Two cards are drawn successively and without replacement from an ordin

Two cards are drawn successively and without replacement from an ordinary deck of playing cards Compute the probability of drawing a. Two hearts. b. A heart on the first draw and a club on the second draw. c. A heart on the first draw and an ace on the second draw. Hint: In part (c), note that a heart can be drawn by getting the ace of hearts or one of the other 12 hearts.

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Daniel Cormack

Step 1
PROBABILITY RULES
A standard deck of cards contains 52 cards, of which 26 are red and 26 are black, 13 are of each suit (hearts, diamonds, spades, clubs) and of which 4 are of each denomination (A, 2 to 10, J, Q, K). The face cards are the jacks J, queens Q and kings K.
General multiplication rule:
$$\displaystyle{P}{\left({A}\cap{B}\right)}={P}{\left({A}\right)}\times{P}{\left({B}{\mid}{A}\right)}={P}{\left({B}\right)}\times{P}{\left({A}{\mid}{B}\right)}$$
Step 2
SOLUTION
a) 13 of the 52 caeds are hearts are hearts
The probability is the number of favorable outcomes divided by the number of possible outcomes:
$$\displaystyle{P}{\left({H}_{{{1}}}\right)}={\frac{{\#\text{of favorable outcomes}}}{{\#\text{of possible outcomes}}}}={\frac{{{13}}}{{{52}}}}={\frac{{{1}}}{{{4}}}}$$
After one heart is selected, there are 12 hearts left among the remaining 51 cards.
$$\displaystyle{P}{\left({H}_{{{2}}}{\mid}{H}_{{{1}}}\right)}={\frac{{\#\text{of favorable outcomes}}}{{\#\text{of possible outcomes}}}}={\frac{{{12}}}{{{51}}}}={\frac{{{4}}}{{{17}}}}$$
Use the general multiplication rule:
$$\displaystyle{P}{\left({H}_{{{1}}}\cap{H}_{{{2}}}\right)}={P}{\left({H}_{{{1}}}\right)}\times{P}{\left({H}_{{{2}}}{\mid}{H}_{{{1}}}\right)}={\frac{{{1}}}{{{4}}}}\times{\frac{{{4}}}{{{17}}}}={\frac{{{1}}}{{{17}}}}\approx{0.05882}$$
Step 3
b) 13 of the 52 cards are hearts
The probability is the number of favorable outcomes divided by the number of possible outcomes:
$$\displaystyle{P}{\left({H}_{{{1}}}\right)}={\frac{{\#\text{of favorable outcomes}}}{{\#\text{of possible outcomes}}}}={\frac{{{13}}}{{{52}}}}={\frac{{{1}}}{{{4}}}}$$
After one heart is selected, there are 13 clubs left among the remaining 51 cards.
$$\displaystyle{P}{\left({C}_{{{2}}}{\mid}{H}_{{{1}}}\right)}={\frac{{\#\text{of favorable outcomes}}}{{\#\text{of possible outcomes}}}}={\frac{{{13}}}{{{51}}}}$$
Use the general multiplication rule:
$$\displaystyle{P}{\left({H}_{{{1}}}\cap{C}_{{{2}}}\right)}={P}{\left({H}_{{{1}}}\right)}\times{P}{\left({C}_{{{2}}}{\mid}{H}_{{{1}}}\right)}={\frac{{{1}}}{{{4}}}}\times{\frac{{{13}}}{{{51}}}}={\frac{{{13}}}{{{204}}}}\approx{0.06373}$$
Step 4
c) 13 of the 52 cards are hearts, while 1 of these cards is the ace of hearts
The probability is the number of favorable outcomes divided by the number of possible outcomes:
$$\displaystyle{P}{\left({H}_{{{A}}}\right)}={\frac{{\#\text{of favorable outcomes}}}{{\#\text{of possible outcomes}}}}={\frac{{{1}}}{{{52}}}}$$
$$\displaystyle{P}{\left({H}_{{\neg\ {A}}})={\frac{{\#\text{of favorable outcomes}}}{{\#\text{of possible outcomes}}}}={\frac{{{12}}}{{{52}}}}={\frac{{{3}}}{{{13}}}}\right.}$$
After the ace of hearts is selected, there are 3 aces left in the remaining 51 cards
$$\displaystyle{P}{\left({A}{\mid}{H}_{{{A}}}\right)}={\frac{{\#\text{of favorable outcomes}}}{{\#\text{of possible outcomes}}}}={\frac{{{3}}}{{{51}}}}$$
After a heart that is not the ace of hearts is selected, there are 4 aces left in the remaining 51 cards
$$\displaystyle{P}{\left({A}{\mid}{H}_{{\neg\ {A}}}\right)}={\frac{{\#\text{of favorable outcomes}}}{{\#\text{of possible outcomes}}}}={\frac{{{4}}}{{{51}}}}$$
Use the general multiplication rule:
$$\displaystyle{P}{\left({H}\cap{A}\right)}={P}{\left({H}_{{{A}}}\cap{A}\right)}+{P}{\left({H}_{{\neg\ {A}}}\cap{A}\right)}$$
$$\displaystyle={P}{\left({H}_{{{A}}}\right)}\times{P}{\left({A}{\mid}{H}_{{{A}}}\right)}+{P}{\left({H}_{{\neg\ {A}}}\right)}\times{P}{\left({A}{\mid}{H}_{{\neg\ {A}}}\right)}$$
$$\displaystyle={\frac{{{1}}}{{{52}}}}\times{\frac{{{3}}}{{{51}}}}+{\frac{{{12}}}{{{52}}}}\times{\frac{{{4}}}{{{51}}}}$$
$$\displaystyle={\frac{{{1}}}{{{52}}}}\approx{0.01923}$$

Not exactly what you’re looking for?
Laura Worden
For part (a) I have:
$$\displaystyle{A}=$$ {draw heart on 1st draw}
$$\displaystyle{B}=$$ {draw heart on 2nd draw}
$$\displaystyle{P}{\left({A}\right)}={\frac{{{13}}}{{{52}}}}$$
$$\displaystyle{P}{\left({B}{\mid}{A}\right)}={\frac{{{12}}}{{{51}}}}$$
Given: P(A intersect B) $$\displaystyle={P}{\left({A}\right)}\cdot{P}{\left({B}{\mid}{A}\right)}$$ Therefore, P(A intersect B) $$\displaystyle={\left({\frac{{{13}}}{{{52}}}}\right)}{\left({\frac{{{12}}}{{{51}}}}\right)}={\frac{{{1}}}{{{17}}}}$$
B)
$$\displaystyle{A}=$$ {draw heart on 1st}
$$\displaystyle{B}=$$ {draw club on 2nd}
$$\displaystyle{P}{\left({A}\right)}={\frac{{{13}}}{{{52}}}}$$ because there are 13 hearts out of 52 cards to choose from
$$\displaystyle{P}{\left({B}{\mid}{A}\right)}={\frac{{{13}}}{{{51}}}}$$ because there are 13 club of which there are 51 cards to choose from.
Therefore,
P(A intersect B) $$\displaystyle={P}{\left({A}\right)}\cdot{P}{\left({B}{\mid}{A}\right)}={\left({\frac{{{13}}}{{{52}}}}\right)}{\left({\frac{{{13}}}{{{51}}}}\right)}={\frac{{{13}}}{{{204}}}}$$
user_27qwe

Step 1
a) There are 13 hearts in a deck
We can draw 2 hearts in 13P2 ways.
We can draw any 2 cards from 52 cards in 52P2 ways
So the probability of getting 2 hearts $$=\frac{13P2}{52P2}=\frac{156}{2652}=\frac{1}{17}$$
Step 2
b) There are 13 hearts and 13 clubs in a deck.
We can draw the first heart in 13 ways and the second club in 13 ways.
We can draw any 2 cards from 52 cards in 52P2 ways
So the probability of getting 1 heart and then 1 club is $$=\frac{13 \times 13}{52P2}=\frac{169}{2652}=\frac{13}{204}$$
Step 3
c) There are 13 hearts and 4 aces in a deck.
And 1 of them is ace of heart.
So there are 2 possible cases:
1) We draw any card from 12 hearts that are not ace, then we draw any ace.
2) We draw the ace of hearts, then we draw an ace from the other 3 aces.
For case 1 total number of outcomes $$=(12C1)(4C1)=12(4)=48$$
For case 2 total number of outcomes $$=1 \times (3C1)=1 \times 3=3$$
So total (48+3)=51 cases.
We can draw any 2 cards from 52 cards in 52P2 ways
So the probability of getting a heart and then an ace is $$=\frac{51}{52P2}=\frac{1}{52}$$