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GCSE Mathematics Probability and Data Handling

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2 Probability The probability of any event lies between 0 and 1. 0 means it will never happen. 1 means it is certain to happen. Between these values, the event may or may not happen. The sum of all probabilities is equal to 1

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3 Probability The probability of one event happening OR another is found by adding the probabilities, being careful of any overlap The probability of one event happening AND another happening is found by multiplying the probabilities, being careful of replacement

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4 B O U R N E M O U T H Find the following probabilities P(B)=1/11P(U)=2/11P(Vowel)=5/11 So P(consonant) =1 - 5/11=6/11 P(U or R) =2/11 + 1/11=3/11 P(Vowel or red) =5/11 +5/11 - 3/11 =7/11

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5 B O U R N E M O U T H One letter is chosen, replaced, then another is chosen. P(2 B’s) =1/11 x1/11 = 1/121 P( M then U) =1/11 x 2/11 = 2/121 P(2 reds) =5/11 x 5/11 = 25/121 Two letters are chosen without replacement P( 2 O's) =2/11 x 1/10=2/110 P( M then U) =1/11 x 2/10 = 2/110 P(2 reds) =5/11 x 4/10 = 2/11

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6 Possibility Space Used to show all the outcomes of an event. Two tetrahedral dice are thrown, one contains the numbers 0 2 4 6, the other contains the numbers 1 2 3 4. The score is found by adding the numbers together

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7 Possibility Space 0 2 4 6 0 2 4 6 11 3 5 7 11 3 5 7 2 2 4 6 8 2 2 4 6 8 3 3 5 7 9 3 3 5 7 9 4 4 6 8 10 4 4 6 8 10 P(8) = 2/16 =1/8 P( odd number) = 8/16 = 1/2 P( less than 5) = 6/16 = 3/8 P( 4 or 9) = 2/16 + 1/16 = 3/16

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8 Probability Trees These are used when more than one event is being described. They conveniently display ALL the possible outcomes. Within a branch the probabilities add up to 1. To calculate the given probability, multiply along the branches.

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9 Example On my way to work I pass two sets of traffic lights. The probability I have to stop at the first set is 0.4 The probability I have to stop at the second set is 0.45 What is the probability I have to stop (i) once only; (ii) at least once?

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10 Example sss s gsg sgs g g gg 1st2nd Outcome 0.4x0.45=0.18 0.4x0.55=0.22 0.6x0.45=0.27 0.6x0.55=0.33

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11 Example Outcome 0.4x0.45=0.18ss 0.4x0.55=0.22sg 0.6x0.45=0.27gs 0.6x0.55=0.33gg P(stop once only) =0.22 +0.27 =0.47 P(Stop at least once) = 0.22+0.27+0.18 =0.67OR 1-0.33 = 0.67

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12 Time taken to get to college

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13 Histogram As the data is continuous there should be no gaps. Each bar should be the same width

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14 Pie Chart Working out: 6/30 x 360 =72 ° 8/30 x360 = 96 ° 14/30 x 360 = 168 ° 2/30 x 360 = 24°

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15 Mean Mean: Take the midpoint as the value of x xffx 5630 158120 2514350 35270 Mean= (30 + 120 + 350 + 70)/30 =570/30 =19 minutes NOT 570/4 = 142.5

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16 Median Use a cumulative frequency curve. USE THE UPPER VALUE x fC.F. 1066 20814 301428 40230

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17 Median and Interquartile Range Median =21 minutes Lower Quartile = 12 Upper Quartile = 26 Interquartile range = 26 - 12 =14

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Applicable Mathematics “Probability”

Applicable Mathematics “Probability”

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