Multiple choice

A bag contains 50 tickets numbered 1, 2, 3, ..., 50 of which five are drawn at random and arranged in ascending order of magnitude $(x_1 < x_2 < x_3 < x_4 < x_5)$. The probability that $x_3=30$ is

  1. $\displaystyle \frac{^{20}C_2}{^{50}C_5}$
  2. $\displaystyle \frac{^{29}C_2}{^{50}C_5}$
  3. $\displaystyle \frac{^{20}C_2\times ^{29}C_2}{^{50}C_5}$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We choose 5 numbers out of 50. For x3 = 30, we must choose 2 numbers from {1, ..., 29} and 2 numbers from {31, ..., 50}. The number of ways is (29C2) * (20C2). Total ways is 50C5.

AI explanation

The total number of ways to draw five tickets from fifty is given by the combination formula 50C5. For the third ticket x3 to be exactly 30, two tickets must be chosen from the numbers smaller than 30 (1 to 29), which can be done in 29C2 ways. The remaining two tickets must be chosen from the numbers larger than 30 (31 to 50), which can be done in 20C2 ways. The required probability is the product of these favorable choices divided by the total choices, which is (29C2 * 20C2) / 50C5.

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