Combined Conditional Probabilities (Edexcel IGCSE Maths B): Revision Note

If a question says "two cards are drawn" then you may assume that they draw 1 card followed by another card without replacement (the maths is the same).

Try not to simplify your probabilities too early as it is easier to add probabilities together when they all have the same denominator!

A bag contains 10 yellow beads, 6 blue beads and 4 green beads.

A bead is taken at random from the bag and not replaced.

A second bead is then taken at random from the bag.

(a) Find the probability that both beads are different colours.

Answer:

Let Y, B and G represent choosing a yellow, blue and green bead

Method 1

The probability of the beads being different colours is equal to 1 minus the probability that the beads are the same colour

Find the probability of both beads being the same colour

P(same colours) = P(YY) + P(BB) + P(GG)

Calculate each conditional probability separately, remembering the number of beads changes after one is drawn and not replaced

For example, P(YY) = $\frac{10}{20}\times \frac{9}{19}$

$\frac{10}{20}\times \frac{9}{19}+\frac{6}{20}\times \frac{5}{19}+\frac{4}{20}\times \frac{3}{19}$

Multiply the pairs of fractions together and add their results

$\frac{232}{380}$

Subtract this from 1

$1-\frac{132}{380}=\frac{248}{380}$

Simplify the answer

$\frac{62}{95}$

Method 2

List all the possibilities of different colours

Remember that YB (yellow first, then blue) is different to BY (blue first, then yellow)

YB, BY, YG, GY, BG, GB

Use the "or" rule to add the cases together

P(different colours) = P(YB) + P(BY) + P(YG) + P(GY) + P(BG) + P(GB)

Calculate each conditional probability separately, remembering the number of beads changes after one is drawn and not replaced

For example, P(YB) = $\frac{10}{20}\times \frac{6}{19}$

$\frac{10}{20}\times \frac{6}{19}+\frac{6}{20}\times \frac{10}{19}+\frac{10}{20}\times \frac{4}{19}+\frac{4}{20}\times \frac{10}{19}+\frac{6}{20}\times \frac{4}{19}+\frac{4}{20}\times \frac{6}{19}$

Multiply the pairs of fractions together and add their results

$\frac{248}{380}$

Simplify the answer

$\frac{62}{95}$

(b) The second bead is not replaced and a third bead is taken at random from the bag.

Find the probability that all three beads are the same colour.

Answer:

List the possibilities

YYY, BBB, GGG

Use the "or" rule to add between cases

P(all the same colour) = P(YYY) + P(BBB) + P(GGG)

Use conditional probabilities in each separate case, remembering the number of beads changes after each one is drawn and not replaced

$\frac{10}{20}\times \frac{9}{19}\times \frac{8}{18}+\frac{6}{20}\times \frac{5}{19}\times \frac{4}{18}+\frac{4}{20}\times \frac{3}{19}\times \frac{2}{18}$

Multiply the triplets of fractions together then add their results

$\frac{720}{6840}+\frac{120}{6840}+\frac{24}{6840}=\frac{864}{6840}$

Simplify the answer

$\frac{12}{95}$

A bag contains only red and blue marbles. There are 3 more red marbles than blue marbles.

Kai takes 2 marbles from the bag at random.

The probability that Kai takes 2 blue marbles is $\frac{1}{7}$.

Work out the number of blue marbles there were in the bag initially.

Show clear algebraic working.

Answer:

Write expressions for the number of each colour

Let $x$ be the number of blue marbles

Then there are $x+3$ red marbles

And there are $x+x+3=2x+3$ marbles in total

Find the probability that the first marble is blue

$\frac{x}{2x+3}$

Given that the first marble is blue, find the probability that the second marble is blue

• After one blue marble is taken, there are $x-1$ blue marbles left out of $2x+2$ marbles in total

$\frac{x-1}{2x+2}$

Find the probability that both marbles are blue by multiplying the probabilities together

• Set this equal to $\frac{1}{7}$

$\frac{x}{2x+3}\times \frac{x-1}{2x+2}=\frac{1}{7}$

Multiply through by $7\left(2x+3\right)\left(2x+2\right)$ to get rid of the fractions

$7x\left(x-1\right)=\left(2x+3\right)\left(2x+2\right)$

Expand and simplify both sides

$$\begin{gathered} 7x^2-7x=4x^2+4x+6x+6 \\ 7x^2-7x=4x^2+10x+6 \end{gathered}$$

Rearrange to make one side equal to zero

$3x^2-17x-6=0$

Factorise the quadratic

• Find two numbers that multiply to give 3×-6=-18 and add to give -17

• 1 and -18

$$\begin{gathered} 3x^2+x-18x-6 \\ x\left(3x+1\right)-6\left(3x+1\right) \\ \left(3x+1\right)\left(x-6\right) \end{gathered}$$

Solve the quadratic equation

$$\begin{gathered} \left(3x+1\right)\left(x-6\right)=0 \\ x=-\frac{1}{3},x=6 \end{gathered}$$

$x$ needs to be a positive whole number

$x=6$ blue marbles initially