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One is asked to say a two-digit number. i. What is the probability of both digits being the same? ii. What is the probability of the first digit being larger? iii. What is the probability of the first digit being smaller? |
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Answer» i. Probability of two digits being same = \(\frac{9}{90}=\frac{1}{10}\) (11, 22, 33, 44, 55, 66, 77, 88, 99) ii. Numbers in which first digit is greater than second digit are 10, 20, 21, 30, 31, 32, 40, 41, 42, 43, 50, 51, 52, 53, 54, 60, 61, 62, 63, 64, 65, 70, 71, 72, 73, 74, 75, 76, 80, 81, 82, 83, 84, 85, 86, 87, 90, 91, 92, 93, 94, 95, 96, 97, 98, 45 outcomes, Required Probability = \(\frac{45}{90}=\frac{1}{2}\) iii. Numbers in which first digit is smaller than second digit are 12, 13, 14, 15, 16, 17, 18, 19, 23, 24, 25, 26, 27, 28, 29, 34, 35, 36, 37, 38, 39, 45, 46, 47, 48, 49, 56, 57, 58, 59, 67, 68, 69, 78, 79, 89, 36 total no.of favourable outcomes = 36 Probability = \(\frac{36}{90}=\frac{18}{45}=\frac{6}{15}=\frac{2}{5}\) |
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