(i) Let "in the L 1 roadway, there is at most one blocking point among the three easy blocking points" as event A.

Then p (a) = c03x (12) 3+c13x12x2 =12.

(2) According to the meaning of the question, the possible value of x is 0, 1, 2.

P(X=0)=( 1？ 34)×( 1? 35)= 1 10P(X = 1)= 34×( 1？ 35)+( 1? 34)×35=920P(X=2)=34×35=920

Therefore, the distribution list of random variable x is:

x 0 1 2P 1 10 920 920 ex = 0× 10+ 1×920+2×920 = 2720

Let L 1 the number of blocking points in the roadway be y, then the possible value of y is 0, 1, 2, 3,

p(Y = 0)= c03×( 12)3 = 18，

p(Y = 1)= c 13× 12×( 12)2 = 38，

p(Y = 2)= c23×( 12)2× 12 = 38，

p(Y = 3)= c33×( 12)3 = 18，

Therefore, the distribution list of random variable y is:

y 0 1 2 3P 18 38 38 18 ey = 0× 18+ 1×38+2×38+3× 18 = 32。

Because EX < EY, it is best to choose L2 roadway as the emergency route.