smc

Question

Ant Amelia starts on the number line at 0 and crawls in the following manner. For n=1,2,3, Amelia chooses a time duration t_n and an increment x_n independently and uniformly at random from the interval (0,1). During the nth step of the process, Amelia moves x_n units in the positive direction, using up t_n minutes. If the total elapsed time has exceeded 1 minute during the nth step, she stops at the end of that step; otherwise, she continues with the next step, taking at most 3 steps in all. What is the probability that Amelia's position when she stops will be greater than 1?

Accepted Answer

Let x and y be random variables that are independently and uniformly distributed in the interval (0,1). Note that P(x+y 1)=121^21^2=12, as shown below: FIGURE Let x,y, and z be random variables that are independently and uniformly distributed in the interval (0,1). Note that P(x+y+z 1)=13(121^2)11^3=16, as shown below: FIGURE We have two cases: 1. Amelia takes exactly 2 steps. We need x_1+x_2>1 and t_1+t_2>1. So, the probability is P(x_1+x_2>1) P(t_1+t_2>1)=(1-12)(1-12)=14. 3. Amelia takes exactly 3 steps. We need x_1+x_2+x_3>1 and t_1+t_21. So, the probability is P(x_1+x_2+x_3>1) P(t_1+t_21)=(1-16)12=512. Together, the answer is 14 + 512 = 23.