Answer: 0.5
Step-by-step explanation:
Binomial distribution
[tex]P(x)=^nC_xp^x(1-p)^{n-x}[/tex], where P(x) is the probability of getting success inn x trials , n is the number of trails and p is the probability of getting success in each trial.
Given : Sarah times her morning commute such that there is an equal likelihood that she will arrive early or late to work on any given day.
Then , the probability that Sarah is late on particular day = [tex]\dfrac{1}{2}=0.50[/tex]
Then , the probability that Sarah will arrive late to work no more than twice during a five day workweek :-
[tex]P(x\leq2)=P(0)+P(1)+P(2)\\\\^5C_0(0.5)^0(0.5)^{5}+^5C_1(0.5)^1(0.5)^{4}+^5C_2(0.5)^2(0.5)^{3}\\\\=(0.5)^5(1+5+\dfrac{5!}{2!3!})=(0.5)^5(16)=0.5[/tex]
Hence, the required probability : 0.5