Answer:
(see attachment for tree diagram)
[tex]\sf Probability\:of\:an\:event\:occurring = \dfrac{Number\:of\:ways\:it\:can\:occur}{Total\:number\:of\:possible\:outcomes}[/tex]
Part (a)
[tex]\textsf{P(Head) and P(3)}=\sf \dfrac{1}{2} \times \dfrac{1}{6}=\dfrac{1}{12}[/tex]
Part (b)
[tex]\textsf{P(Tail) and P(even)}=\sf \dfrac{1}{2} \times \dfrac{3}{6}=\dfrac{3}{12}=\dfrac{1}{4}[/tex]
Part (c)
[tex]\textsf{P(not 6)}=\sf 1-\textsf{P(6)}=1-\dfrac{1}{6}=\dfrac{5}{6}[/tex]
[tex]\implies \textsf{P(Head) and P(not 6)}=\sf \dfrac{1}{2} \times \dfrac{5}{6}=\dfrac{5}{12}[/tex]
Part (d)
As the six-sided die does not have a side labelled "7", the probability of rolling a 7 is zero.
[tex]\implies \textsf{P(Head) and P(7)}=\sf \dfrac{1}{2} \times 0=0[/tex]
