Multiple choice

A bag contains $17$ tickets numbered from $1$ to $17$. A ticket is drawn at random, then another ticket is drawn without replacing the first one. The probability that both the tickets may show even numbers is?

  1. $\dfrac{7}{34}$
  2. $\dfrac{8}{17}$
  3. $\dfrac{7}{16}$
  4. $\dfrac{7}{17}$
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A Correct answer
Explanation

Total tickets = 17. Even numbers are {2, 4, 6, 8, 10, 12, 14, 16} (8 total). Probability of first even = 8/17. Probability of second even = 7/16. Total probability = (8/17) * (7/16) = 56 / 272 = 7/34.

AI explanation

From the numbers 1 to 17, there are 8 even numbers (2, 4, 6, 8, 10, 12, 14, 16). Using the multiplication rule for dependent events, the probability is (8/17) multiplied by (7/16). This gives 56/272, which simplifies by dividing the numerator and denominator by 8. The resulting probability is 7/34.

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