Three integers are chosen at random from the first 20 integers. The pr

1.	Three integers are chosen at random from the first 20 integers. The probability that their product is even is A. \(\frac{2}{19}\)B. \(\frac{2}{29}\)C. \(\frac{17}{19}\)D. \(\frac{4}{19}\)
Answer» 3 integers out of 20 can be chosen in ²⁰C₃ ways. As in first 20 integers, 10 are even. To have the product of 3 integer even the chosen integers must be even. ∴ 3 integers that are even can be chosen from first 20 integers in ¹⁰C₃ ways. ∴ Probability = \(\frac{^{10}C_2}{^{20}C_2}\) = \(\frac{10\times9\times8}{20\times19\times18} = \frac{2}{19}\) Our answer matches with option (a) ∴ Option (a) is the only correct choice

Three integers are chosen at random from the first 20 integers. The probability that their product is even is A. \(\frac{2}{19}\)B. \(\frac{2}{29}\)C. \(\frac{17}{19}\)D. \(\frac{4}{19}\)

Answer»

3 integers out of 20 can be chosen in ²⁰C₃ ways.

As in first 20 integers, 10 are even.

To have the product of 3 integer even the chosen integers must be even.

∴ 3 integers that are even can be chosen from first 20 integers in ¹⁰C₃ ways.

∴ Probability = \(\frac{^{10}C_2}{^{20}C_2}\) = \(\frac{10\times9\times8}{20\times19\times18} = \frac{2}{19}\)

Our answer matches with option (a)

∴ Option (a) is the only correct choice

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Three integers are chosen at random from the first 20 integers. The probability that their product is even is A. \(\frac{2}{19}\)B. \(\frac{2}{29}\)C. \(\frac{17}{19}\)D. \(\frac{4}{19}\)

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