91Ó°ÊÓ

We will start by evaluating the given probabilities one by one using the formula for combinations. a. \(\frac{C_{1}^{3} C_{1}^{2}}{C_{2}^{5}}\) First, we need to find the values of the combinations in the expression. $$C_{1}^{3} = \frac{3!}{1!(3-1)!} = \frac{3 \cdot 2}{1 \cdot 2} = 3$$ $$C_{1}^{2} = \frac{2!}{1!(2-1)!} = \frac{2}{1} = 2$$ $$C_{2}^{5} = \frac{5!}{2!(5-2)!} = \frac{5 \cdot 4}{2 \cdot 2} = 10$$ Now, we substitute these values back into the expression and simplify: $$\frac{C_{1}^{3} C_{1}^{2}}{C_{2}^{5}} = \frac{3 \cdot 2}{10} = \frac{6}{10} = \frac{3}{5}$$ Therefore, probability (a) is \(\frac{3}{5}\). b. \(\frac{C_{2}^{4} C_{1}^{3}}{C_{3}^{7}}\) Following the same steps as above: $$C_{2}^{4} = \frac{4!}{2!(4-2)!} = \frac{4 \cdot 3}{2} = 6$$ $$C_{1}^{3} = \frac{3!}{1!(3-1)!} = \frac{3 \cdot 2}{1 \cdot 2} = 3$$ $$C_{3}^{7} = \frac{7!}{3!(7-3)!} = \frac{7 \cdot 6 \cdot 5}{3 \cdot 2} = 35$$ Now substitute the values back into the expression and simplify: $$\frac{C_{2}^{4} C_{1}^{3}}{C_{3}^{7}} = \frac{6 \cdot 3}{35} = \frac{18}{35}$$ Therefore, probability (b) is \(\frac{18}{35}\). c. \(\frac{C_{4}^{5} C_{0}^{3}}{C_{4}^{8}}\) Following the same steps as above: $$C_{4}^{5} = \frac{5!}{4!(5-4)!} = 5$$ $$C_{0}^{3} = \frac{3!}{0!(3-0)!} = \frac{3!}{3!} = 1$$ $$C_{4}^{8} = \frac{8!}{4!(8-4)!} = \frac{8 \cdot 7 \cdot 6 \cdot 5}{4 \cdot 3 \cdot 2} = 70$$ Now substitute the values back into the expression and simplify: $$\frac{C_{4}^{5} C_{0}^{3}}{C_{4}^{8}} = \frac{5 \cdot 1}{70} = \frac{5}{70} = \frac{1}{14}$$ Therefore, probability (c) is \(\frac{1}{14}\).

Now, let's present each probability in their simplest form: a. The probability is \(\frac{3}{5}\) b. The probability is \(\frac{18}{35}\) c. The probability is \(\frac{1}{14}\)