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