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If a two digit number is chosen at random, then the probability that the number chosen is a multiple of 3, isA. \(\frac{3}{10}\)B. \(\frac{29}{100}\)C. \(\frac{1}{3}\)D.\(\frac{7}{25}\) |
Answer» Total numbers of elementary event are: 90 Let E be the event of choosing a multiple of 3 Favorable outcomes are: 3 × 4 =12 , 3 × 5 = 15, 3 × 6 = 18, 3 × 7= 21,….. 3 × 33 = 99 Numbers of favorable outcomes are = 30 P( multiple of 3) = P (E) = \(\frac{30}{90}\) = \(\frac{1}{3}\) |
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