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Probability that two drawn cards are face cards

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2

$begingroup$

Two cards are randomly drawn from the standard deck of cards. Given that the two cards drawn are of different colors(Black and Red), Find the conditional probability that both the cards are face cards.

My try:

Let $A$ be an event of drawing two face cards and $B$ be an event of drawing two cards of different colors.

We need to find $$Pleft(A/Bright)$$

By definition we have

$$Pleft(A/Bright)=fracP(A cap B)P(B)$$

$$P(Acap B)=binom122-2binom62=36$$

$$P(B)=binom522-2binom262=676$$

Hence $$Pleft(A/Bright)=frac36676=frac9169$$

But the answer is $frac18169$

Anything went wrong?

probability algebra-precalculus conditional-probability

share|cite|improve this question

asked May 5 at 13:14

Ekaveera Kumar SharmaEkaveera Kumar Sharma

5,68811533

$endgroup$

add a comment | 

2

$begingroup$

Two cards are randomly drawn from the standard deck of cards. Given that the two cards drawn are of different colors(Black and Red), Find the conditional probability that both the cards are face cards.

My try:

Let $A$ be an event of drawing two face cards and $B$ be an event of drawing two cards of different colors.

We need to find $$Pleft(A/Bright)$$

By definition we have

$$Pleft(A/Bright)=fracP(A cap B)P(B)$$

$$P(Acap B)=binom122-2binom62=36$$

$$P(B)=binom522-2binom262=676$$

Hence $$Pleft(A/Bright)=frac36676=frac9169$$

But the answer is $frac18169$

Anything went wrong?

probability algebra-precalculus conditional-probability

share|cite|improve this question

asked May 5 at 13:14

Ekaveera Kumar SharmaEkaveera Kumar Sharma

5,68811533

$endgroup$

add a comment | 

$begingroup$

Two cards are randomly drawn from the standard deck of cards. Given that the two cards drawn are of different colors(Black and Red), Find the conditional probability that both the cards are face cards.

My try:

Let $A$ be an event of drawing two face cards and $B$ be an event of drawing two cards of different colors.

We need to find $$Pleft(A/Bright)$$

By definition we have

$$Pleft(A/Bright)=fracP(A cap B)P(B)$$

$$P(Acap B)=binom122-2binom62=36$$

$$P(B)=binom522-2binom262=676$$

Hence $$Pleft(A/Bright)=frac36676=frac9169$$

But the answer is $frac18169$

Anything went wrong?

probability algebra-precalculus conditional-probability

share|cite|improve this question

asked May 5 at 13:14

Ekaveera Kumar SharmaEkaveera Kumar Sharma

5,68811533

$endgroup$

share|cite|improve this question

asked May 5 at 13:14

Ekaveera Kumar SharmaEkaveera Kumar Sharma

5,68811533

share|cite|improve this question

asked May 5 at 13:14

Ekaveera Kumar SharmaEkaveera Kumar Sharma

5,68811533

asked May 5 at 13:14

Ekaveera Kumar SharmaEkaveera Kumar Sharma

5,68811533

asked May 5 at 13:14

Ekaveera Kumar SharmaEkaveera Kumar Sharma

5,68811533

2 Answers
2

active

oldest

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3

$begingroup$

Another way to look at it is to not consider the problem as a conditional probability problem, but frame it differently. What you are really asking is, given 2 half decks of cards, one with all the red ones, one with all the black ones, what is the chance that I get 2 faces if I draw one card randomly from each of the half decks.
So this is simply:
$$
left(frac2 times 326right)^2 = frac36676
$$

I also think that your ground truth answer is wrong.

share|cite|improve this answer

edited May 5 at 13:46

answered May 5 at 13:35

Zaccharie RamziZaccharie Ramzi

1548

$endgroup$

add a comment | 

2

$begingroup$

$$P(A|B) = fracP(A cap B)P(B)$$
with $A$ and $B$ as defined in your question.

$$P(A cap B) = fracbinom61 times binom61binom522$$

$$P(B) = fracbinom261 times binom261binom522$$

Therefore,

$$P(A|B) = frac36676$$

The answer given is certainly wrong.

share|cite|improve this answer

answered May 5 at 13:25

VizagVizag

1,142313

$endgroup$

add a comment | 

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3

$begingroup$

Another way to look at it is to not consider the problem as a conditional probability problem, but frame it differently. What you are really asking is, given 2 half decks of cards, one with all the red ones, one with all the black ones, what is the chance that I get 2 faces if I draw one card randomly from each of the half decks.
So this is simply:
$$
left(frac2 times 326right)^2 = frac36676
$$

I also think that your ground truth answer is wrong.

share|cite|improve this answer

edited May 5 at 13:46

answered May 5 at 13:35

Zaccharie RamziZaccharie Ramzi

1548

$endgroup$

add a comment | 

3

$begingroup$

Another way to look at it is to not consider the problem as a conditional probability problem, but frame it differently. What you are really asking is, given 2 half decks of cards, one with all the red ones, one with all the black ones, what is the chance that I get 2 faces if I draw one card randomly from each of the half decks.
So this is simply:
$$
left(frac2 times 326right)^2 = frac36676
$$

I also think that your ground truth answer is wrong.

share|cite|improve this answer

edited May 5 at 13:46

answered May 5 at 13:35

Zaccharie RamziZaccharie Ramzi

1548

$endgroup$

add a comment | 

$begingroup$

Another way to look at it is to not consider the problem as a conditional probability problem, but frame it differently. What you are really asking is, given 2 half decks of cards, one with all the red ones, one with all the black ones, what is the chance that I get 2 faces if I draw one card randomly from each of the half decks.
So this is simply:
$$
left(frac2 times 326right)^2 = frac36676
$$

I also think that your ground truth answer is wrong.

share|cite|improve this answer

edited May 5 at 13:46

answered May 5 at 13:35

Zaccharie RamziZaccharie Ramzi

1548

$endgroup$

share|cite|improve this answer

edited May 5 at 13:46

answered May 5 at 13:35

Zaccharie RamziZaccharie Ramzi

1548

answered May 5 at 13:35

Zaccharie RamziZaccharie Ramzi

1548

answered May 5 at 13:35

Zaccharie RamziZaccharie Ramzi

1548

2

$begingroup$

$$P(A|B) = fracP(A cap B)P(B)$$
with $A$ and $B$ as defined in your question.

$$P(A cap B) = fracbinom61 times binom61binom522$$

$$P(B) = fracbinom261 times binom261binom522$$

Therefore,

$$P(A|B) = frac36676$$

The answer given is certainly wrong.

share|cite|improve this answer

answered May 5 at 13:25

VizagVizag

1,142313

$endgroup$

add a comment | 

2

$begingroup$

$$P(A|B) = fracP(A cap B)P(B)$$
with $A$ and $B$ as defined in your question.

$$P(A cap B) = fracbinom61 times binom61binom522$$

$$P(B) = fracbinom261 times binom261binom522$$

Therefore,

$$P(A|B) = frac36676$$

The answer given is certainly wrong.

share|cite|improve this answer

answered May 5 at 13:25

VizagVizag

1,142313

$endgroup$

add a comment | 

$begingroup$

$$P(A|B) = fracP(A cap B)P(B)$$
with $A$ and $B$ as defined in your question.

$$P(A cap B) = fracbinom61 times binom61binom522$$

$$P(B) = fracbinom261 times binom261binom522$$

Therefore,

$$P(A|B) = frac36676$$

The answer given is certainly wrong.

share|cite|improve this answer

answered May 5 at 13:25

VizagVizag

1,142313

$endgroup$

share|cite|improve this answer

answered May 5 at 13:25

VizagVizag

1,142313

answered May 5 at 13:25

VizagVizag

1,142313

answered May 5 at 13:25

VizagVizag

1,142313