Probability • Topic 2 of 4

Conditional Probability

Conditional probability answers "what is the chance of A given that B has already happened?" The condition shrinks the sample space to only the outcomes where B is true, so P(A|B) = P(A∩B)/P(B). The clearest CAT signal is the phrase "given that" or any setup where one event is already known to have occurred. With cards and dice the fast method is to recount directly inside the reduced space: if a die shows an even number, there are only 3 outcomes left {2,4,6}, so the chance it is also greater than 3 is just 2/3. Drawing "without replacement" is conditional by nature — the second draw’s probabilities depend on what the first draw removed. Rearranging the formula gives the multiplication rule P(A∩B) = P(B)·P(A|B), the engine behind sequential draws.

✅ Solved examples

1. A die is rolled and the result is known to be even. Find the probability that it is greater than 3.
Reduced space = {2,4,6}. Greater than 3 among these = {4,6} = 2. P = 2/3.
2. Two cards are drawn without replacement. Find the probability both are kings.
P(1st king) = 4/52. Given that, P(2nd king) = 3/51. P(both) = (4/52)(3/51) = 12/2652 = 1/221.
3. In a class, 60% study Maths and 30% study both Maths and Stats. Of those who study Maths, what fraction also study Stats?
P(Stats|Maths) = P(both)/P(Maths) = 0.30/0.60 = 1/2.
4. Two dice are rolled; the sum is known to be 8. Find the probability that one die shows a 5.
Sum 8: (2,6),(3,5),(4,4),(5,3),(6,2) = 5 outcomes. A 5 appears in (3,5) and (5,3) = 2. P = 2/5.

✏️ Practice — try these, take hints as needed

1. A die shows an odd number. Probability it is a prime?
Odd space = {1,3,5}.
Primes there: 3, 5.
That is 2 of 3.
2/3
2. Two cards drawn without replacement. Probability both are red?
First red = 26/52.
Then 25/51.
Multiply.
25/102
3. 70% of people like tea, 40% like both tea and coffee. Given a tea-liker, probability they like coffee?
P(coffee|tea) = both/tea.
0.40/0.70.
Simplify.
4/7
4. Two dice rolled; sum is 6. Probability one die shows 2?
Sum 6: (1,5),(2,4),(3,3),(4,2),(5,1) = 5.
A 2 appears in (2,4),(4,2).
2 of 5.
2/5
5. A bag has 3 red, 2 blue. Two drawn without replacement. Probability second is red given first was red?
One red already removed.
Now 2 red of 4 left.
Ratio.
1/2

📝 Topic test — 8 questions

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