Answer:
P(that his schedule this week incldes working on friday) = 1/35
Explanation:
You work 3 evenings each week
Step1: find the total outcomes
[tex]\begin{gathered} \text{The total outcomes can be calculated by using combination} \\ ^nC_{^{}r}\text{ = }\frac{n!}{(n\text{ - r)!r!}} \\ \text{n = 7 and r = 3} \\ ^7C_3\text{ = }\frac{7!}{(7\text{ - 3)!3!}} \\ ^7C_3\text{ = }\frac{7!}{4!3!} \\ ^7C_3\text{ = }\frac{7\text{ x 6 x 5 x 4 x 3 x 2 x 1}}{4\text{ x 3 x 2 x1! 3 x 2 x 1}} \\ ^7C_3\text{ = }\frac{7\text{ x 6 x 5}}{3\text{ x 2 x1}} \\ ^7C_3\text{ = }\frac{210}{6} \\ ^7C_3\text{ = 35} \\ \text{Therefore, the total possible outcomes is 35} \end{gathered}[/tex]Probability that his schedule will include friday is
Possible outcome of including a friday is 1
Probability of including a friday = 1 / 35
Therefore, the probability of including a friday is 1/35
