Probability problem I can not solve. Please help.

There are 3 very tardy students who are usually late for class. The first student has 0.20 (20%) chance of making it on time, the second student has 0.15 (15%) chance of making it on time, and the third student has 0.09 (9%) chance of making it on time for class.

a) What is the probability of ANY one of them making it on time for class?

b) what is the probability of ALL of them making it on time for class?

a) P(ONLY 1) = P(A)P(~B)P(~C) + P(~A)P(B)P(~C) + P(~A)P(~B)P(C)
P(ONLY 1) = (0.2)(1 - 0.15)(1 - 0.09) + (1 - 0.2)(0.15)(1 - 0.09) + (1 - 0.2)(1 - 0.15)(0.09)
P(ONLY 1) ≈ 0.3251

b) P(ALL 3) = P(A)P(B)P(C)
P(ALL 3) = (0.2)(0.15)(0.09)
P(ALL 3) = 0.0027

a.)one of them making on the time
x=0.2
y=0.15
z=0.09
making x on the time=0.2*(1-0.15)*(1-0.09)=0.1547
making y on the time=(1-0.2)*(0.15)*(1-0.09)=0.1092
making z on the time=(1-0.2)*(1-0.15)*(0.09)=0.0612
therefore the probability of ANY one of them making it on time for class=0.1547+0.1092+0.0612=0.3251

b.) the probability of ALL of them making it on time for class=0.2*0.15*0.09=0.0027