IIt: Probability Using warehouse employs 21 workers on first shift, 16 workers on second shift, and 13 workers on third shift. Eight workers are chosen at random to be interviewed about the work environment. Find the \( 42.86 \%, 3 \) of 7 points hoosing exactly five first.shift workers.

Let - \( N_1 = 21 \) be the number of first‐shift workers, - \( N_2 = 16 \) be the number of second‐shift workers, - \( N_3 = 13 \) be the number of third‐shift workers. The total number of workers is \[ N = 21+16+13=50. \] When 8 workers are chosen at random, the total number of selections is \[ \binom{50}{8}. \] We wish to find the probability that exactly 5 of the selected workers are from the first shift. This means we choose: - 5 workers out of the 21 first‐shift workers, and - the remaining 3 workers out of the other \(50-21 = 29\) workers (those from the second and third shifts). Thus, the number of favorable outcomes is \[ \binom{21}{5}\binom{29}{3}. \] The required probability is therefore \[ P = \frac{\binom{21}{5}\binom{29}{3}}{\binom{50}{8}}. \] For completeness, we can compute each combination: 1. First‐shift choice: \[ \binom{21}{5} = \frac{21\times20\times19\times18\times17}{5\times4\times3\times2\times1} = 20349. \] 2. Other workers: \[ \binom{29}{3} = \frac{29\times28\times27}{3\times2\times1} = 3654. \] 3. Total selections: \[ \binom{50}{8} = 536878650. \] Thus, \[ P = \frac{20349 \times 3654}{536878650}. \] Multiplying the numerator, \[ 20349 \times 3654 = 74355246. \] So, \[ P = \frac{74355246}{536878650}\approx 0.1385. \] Hence, the probability of choosing exactly five first‐shift workers is approximately \(13.85\%\).