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Published byTerence West Modified over 3 years ago

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Probability 2 Compound Probability

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Now lets consider the following: 2 dice are rolled and the numbers are added together. What are the numbers we can obtain? This is best done as a table (called a SPACE diagram). The table will give the total score.

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123456 1 2 3 4 5 6 1 st Die 2 nd Die Now fill in the numbers – we need the sum of the 2 numbers on each die 2 3 4 5 6 7 3 4 5 6 7 8 4 5 6 7 8 9 5 6 7 8 9 10 6 7 8 9 11 7 8 9 10 11 12 Space Diagram

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Remember that the probability of an event happening is: P(E) = Number of desired outcomes Total number of possible outcomes What is the total number of possible outcomes from the table? 36 as there are 36 filled in squares.

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What is the probability of scoring: 1 ? 5 ? 7 ? an odd number ? an even number ? a number less than 4 ? a number greater than 6 ? 0 4 / 36 6 / 36 18 / 36 3/36 21 / 36

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A similar question: 2 dice are rolled and the numbers are multiplied together. What are the numbers we can obtain? This is best done as a table (space diagram). See next slide. The table will give the total score.

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123456 1 2 3 4 5 6 1 st Die 2 nd Die Now fill in the numbers – we need the product of the 2 numbers on each die 1 2 3 4 5 6 2 4 6 8 10 12 3 6 9 15 18 4 8 12 16 20 24 5 10 15 20 25 30 6 12 18 24 30 36 Space Diagram

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What is the probability of scoring: 8 ? 12 ? a multiple of 3 ? a multiple of 4 ? an even number ? a number less than 9 ? a number greater than 20 ? 2 / 36 4 / 36 20 / 36 15 / 36 27 / 36 16 / 36 6 / 36

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So far we have looked at simple probabilities by selection and replacement. What about if the removed object is kept, how does this affect the outcomes of subsequent selection? Consider the initial 6 red and 5 black balls in an earlier problem.

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If a red ball is chosen P(Red) = 6 / 11 this then leaves only 5 red and a total of 10 balls remaining. So P(Red) as a second choice is 5 / 10 This idea of reducing numbers continues as we remove and keep the chosen object !! This is better viewed as a tree diagram.

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R B 6 R 5 B 5/11 Red chosen first Black chosen first Second choice R B R B 6/11 5/10 6/10 4/10 5 reds left 10 to choose from 5 blacks left 10 to choose from 6 reds left 10 to choose from 4 blacks left 10 to choose from

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To work out the probability of choosing a red followed by a red, we look along the tree diagram collecting the probabilities. In this case 6 / 11 and 5 / 10 To work out the probability of selecting 2 reds, we MULTIPLY the two probabilities together. P(Red,Red)= 6 / 11 × 5 / 10 = 30 / 110 = 3 / 11

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To work out the probability of choosing a black followed by a black, we look along the tree diagram collecting the probabilities. In this case 5 / 11 and 4 / 10 To work out the probability of selecting 2 reds, we MULTIPLY the two probabilities together. P(Black,Black) = 5 / 11 × 4 / 10 = 20 / 110 = 2 / 11

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To work out the probability of choosing one of each colour, we look along the tree diagram collecting the probabilities. In this case 6 / 11 and 5 / 10 = red black 5 / 11 and 6 / 10 = black red To work out the probability of selecting 1 of each, we MULTIPLY the two probabilities together and add the answers.

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In this case 6 / 11 and 5 / 10 = red black 5 / 11 and 6 / 10 = black red To work out the probability of selecting 1 of each, we MULTIPLY the two probabilities together and add the answers. P(Red,Black) = 6 / 11 × 5 / 10 = 30 / 110 = 3 / 11 P(Black,Red) = 5 / 11 × 6 / 10 = 30 / 110 = 3 / 11 Total = 3 / 11 + 3 / 11 = 6 / 11

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Notice that P(Red,Red) = 3 / 11 P(Black,Black) = 2 / 11 P(Red,Black Or Black,Red) = 6 / 11 Total = 3 / 11 + 2 / 11 + 6 / 11 = 11 / 11 = 1 So we are certain to get RR or BB or 1 of each

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These are called dependent events since the outcome of the second event depends on the outcome of the first event.

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