10-x & gt； 0

x & lt 10 ( 1)

x-2≥0

x≥2 (2)

(1) and (2)

2≤x & lt； 10

a = { x | 2≤x & lt； 10 }

b = { x | x^2-6x+5≤0 } = { x |(x- 1)(x-5)≤0 } = { x | 1≤x≤5 }

AUB = { x | 1≤x & lt； 10}

A∩B = { x| 2≤x≤5)

P(x∈A∩B)

=(5-2)/( 10- 1)

=3/9

= 1/3

(2)

P(x？ answer

= 1-P(x∈A)

= 1- ( 10-2)/( 10- 1)

= 1- 8/9

= 1/9