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A die is rolled twice. What is the probability of getting a sum equal to 9?

A die is rolled twice. What is the probability of getting a sum equal to 9? Correct Answer 1/9

Total number of out comes possible when a die is rolled = 6 Hence, total number of out comes possible when a die is rolled twice, n(s) = 6 <math xmlns = "http://www.w3.org/1998/Math/MathML"><mo>&#xD7;</mo></math> 6 = 36 E = Geting a sum of 9 when the two dice fell = {(3,6), (4,5), (5,4), (6,3)} Hence n (E) = 4 P(E) = <math xmlns = "http://www.w3.org/1998/Math/MathML"><mfrac><mrow><mi>n</mi><mo>(</mo><mi>E</mi><mo>)</mo></mrow><mrow><mi>n</mi><mo>(</mo><mi>S</mi><mo>)</mo></mrow></mfrac><mo> = </mo><mo>&#xA0;</mo><mfrac><mn>4</mn><mn>36</mn></mfrac><mo>&#xA0;</mo><mo> = </mo><mo>&#xA0;</mo><mfrac><mn>1</mn><mn>9</mn></mfrac></math>

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