1. Conditional Probability Lesson objectives: Work out the probability of combined events when the probabilities change after each event.

2. Starter Get ready for SINGLE, DOUBLE, TRIPLE! Put the numbers 1 – 15 down the side of your book and then three columns of Single, Double and Triple

3. Can you think of an example of where this may be the case? Refers to events where the probability of an event is dependent on the outcome of another event. Conditional Probability

4. Conditional Probability – Question 1 There are four blue and six green balls in a bag. One is taken out at random and not replaced. This is repeated three times. What is the probability that: a.) No green balls are selected b.) At least one green ball is selected? SOLUTION a.) p(no green) = p(B and B and B) SOLUTION b.) p(at least one green) = 1 – p(no green) = 1 – p(B and B and B) = 1 –

5. It is often useful to use a tree diagram to help answer these questions. Conditional Probability – Question 2 Sandra has a bag of coloured sweets. 7 are green and 3 are red. She takes one out and eats it before offering the bag to Jim. What is the probability that they select: a.) Two green sweetsb.) Two red sweets c.) One green sweet and one red sweet?

6. Conditional Probability - Question Sandra has a bag of coloured sweets. 7 are green and 3 are red. She takes one out and eats it before offering the bag to Jim. What are the two events? Sandra Jim G R 7 10 3 G G R R 6 9 3 9 7 9 2 9 Outcomes GG GR RG RR Now we can use this diagram to help us answer the question. That is the hardest part.

7. Conditional Probability - Question Sandra Jim G R 7 10 3 G G R R 6 9 3 9 7 9 2 9 Outcomes GG GR RG RR a.) Two green sweets p(GG) = top line 7 10 G 6 9 G GG p(GG) = 7 × 6 = 7 10 9 15 b.) Two red sweets p(RR) = bottom line 3 10 R R 2 9 RR p(RR) = 3 × 2 = 1 10 9 15 c.) One green and one red How many ways can we select this? G R 3 9 GRGR 3 10 R G 7 9 RGRG p(one G one R) = p(GR) + p(RG) = 7 × 3 + 3 × 7 10 9 10 9 = 7 + 7 30 30 = 7 15

8. Conditional Probability – Question 3 On my way to work I pass two sets of traffic Have a go at setting up the tree diagram for this question. lights. The probability that the first is green is 1/5. If the first is green the probability that the second is green is 2/3. If the first is red, the probability that the second is green is 1/4. What is the probability of each of these? a.) both are greenb.) none are green c.) exactly one is greend.) at least one is green

9. Conditional Probability – Question 3 1 st Set2 nd Set G R 1 5 4 5 G G R R 2 3 1 3 1 4 3 4 Outcomes GG GR RG RR

10. Conditional Probability – Question 3 1 st Set2 nd Set G R 1 5 4 5 G G R R 2 3 1 3 1 4 3 4 Outcomes GG GR RG RR a.) p(GG) = 2 15 b.) p(RR) = 3 5 c.) p(one G) = p(GR) + p(RG) = 1 + 1 = 4 15 5 15 d.) p(at least 1 G) =1 – p(no G) = 1 – p(RR) = 1 – 3 = 2 5 5

11. Question Practice Work through exercise 19K - page 479 If you are using a tree diagram – take extra care when working out the probability for the second event