You roll a number cube and flip a coin. Which probability does not belong with The other three

Given:
We roll a number cube and flip a coin.
The sample space is
[tex]S=\mleft\lbrace(1,T),(2,T),(3,T),(4,T),(5,T),(6,T),(1,H),(2,H),(3,H),(4,H),(5,H),(6,TH)\mright\rbrace[/tex][tex]n(S)=12[/tex]1)
Let A be the event getting the number less than 2 and heads.
[tex]A=\lbrack(1,H)\}[/tex][tex]n(A)=1[/tex]The probability is
[tex]P(A)=\frac{n(A)}{n(S)}=\frac{1}{12}[/tex]2)
Let B be the event getting the number less than 2 and tails.
[tex]B=\lbrack(1,T)\}[/tex][tex]n(B)=1[/tex]The probability is
[tex]P(B)=\frac{n(B)}{n(S)}=\frac{1}{12}[/tex]3)
Let C be the event getting the number greater than 2 and tails.
[tex]C=\mleft\lbrace(3,T),(4,T),(5,T),(6,T)\mright\rbrace[/tex][tex]n(C)=4[/tex]The probability is
[tex]P(C)=\frac{n(C)}{n(S)}=\frac{4}{12}=\frac{1}{3}[/tex]4)
Let D be the event getting the number greater than 5 and heads.
[tex]D=\mleft\lbrace(6,H\mright)\}[/tex][tex]n(D)=1[/tex]The probability is
[tex]P(D)=\frac{n(D)}{n(S)}=\frac{1}{12}[/tex]Hence we get
P(less than 2 and tails)=1/12
P(less than 2 and heads)=1/12
P(greater than 2 and tails)=1/3
P(greater than 5 and heads)=1/12
P(greater than 2 and tails) is not belonging to others.