Show that $P(A_1 \cup A_2 \cup A_3)=1-(1-1/3)^3$ Roll a fiar six-side die three times. Let: $A_1=${ 1 or 2 on the 1st roll} $A_2=${ 3 or 4 on the 2nd roll} $A_3=${ 5 or 6 on the 3rd roll} Given that: $P(A_i)=1/3, i=1,2,3$ $P(A_i \cap A_i)=(1/3)^2, i\neq j$ $P(A_1 \cap A_2 \cap A_3)=(1/3)^3 $ Show that $P(A_1 \cup A_2 \cup A_3)=1-(1-1/3)^3$ This is what I have done so far: $P(A_1 \cup A_2 \cup A_3)=P(A_1)+P(A_2)+P(A_3)-P(A_1 \cap A_2)- P(A_1 \cap A_3)-P(A_2 \cap A_3)+ P(A_1 \cap A_2 \cap A_3)=3(1/3)-3(1/3)^2+(1/3)^3=3(1/3)(1-1/3)+(1/3)^3=1-1/3+(1/3)^3=1-(1/3(1+(1/3)^2)$ This is where I got stuck. Any help is appreciated. Thank you!

$$3(1/3)-3(1/3)^2+(1/3)^3=$$

$$3(1/3)-3*3*(1/3)(1/3)^2+(1/3)^3=$$

$$1-9(1/3)^3+(1/3)^3=$$

$$1-8(1/3)^3=$$

$$1-2^3(1/3)^3=$$

$$1-(2/3)^3=$$

$$1-(1-1/3)^3$$