Conditional probability with choosing two cards from a deck
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Two cards are picked randomly from a deck. What is the probability that if the first card is Spades, the second card will be a Queen?
What I have so far:
Let $A$ be the event the first card is Spades.
Let $B$ be the event the second card is a Queen
$$P(B|A) = frac{P(Acap B)}{P(A)}$$
$$P(B|A) = frac{P(A)P(B|A)}{P(A)}$$
$$P(B|A) = frac{frac{13}{52}P(B|A)}{frac{13}{52}}$$
But now i get confused for $P(B|A)$ since its the problem im trying to solve in the first place... I know that $A$ and $B$ are dependent events since if a spade is drawn it could be the Queen spade.
So we we remove that possibility, which would make the probability of drawing the second queen be $3/51$. But, if that $P(B|A) = 3/51$ wouldnt that be the solution to the question in the first place? What happend to the $13/52$ parts?Also, if the first card is not a queen then shouldn't we consider that case as well?
statistics conditional-probability
edited Jan 10 at 0:05
David G. Stork
11k41432
asked Jan 9 at 23:57
shockwave88shockwave88
122
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add a comment |
$begingroup$
Two cards are picked randomly from a deck. What is the probability that if the first card is Spades, the second card will be a Queen?
What I have so far:
Let $A$ be the event the first card is Spades.
Let $B$ be the event the second card is a Queen
$$P(B|A) = frac{P(Acap B)}{P(A)}$$
$$P(B|A) = frac{P(A)P(B|A)}{P(A)}$$
$$P(B|A) = frac{frac{13}{52}P(B|A)}{frac{13}{52}}$$
But now i get confused for $P(B|A)$ since its the problem im trying to solve in the first place... I know that $A$ and $B$ are dependent events since if a spade is drawn it could be the Queen spade.
So we we remove that possibility, which would make the probability of drawing the second queen be $3/51$. But, if that $P(B|A) = 3/51$ wouldnt that be the solution to the question in the first place? What happend to the $13/52$ parts?Also, if the first card is not a queen then shouldn't we consider that case as well?
statistics conditional-probability
edited Jan 10 at 0:05
David G. Stork
11k41432
asked Jan 9 at 23:57
shockwave88shockwave88
122
$endgroup$
add a comment |
$begingroup$
Two cards are picked randomly from a deck. What is the probability that if the first card is Spades, the second card will be a Queen?
What I have so far:
Let $A$ be the event the first card is Spades.
Let $B$ be the event the second card is a Queen
$$P(B|A) = frac{P(Acap B)}{P(A)}$$
$$P(B|A) = frac{P(A)P(B|A)}{P(A)}$$
$$P(B|A) = frac{frac{13}{52}P(B|A)}{frac{13}{52}}$$
But now i get confused for $P(B|A)$ since its the problem im trying to solve in the first place... I know that $A$ and $B$ are dependent events since if a spade is drawn it could be the Queen spade.
So we we remove that possibility, which would make the probability of drawing the second queen be $3/51$. But, if that $P(B|A) = 3/51$ wouldnt that be the solution to the question in the first place? What happend to the $13/52$ parts?Also, if the first card is not a queen then shouldn't we consider that case as well?
statistics conditional-probability
edited Jan 10 at 0:05
David G. Stork
11k41432
asked Jan 9 at 23:57
shockwave88shockwave88
122
$endgroup$
Two cards are picked randomly from a deck. What is the probability that if the first card is Spades, the second card will be a Queen?
What I have so far:
Let $A$ be the event the first card is Spades.
Let $B$ be the event the second card is a Queen
$$P(B|A) = frac{P(Acap B)}{P(A)}$$
$$P(B|A) = frac{P(A)P(B|A)}{P(A)}$$
$$P(B|A) = frac{frac{13}{52}P(B|A)}{frac{13}{52}}$$
But now i get confused for $P(B|A)$ since its the problem im trying to solve in the first place... I know that $A$ and $B$ are dependent events since if a spade is drawn it could be the Queen spade.
So we we remove that possibility, which would make the probability of drawing the second queen be $3/51$. But, if that $P(B|A) = 3/51$ wouldnt that be the solution to the question in the first place? What happend to the $13/52$ parts?Also, if the first card is not a queen then shouldn't we consider that case as well?
statistics conditional-probability
statistics conditional-probability
edited Jan 10 at 0:05
David G. Stork
11k41432
asked Jan 9 at 23:57
shockwave88shockwave88
122
edited Jan 10 at 0:05
David G. Stork
11k41432
asked Jan 9 at 23:57
shockwave88shockwave88
122
edited Jan 10 at 0:05
David G. Stork
11k41432
edited Jan 10 at 0:05
David G. Stork
11k41432
edited Jan 10 at 0:05
David G. Stork
11k41432
11k41432
asked Jan 9 at 23:57
shockwave88shockwave88
122
asked Jan 9 at 23:57
shockwave88shockwave88
122
asked Jan 9 at 23:57
shockwave88shockwave88
122
122
add a comment |
add a comment |
1 Answer
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$begingroup$
Given the first card is a spade, there are two relevant cases: either the first card is the queen of spades (probability ${1 over 13}$), or it is a different spade (probability ${12 over 13}$).
In the first case, the probability the second card is a queen is ${3 over 51}$. In the second case, the probability the second card is a queen is ${4 over 51}$. Hence:
$$P({rm queen}|{rm spade}) = {1 over 13}{3 over 51}+ {12 over 13}{4 over 51} = {1 over 13}$$
answered Jan 10 at 0:10
David G. StorkDavid G. Stork
11k41432
$endgroup$
$begingroup$
Intuitively your answer makes perfect sense to me. However, in terms of the defined probability equation, i dont understand why it isnt $frac{frac{1}{13}frac{3}{51}}{frac{1}{13}}frac{frac{12}{13}frac{4}{51}}{frac{12}{13}} $
$endgroup$
– shockwave88
Jan 10 at 1:08
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@shockwave88: Why it isn't what?
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– David G. Stork
Jan 10 at 1:11
$begingroup$
I updated the answer, if u dont mind answering it. I think its that way because of the denominator in the conditional probabilty equation.
$endgroup$
– shockwave88
Jan 10 at 1:28
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
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active
oldest
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active
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votes
$begingroup$
Given the first card is a spade, there are two relevant cases: either the first card is the queen of spades (probability ${1 over 13}$), or it is a different spade (probability ${12 over 13}$).
In the first case, the probability the second card is a queen is ${3 over 51}$. In the second case, the probability the second card is a queen is ${4 over 51}$. Hence:
$$P({rm queen}|{rm spade}) = {1 over 13}{3 over 51}+ {12 over 13}{4 over 51} = {1 over 13}$$
answered Jan 10 at 0:10
David G. StorkDavid G. Stork
11k41432
$endgroup$
$begingroup$
Intuitively your answer makes perfect sense to me. However, in terms of the defined probability equation, i dont understand why it isnt $frac{frac{1}{13}frac{3}{51}}{frac{1}{13}}frac{frac{12}{13}frac{4}{51}}{frac{12}{13}} $
$endgroup$
– shockwave88
Jan 10 at 1:08
$begingroup$
@shockwave88: Why it isn't what?
$endgroup$
– David G. Stork
Jan 10 at 1:11
$begingroup$
I updated the answer, if u dont mind answering it. I think its that way because of the denominator in the conditional probabilty equation.
$endgroup$
– shockwave88
Jan 10 at 1:28
add a comment |
$begingroup$
Given the first card is a spade, there are two relevant cases: either the first card is the queen of spades (probability ${1 over 13}$), or it is a different spade (probability ${12 over 13}$).
In the first case, the probability the second card is a queen is ${3 over 51}$. In the second case, the probability the second card is a queen is ${4 over 51}$. Hence:
$$P({rm queen}|{rm spade}) = {1 over 13}{3 over 51}+ {12 over 13}{4 over 51} = {1 over 13}$$
answered Jan 10 at 0:10
David G. StorkDavid G. Stork
11k41432
$endgroup$
$begingroup$
Intuitively your answer makes perfect sense to me. However, in terms of the defined probability equation, i dont understand why it isnt $frac{frac{1}{13}frac{3}{51}}{frac{1}{13}}frac{frac{12}{13}frac{4}{51}}{frac{12}{13}} $
$endgroup$
– shockwave88
Jan 10 at 1:08
$begingroup$
@shockwave88: Why it isn't what?
$endgroup$
– David G. Stork
Jan 10 at 1:11
$begingroup$
I updated the answer, if u dont mind answering it. I think its that way because of the denominator in the conditional probabilty equation.
$endgroup$
– shockwave88
Jan 10 at 1:28
add a comment |
$begingroup$
Given the first card is a spade, there are two relevant cases: either the first card is the queen of spades (probability ${1 over 13}$), or it is a different spade (probability ${12 over 13}$).
In the first case, the probability the second card is a queen is ${3 over 51}$. In the second case, the probability the second card is a queen is ${4 over 51}$. Hence:
$$P({rm queen}|{rm spade}) = {1 over 13}{3 over 51}+ {12 over 13}{4 over 51} = {1 over 13}$$
answered Jan 10 at 0:10
David G. StorkDavid G. Stork
11k41432
$endgroup$
Given the first card is a spade, there are two relevant cases: either the first card is the queen of spades (probability ${1 over 13}$), or it is a different spade (probability ${12 over 13}$).
In the first case, the probability the second card is a queen is ${3 over 51}$. In the second case, the probability the second card is a queen is ${4 over 51}$. Hence:
$$P({rm queen}|{rm spade}) = {1 over 13}{3 over 51}+ {12 over 13}{4 over 51} = {1 over 13}$$
answered Jan 10 at 0:10
David G. StorkDavid G. Stork
11k41432
answered Jan 10 at 0:10
David G. StorkDavid G. Stork
11k41432
answered Jan 10 at 0:10
David G. StorkDavid G. Stork
11k41432
answered Jan 10 at 0:10
David G. StorkDavid G. Stork
11k41432
11k41432
$begingroup$
Intuitively your answer makes perfect sense to me. However, in terms of the defined probability equation, i dont understand why it isnt $frac{frac{1}{13}frac{3}{51}}{frac{1}{13}}frac{frac{12}{13}frac{4}{51}}{frac{12}{13}} $
$endgroup$
– shockwave88
Jan 10 at 1:08
$begingroup$
@shockwave88: Why it isn't what?
$endgroup$
– David G. Stork
Jan 10 at 1:11
$begingroup$
I updated the answer, if u dont mind answering it. I think its that way because of the denominator in the conditional probabilty equation.
$endgroup$
– shockwave88
Jan 10 at 1:28
add a comment |
$begingroup$
Intuitively your answer makes perfect sense to me. However, in terms of the defined probability equation, i dont understand why it isnt $frac{frac{1}{13}frac{3}{51}}{frac{1}{13}}frac{frac{12}{13}frac{4}{51}}{frac{12}{13}} $
$endgroup$
– shockwave88
Jan 10 at 1:08
$begingroup$
@shockwave88: Why it isn't what?
$endgroup$
– David G. Stork
Jan 10 at 1:11
$begingroup$
I updated the answer, if u dont mind answering it. I think its that way because of the denominator in the conditional probabilty equation.
$endgroup$
– shockwave88
Jan 10 at 1:28
$begingroup$
Intuitively your answer makes perfect sense to me. However, in terms of the defined probability equation, i dont understand why it isnt $frac{frac{1}{13}frac{3}{51}}{frac{1}{13}}frac{frac{12}{13}frac{4}{51}}{frac{12}{13}} $
$endgroup$
– shockwave88
Jan 10 at 1:08
$begingroup$
Intuitively your answer makes perfect sense to me. However, in terms of the defined probability equation, i dont understand why it isnt $frac{frac{1}{13}frac{3}{51}}{frac{1}{13}}frac{frac{12}{13}frac{4}{51}}{frac{12}{13}} $
$endgroup$
– shockwave88
Jan 10 at 1:08
$begingroup$
@shockwave88: Why it isn't what?
$endgroup$
– David G. Stork
Jan 10 at 1:11
$begingroup$
@shockwave88: Why it isn't what?
$endgroup$
– David G. Stork
Jan 10 at 1:11
$begingroup$
I updated the answer, if u dont mind answering it. I think its that way because of the denominator in the conditional probabilty equation.
$endgroup$
– shockwave88
Jan 10 at 1:28
$begingroup$
I updated the answer, if u dont mind answering it. I think its that way because of the denominator in the conditional probabilty equation.
$endgroup$
– shockwave88
Jan 10 at 1:28
add a comment |
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