text{different}) = 1 - P(\text{same})

Worked example

A bag has 2 red, 4 blue and 7 white balls. Two are drawn together. Find the probability that they are different colours.

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionHard

Example 3 · Probability · Selecting Balls with Combinations

A box contains

5

red,

3

green, and

2

yellow balls. If two balls are drawn without replacement, what is the probability that they are of different colors?

One of each is a product, not a sum

One red, one blue and one green from 2, 3 and 4 balls is

2 \times 3 \times 4 = 24

choices, not

2 + 3 + 4

More colours than balls

Three balls from four colours: first choose which three colours, then one ball of each, and add over all four choices of colours.

Drill 6 more on different colours: one minus same, or one of each

Concept 4 of 4: At least: add the exact cases, or take away the rest

'At least 2 red' among 3 chosen is exactly 2 red or exactly 3 red. Count each case with the mix rule and add. When the cases are many, count what is not wanted and subtract.

Definition

At least k of a colour: add exactly k, k + 1, and so on up to the most possible.
At least one of a colour with a balls: $1 - \dfrac{^{n - a}C_r}{^nC_r}$ .
Pick whichever list of cases is shorter.

At least one of a colour

P(\text{at least one}) = 1 - \dfrac{{}^{n - a}C_r}{{}^nC_r}

Worked example

A bag has 5 red and 4 blue balls. Three are drawn together. Find the probability that at least 2 are red.

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionHard

Example 4 · Probability · Selecting Balls with Combinations

A bag contains

7

white and

5

black balls. What is the probability of drawing

4

balls such that at least

3

are white?

At least includes the top case

At least 2 red among 3 drawn includes 3 red. Counting only exactly 2 red leaves that case out.

The opposite of at least one

The complement of 'at least one red' is 'no red', not 'exactly one red'.

Drill 2 more on at least: add the exact cases, or take away the rest

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (4)

A given mix of colours: choose from each colour and multiply
A mix of colours
$P = \dfrac{{}^aC_x \times {}^bC_y}{{}^nC_r}$
Same colour: add over every colour
Same colour
$P(\text{same}) = \dfrac{{}^aC_r + {}^bC_r + {}^cC_r}{{}^nC_r}$
Different colours: one minus same, or one of each
Different colours
$P(\text{different}) = 1 - P(\text{same})$
At least: add the exact cases, or take away the rest
At least one of a colour
$P(\text{at least one}) = 1 - \dfrac{{}^{n - a}C_r}{{}^nC_r}$

Watch out for (7)

Every ball is in the total
→ A given mix of colours: choose from each colour and multiply
Exactly two red needs the rest non-red
→ A given mix of colours: choose from each colour and multiply
Do not stop at one colour
→ Same colour: add over every colour
One of each is a product, not a sum
→ Different colours: one minus same, or one of each
More colours than balls
→ Different colours: one minus same, or one of each
At least includes the top case
→ At least: add the exact cases, or take away the rest
The opposite of at least one
→ At least: add the exact cases, or take away the rest

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