Multiple choice

A bag contains $12$ tickets numbered $1,2,3,...,12$.Five of them are drawn at random and arranged in the ascending order of their numbers . What is the probability that the third in the order is $5$?

  1. $\displaystyle \frac{7}{44}$
  2. $\displaystyle \frac{5}{12}$
  3. $\displaystyle \frac{2}{3}$
  4. $\displaystyle \frac{1}{2}$
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A Correct answer
Explanation

The total number of ways to choose 5 tickets out of 12 is 12C5 = 792. For the third ticket to be 5, we must choose 2 tickets from the 4 smaller than 5 (1, 2, 3, 4) and 2 tickets from the 7 larger than 5 (6, 7, 8, 9, 10, 11, 12). The number of ways is 4C2 * 7C2 = 6 * 21 = 126. The probability is 126/792 = 7/44.

AI explanation

For the third card in ascending order to be 5, exactly two cards must be chosen from the numbers less than 5 (1, 2, 3, 4) and exactly two cards must be chosen from the numbers greater than 5 (6, 7, 8, 9, 10, 11, 12). The number of favorable ways is 4C2 * 7C2, which equals 6 * 21 = 126. The total number of ways to draw 5 cards from 12 is 12C5 = 792, making the probability 126/792, which simplifies to 7/44.

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