Two cards are drawn successively and without replacement from an ordin

Donald Johnson 2021-12-24 Answered
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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Expert Answer

Daniel Cormack
Answered 2021-12-25 Author has 1149 answers

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:
\(\displaystyle{P}{\left({H}\cap{A}\right)}={P}{\left({H}_{{{A}}}\cap{A}\right)}+{P}{\left({H}_{{\neg\ {A}}}\cap{A}\right)}\)
\(\displaystyle={P}{\left({H}_{{{A}}}\right)}\times{P}{\left({A}{\mid}{H}_{{{A}}}\right)}+{P}{\left({H}_{{\neg\ {A}}}\right)}\times{P}{\left({A}{\mid}{H}_{{\neg\ {A}}}\right)}\)
\(\displaystyle={\frac{{{1}}}{{{52}}}}\times{\frac{{{3}}}{{{51}}}}+{\frac{{{12}}}{{{52}}}}\times{\frac{{{4}}}{{{51}}}}\)
\(\displaystyle={\frac{{{1}}}{{{52}}}}\approx{0.01923}\)

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Laura Worden
Answered 2021-12-26 Author has 2457 answers
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) \(\displaystyle={P}{\left({A}\right)}\cdot{P}{\left({B}{\mid}{A}\right)}={\left({\frac{{{13}}}{{{52}}}}\right)}{\left({\frac{{{13}}}{{{51}}}}\right)}={\frac{{{13}}}{{{204}}}}\)
0
user_27qwe
Answered 2021-12-30 Author has 10046 answers

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 \(=\frac{51}{52P2}=\frac{1}{52}\)

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