Question

Randomly select one card from a well shuffled deck of cards.Let A: the selected card is a Ace; H: the selected card is a Heart Card;B: the selected card is a Black Card; R: the selected card is a Red Card.

Answer 1

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,

`P(A)=(4)/(52)=(1)/(13)`

b)

`P(A\cap B)=P(ace-and-black)=(2)/(52)=(1)/(26)`

c)

`\begin{gathered} P(A\cup B)=P(A)+P(B)-P(A\cap B)=(4)/(52)+(26)/(52)-(2)/(52)=(7)/(13) \\ \Rightarrow P(A\cup B)=(7)/(13) \end{gathered}`

d)

`P(H|A)=(P(H\cap A))/(P(A))`

and

`\begin{gathered} P(H\cap A)=(1)/(52) \\ \Rightarrow P(H|A)=((1)/(52))/((1)/(13))=(13)/(52)=(1)/(4) \\ \Rightarrow P(H|A)=(1)/(4) \end{gathered}`

e)

`\begin{gathered} P(H\cap R)=P(heart-and-red)=P(H)=(13)/(52)=(1)/(4) \\ \Rightarrow P(H|R)=(P(H\cap R))/(P(R))=((1)/(4))/((1)/(2))=(2)/(4)=(1)/(2) \\ \Rightarrow P(H|R)=(1)/(2) \end{gathered}`

f)

`\begin{gathered} P(A|H)=(P(A\cap H))/(P(H))=((1)/(52))/((1)/(4))=(4)/(52)=(1)/(13) \\ \Rightarrow P(A|H)=(1)/(13) \end{gathered}`

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

`(1)/(52)\cdot(1)/(51)=(1)/(2652)`