Combined Conditional Probability

 

What Is Combined Conditional Probability?

Combined conditional probability is about finding the chance that multiple events happen in order, without replacement.

When something is removed and not put back, the probabilities change on the second draw. This is why it’s conditional — the second event depends on what happened first.

 
Key Rules to Remember

If events A and B happen one after the other: $P(A \text{ and } B) = P(A) \times P(B | A)$

If you're calculating the probability of getting different combinations (like blue then green or green then blue), add the separate probabilities: $P(BG \text{ or } GB) = P(BG) + P(GB)$

 

Example 1: Two Balls, Different Colours

A bag contains:

• 4 blue balls
• 5 red balls
• 1 green ball

Two balls are picked one after the other without replacement.

Question: What is the probability that the two balls are different colours?

 

Method 1: Subtracting from 1

We find the probability that both balls are the same colour and subtract it from 1.

$P(BB) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90}$

$P(RR) = \frac{5}{10} \times \frac{4}{9} = \frac{20}{90}$

$P(GG) = \frac{1}{10} \times \frac{0}{9} = 0$

So: $P(\text{same colour}) = \frac{12 + 20 + 0}{90} = \frac{32}{90} \quad P(\text{different colours}) = 1 - \frac{32}{90} = \frac{58}{90} = \frac{29}{45}$

 

Method 2: Listing All Different Colour Combos

Possible orders for different colours:

• BR, RB
• BG, GB
• RG, GR

Calculate each:

• $P(BR) = \frac{4}{10} \times \frac{5}{9} = \frac{20}{90}$
• $P(RB) = \frac{5}{10} \times \frac{4}{9} = \frac{20}{90}$
• $P(BG) = \frac{4}{10} \times \frac{1}{9} = \frac{4}{90}$
• $P(GB) = \frac{1}{10} \times \frac{4}{9} = \frac{4}{90}$
• $P(RG) = \frac{5}{10} \times \frac{1}{9} = \frac{5}{90}$
• $P(GR) = \frac{1}{10} \times \frac{5}{9} = \frac{5}{90}$

Add all of them: $P(\text{different}) = \frac{20 + 20 + 4 + 4 + 5 + 5}{90} = \frac{58}{90} = \frac{29}{45}$

 

 

Example 2: Three Balls, All the Same Colour

Same bag: 4 blue, 5 red, 1 green.

Question: What is the probability of drawing three balls all the same colour (without replacement)?

Possible sets:

BBB

RRR

GGG (not possible, only one green ball)

Calculate:

• $P(BBB) = \frac{4}{10} \times \frac{3}{9} \times \frac{2}{8} = \frac{24}{720}$
• $P(RRR) = \frac{5}{10} \times \frac{4}{9} \times \frac{3}{8} = \frac{60}{720}$

Add them: $P(\text{same}) = \frac{24 + 60}{720} = \frac{84}{720} = \frac{7}{60}$

Final Answers:

P(two different colours): $\frac{29}{45}$

P(three same colour): $\frac{7}{60}$