Questions: In a group of seven Olympic track stars, six are hurdlers. If three track stars are selected at random without replacement, find the probability that they are all hurdlers. Express your answer as a decimal rounded to four places if necessary.

Solution Steps

We need to find the probability that all three selected track stars from a group of seven Olympic track stars are hurdlers. In this group, there are six hurdlers and one non-hurdler.

The probability of selecting \( k \) successes (hurdlers) in \( n \) draws (selected track stars) from a population of \( N \) items (total track stars) containing \( K \) successes (total hurdlers) is given by the hypergeometric distribution:

\[ P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}} \]

Where:

• \( N = 7 \) (total track stars)
• \( K = 6 \) (total hurdlers)
• \( n = 3 \) (number of selected track stars)
• \( k = 3 \) (number of hurdlers we want in the selection)

We calculate the combinations as follows:

1. Calculate \( \binom{K}{k} = \binom{6}{3} \): \[ \binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]

2. Calculate \( \binom{N-K}{n-k} = \binom{1}{0} \): \[ \binom{1}{0} = 1 \]

3. Calculate \( \binom{N}{n} = \binom{7}{3} \): \[ \binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \]

Now we substitute the values into the probability formula:

\[ P(X = 3) = \frac{\binom{6}{3} \binom{1}{0}}{\binom{7}{3}} = \frac{20 \times 1}{35} = \frac{20}{35} = \frac{4}{7} \]

Calculating the decimal value of \( \frac{4}{7} \):

\[ \frac{4}{7} \approx 0.5714 \]

Final Answer