Probability with Marbles, Spinners and Numbers

Drawing Balls or Marbles from a Bag

If a bag holds coloured balls, each ball is equally likely to be drawn. The probability of a colour is $$P(\text{colour}) = \dfrac{\text{number of balls of that colour}}{\text{total number of balls}}.$$

For a bag with $r$ red and $b$ blue balls (total $r+b$): $P(\text{red}) = \dfrac{r}{r+b}$, and $P(\text{blue}) = \dfrac{b}{r+b}$; these add to 1.

Key Point: Total $=$ all the balls, not just one colour. The colour counts are the numerators.

Spinners

A spinner divided into equal sectors is like a die with as many faces as sectors. If a spinner has $n$ equal sectors, each is equally likely, and $$P(\text{a particular sector or colour}) = \dfrac{\text{number of favourable sectors}}{n}.$$

If the sectors are not equal, the outcomes are not equally likely and this simple formula does not apply directly.

Key Point: Equal sectors $\Rightarrow$ equally likely. Count favourable sectors over total sectors.

Choosing a Number

When a number is picked at random from $1$ to $n$, each is equally likely. Then, for example:

• $P(\text{even})$, $P(\text{prime})$, $P(\text{a perfect square})$, $P(\text{a multiple of 5})$ are each (how many such numbers) $\div\ n$.

Count the favourable numbers carefully — for instance, the perfect squares from 1 to 100 are $1,4,9,\dots,100$ (ten of them), so $P(\text{perfect square}) = \dfrac{10}{100} = \dfrac{1}{10}$.

Key Point: List or count the favourable numbers in the given range, then divide by how many numbers there are.

Defective Items and Real-Life Counts

Everyday problems ("a lot of bulbs, some defective", "tickets in a box") are the same idea: $$P = \dfrac{\text{favourable count}}{\text{total count}}.$$ For 12 bulbs of which 3 are defective, $P(\text{defective}) = \dfrac{3}{12} = \dfrac14$ and $P(\text{good}) = \dfrac{9}{12} = \dfrac34$.

Key Point: Identify the total and the favourable count from the wording — the formula never changes.

Solved Examples

Example 1: Coloured balls

A bag contains 5 red and 3 blue balls. One ball is drawn at random. Find $P(\text{red})$ and $P(\text{blue})$.

Solution:

1. Total $= 5+3 = 8$.
2. $P(\text{red}) = \dfrac58$, $P(\text{blue}) = \dfrac38$.

Final Answer: $\dfrac58$ and $\dfrac38$ (they add to 1).

Takeaway: Colour count over total count.

Example 2: Not a particular colour

A bag has 4 red, 5 green and 6 blue balls. Find the probability that a drawn ball is not green.

Solution:

1. Total $= 4+5+6 = 15$; green $= 5$.
2. $P(\text{not green}) = 1 - \dfrac{5}{15} = 1 - \dfrac13 = \dfrac23$.

Final Answer: $\dfrac23$.

Takeaway: Complement, or count the $4+6=10$ non-green directly.

Example 3: Choosing a number

A number is chosen at random from 1 to 20. Find the probability that it is a multiple of 3.

Solution:

1. Multiples of 3 from 1 to 20: $3,6,9,12,15,18$ — six of them.
2. $P = \dfrac{6}{20} = \dfrac{3}{10}$.

Final Answer: $\dfrac{3}{10}$.

Takeaway: Count the favourable numbers in the range.

Example 4: Defective bulbs

A carton has 20 bulbs, 4 of them defective. One bulb is drawn. Find the probability it is (i) defective, (ii) good.

Solution:

1. (i) $P(\text{defective}) = \dfrac{4}{20} = \dfrac15$.
2. (ii) $P(\text{good}) = 1 - \dfrac15 = \dfrac45$ (or $\dfrac{16}{20}$).

Final Answer: (i) $\dfrac15$; (ii) $\dfrac45$.

Takeaway: Good $+$ defective $=$ all, so their probabilities add to 1.