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Conditional Probability

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

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Conditional Probability

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

  4. Conditional Probability – Question 1 = 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)

  5. 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 sweets b.) Two red sweets c.) One green sweet and one red sweet? It is often useful to use a tree diagram to help answer these questions.

  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? Outcomes Jim Sandra GG G That is the hardest part. 6 9 G 7 3 10 GR Now we can use this diagram to help us answer the question. 9 R 7 G RG 9 3 R 10 2 R RR 9

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

  8. Conditional Probability – Question 3 On my way to work I pass two sets of traffic 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 green b.) none are green c.) exactly one is green d.) at least one is green Have a go at setting up the tree diagram for this question.

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

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

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