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A man is asked to say a 3 digit number, a. What is the probability that the first and last digits be equal?b. What is the probability that the last two digits be ‘O’?c. What is the probability that the last digits being greater than the first? |
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Answer» Total 3 digit numbers = 900 a. Numbers with first and last digits are equal 101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212, 222, 232, 242, 252, 262, 272, 282, 292, 303, 313, 323, 333, 343, 353, 363, 373, 383, 393 There are 90 such numbers. ∴ Probability = \(\frac{90}{900}=\frac{1}{10}\) b. The numbers with last two digits zero are 100, 200, 300,……… 900 total numbers 9. ∴ Probability = \(\frac{90}{900}=\frac{1}{100}\) c. The numbers with the last digit greater than the first digit is 36 ∴ probability = \(\frac{36}{90}=\frac{1}{5}\) |
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