In a deck of cards, there are 4 aces, 4 ones, 4 twos,... 4 Jacks, 4 queens, and 4 kings; 52 cards in total.
a) There are 4 aces in total in the deck and 52 total cards; therefore,
[tex]P(A)=\frac{4}{52}=\frac{1}{13}[/tex]b)
[tex]P(A\cap B)=P(ace-and-black)=\frac{2}{52}=\frac{1}{26}[/tex]c)
[tex]\begin{gathered} P(A\cup B)=P(A)+P(B)-P(A\cap B)=\frac{4}{52}+\frac{26}{52}-\frac{2}{52}=\frac{7}{13} \\ \Rightarrow P(A\cup B)=\frac{7}{13} \end{gathered}[/tex]d)
[tex]P(H|A)=\frac{P(H\cap A)}{P(A)}[/tex]and
[tex]\begin{gathered} P(H\cap A)=\frac{1}{52} \\ \Rightarrow P(H|A)=\frac{\frac{1}{52}}{\frac{1}{13}}=\frac{13}{52}=\frac{1}{4} \\ \Rightarrow P(H|A)=\frac{1}{4} \end{gathered}[/tex]e)
[tex]\begin{gathered} P(H\cap R)=P(heart-and-red)=P(H)=\frac{13}{52}=\frac{1}{4} \\ \Rightarrow P(H|R)=\frac{P(H\cap R)}{P(R)}=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{2}{4}=\frac{1}{2} \\ \Rightarrow P(H|R)=\frac{1}{2} \end{gathered}[/tex]f)
[tex]\begin{gathered} P(A|H)=\frac{P(A\cap H)}{P(H)}=\frac{\frac{1}{52}}{\frac{1}{4}}=\frac{4}{52}=\frac{1}{13} \\ \Rightarrow P(A|H)=\frac{1}{13} \end{gathered}[/tex]The only event with those characteristics (picking a red ace first and then a heart ace) is when we draw a diamond ace first and then a heart ace. The probability of this event is
[tex]\frac{1}{52}\cdot\frac{1}{51}=\frac{1}{2652}[/tex]