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Published byAbel Blair Modified over 8 years ago
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ROLL A PAIR OF DICE AND ADD THE NUMBERS
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Possible Outcomes: There are 6 x 6 = 36 equally likely
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A more compact sample space might list only the possible sums. These outcomes are not equiprobable and must be weighted as follows:
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{ 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 } The SAMPLE SPACE = 661121 12 65 56 41 32 23 14 63 54 45 36 31 22 13 64 55 46 61 52 43 34 25 16 51 42 33 24 15 62 53 44 35 26 1/36 2/36 6/365/36 4/36 3/36
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{ 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 } The SAMPLE SPACE = 1/36 2/36 6/365/36 4/36 3/36 T = the event the sum is a multiple of 3 T = { 3, 6, 9, 12 } 2/36 5/36 4/36 1/36 P(T) = 2/36 + 5/36 + 4/36 + 1/36 = 12/36
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{ 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 } The SAMPLE SPACE = 1/36 2/36 6/365/36 4/36 3/36 E = the event the sum is greater than 8 E = { 9, 10, 11, 12 } 4/36 3/36 2/36 1/36 P(E) = 4/36 + 3/36 + 2/36 + 1/36 = 10/36
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{ 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 } The SAMPLE SPACE = 1/36 2/36 6/365/36 4/36 3/36 T = the event the sum is a multiple of 3 T = { 3, 6, 9, 12 } P(T) = 12/36 2/36 5/36 4/36 1/36 E = the event the sum is greater than 8 E = { 9, 10, 11, 12 } P(E) = 10/36 4/36 3/36 2/36 1/36
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{ 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 } The SAMPLE SPACE = 1/36 2/36 6/365/36 4/36 3/36 T = { 3, 6, 9, 12 } P(T) = 12/36 2/36 5/36 4/36 1/36 E = { 9, 10, 11, 12 } P(E) = 10/36 4/36 3/36 2/36 1/36 T E= the event the sum is a multiple of 3 AND greater than 8 = { 9, 12 } P(T E) = 5/36 T E= the event the sum is a multiple of 3 OR greater than 8 P(T E) = P(T) + P(E) - P(T E) = 12/36 + 10/36 - 5/36 = 17/36
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{ 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 } The SAMPLE SPACE = 1/36 2/36 6/365/36 4/36 3/36 T = { 3, 6, 9, 12 } P(T) = 12/36 2/36 5/36 4/36 1/36 E = { 9, 10, 11, 12 } P(E) = 10/36 4/36 3/36 2/36 1/36 T E= the event the sum is a multiple of 3 AND greater than 8 = { 9, 12 } P(T E) = 5/36 T / E= the event the sum is a multiple of 3 IF greater than 8 P(T / E) = P(T E) = 5/36 = 5 P( E ) 10/36 10
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{ 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 } The SAMPLE SPACE = 1/36 2/36 6/365/36 4/36 3/36
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