Calculating Conditional Probability
This topic covers the core rules for combining probabilities of events, focusing on when to add (for 'OR' scenarios) and when to multiply (for 'AND' scenarios). It also introduces conditional probability, where the chance of one event happening is affected by the outcome of another.
Part of the ESAT Mathematics 1 syllabus — revision for the Engineering and Science Admissions Test (ESAT), the UAT-UK admissions test for Cambridge, Imperial, Oxford and UCL.
Key points
- Use addition for the probability of 'A OR B'. If the events are mutually exclusive (cannot happen together), simply add their probabilities. If they can, you must subtract the probability of them both happening to avoid double-counting.
- Use multiplication for the probability of 'A AND B'. If the events are independent, multiply their individual probabilities. If they are dependent, the probability of the second event will change after the first has occurred.
- Conditional probability, P(B|A), is the probability of B happening, given that A has already happened. The key is that the set of possible outcomes has been restricted by event A.
- Tree diagrams are essential for visualising sequences of events, especially when probabilities are dependent (e.g., drawing items 'without replacement'). Multiply along the branches to find the probability of a sequence, and add the results of different branches that meet the condition.
- Venn diagrams and two-way tables are useful for organising data about overlapping groups, making it easier to calculate conditional probabilities by focusing on the relevant row, column, or circle.
- The probability of 'at least one' of something happening is often most easily calculated as `1 - P(none of them happening)`.
Formulae
P(A or B) = P(A) + P(B) To find the probability of event A or event B occurring, when A and B are mutually exclusive.
P(A or B) = P(A) + P(B) - P(A and B) To find the probability of event A or event B occurring, when they are not mutually exclusive (i.e., they can happen together).
P(A and B) = P(A) × P(B) To find the probability of both independent events A and B occurring.
P(B|A) = P(A and B) / P(A) To calculate the conditional probability of event B, given that event A has already occurred.
Definitions
- Mutually Exclusive Events
- Two or more events that cannot occur at the same time. For example, a single coin toss cannot be both heads and tails.
- Independent Events
- Events where the outcome of one has no effect on the outcome of the other. For example, rolling a die and then tossing a coin.
- Conditional Probability
- The probability of an event occurring, given that another event is known to have already occurred. This is relevant for dependent events.
Worked example
A box contains 6 lemon sweets and 4 strawberry sweets. One sweet is taken at random and eaten. Then, a second sweet is taken at random. What is the probability that the two sweets are of different flavours?
- 1
This is a 'without replacement' problem, so the events are dependent.
- 2
The condition 'different flavours' can be met in two mutually exclusive ways:
(Lemon then Strawberry) OR (Strawberry then Lemon).
We need to calculate the probability of each path and add them.
- 3
Path 1:
Lemon then Strawberry.
P(1st is Lemon) = 6/10.
After one lemon is eaten, there are 9 sweets left, 4 of which are strawberry.
P(2nd is Strawberry | 1st was Lemon) = 4/9.
The probability of this path is (6/10) × (4/9) = 24/90.
- 4
Path 2:
Strawberry then Lemon.
P(1st is Strawberry) = 4/10.
After one strawberry is eaten, there are 9 sweets left, 6 of which are lemon.
P(2nd is Lemon | 1st was Strawberry) = 6/9.
The probability of this path is (4/10) × (6/9) = 24/90.
- 5
Add the probabilities of the two paths:
P(different) = P(L then S) + P(S then L) = 24/90 + 24/90 = 48/90.
- 6
Simplify the final fraction:
48/90 = 24/45 = 8/15
Answer: 8/15
Common mistakes
- ×Making arithmetic errors with fractions is the most frequent mistake. Take care when finding common denominators or multiplying numerators and denominators.
- ×Forgetting to update the probabilities for the second event in 'without replacement' scenarios. Remember to decrease both the number of the item taken and the total number of items by one.
- ×Confusing the rules for 'AND' and 'OR'. 'AND' implies a sequence of events, so you multiply along the branches of a tree diagram. 'OR' implies a choice between outcomes, so you add the probabilities of the relevant final branches.
No-calculator tips
- ✓When multiplying fractions, simplify before you multiply. In (6/10) × (4/9), you can simplify 6/9 to 2/3, giving (6/10) × (2/3) = 12/30. Or simplify 4/10 to 2/5, giving (6/9) × (2/5) = 12/45. Both simplify to 4/15. Wait, 24/90 is 8/15⋯ let's recheck. (6/10)*(4/9) → simplify 6 and 9 by 3 → (2/10)*(4/3) = 8/30 = 4/15. My example calculation had an error. Let me fix the worked example. (6/10) × (4/9) = 24/90. (4/10) × (6/9) = 24/90. Total 48/90. 48/90 simplifies to 8/15. Correct. My no-calc tip should be about simplifying diagonally. For (6/10)*(4/9), simplify 6 and 9 to 2 and 3. Simplify 4 and 10 to 2 and 5. This gives (2/5)*(2/3)=4/15. Ok this is much better.
- ✓When multiplying fractions like (6/10) × (4/9), look for common factors diagonally or vertically to simplify before multiplying. Here, 6 and 9 share a factor of 3, and 4 and 10 share a factor of 2. This becomes (2/5) × (2/3) = 4/15. This makes the calculation much simpler.
- ✓In problems with tables of data, don't use the grand total for conditional probabilities. If an event is 'given', that group becomes your new denominator. For example, 'probability they drive, given they work in the warehouse' means the denominator is the total number of warehouse staff, not the total number of all staff.