Mixing
Conditional Probability on Cards of the Same Suit
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A normal deck of 52 playing cards is well shuffled and then 4 cards are dealt to Ann and 4 cards ar dealt to Bob. Ann looks at her c that all 4 of them are from the same suit, that suit being hearts. She is interested in the probability that Bob also has 4 cards that belong to a single suit, allowing for her knowledge of the cards she holds.
Please correct my answer if it is incorrect:
$$text{Probability}=P(text{Bob has all hearts} |text{Ann has all hearts})+P(text{Bob has all clubs/spades/diamonds}|text{Ann has all hearts})$$
$$=frac{binom{9}{4}}{binom{48}{4}}+frac{binom{3}{1}timesbinom{13}{4}}{binom{48}{4}}$$
$$=frac{757}{64860}$$
I am aware that this is a conditional probability, but I was unsure how to apply Bayes' Theorem in this scenario.
probability combinations
asked 21 mins ago
An Invisible Carrot
698
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A normal deck of 52 playing cards is well shuffled and then 4 cards are dealt to Ann and 4 cards ar dealt to Bob. Ann looks at her c that all 4 of them are from the same suit, that suit being hearts. She is interested in the probability that Bob also has 4 cards that belong to a single suit, allowing for her knowledge of the cards she holds.
Please correct my answer if it is incorrect:
$$text{Probability}=P(text{Bob has all hearts} |text{Ann has all hearts})+P(text{Bob has all clubs/spades/diamonds}|text{Ann has all hearts})$$
$$=frac{binom{9}{4}}{binom{48}{4}}+frac{binom{3}{1}timesbinom{13}{4}}{binom{48}{4}}$$
$$=frac{757}{64860}$$
I am aware that this is a conditional probability, but I was unsure how to apply Bayes' Theorem in this scenario.
probability combinations
asked 21 mins ago
An Invisible Carrot
698
correct except $^{48}C_4$ in denominator
– idea
6 mins ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
A normal deck of 52 playing cards is well shuffled and then 4 cards are dealt to Ann and 4 cards ar dealt to Bob. Ann looks at her c that all 4 of them are from the same suit, that suit being hearts. She is interested in the probability that Bob also has 4 cards that belong to a single suit, allowing for her knowledge of the cards she holds.
Please correct my answer if it is incorrect:
$$text{Probability}=P(text{Bob has all hearts} |text{Ann has all hearts})+P(text{Bob has all clubs/spades/diamonds}|text{Ann has all hearts})$$
$$=frac{binom{9}{4}}{binom{48}{4}}+frac{binom{3}{1}timesbinom{13}{4}}{binom{48}{4}}$$
$$=frac{757}{64860}$$
I am aware that this is a conditional probability, but I was unsure how to apply Bayes' Theorem in this scenario.
probability combinations
asked 21 mins ago
An Invisible Carrot
698
A normal deck of 52 playing cards is well shuffled and then 4 cards are dealt to Ann and 4 cards ar dealt to Bob. Ann looks at her c that all 4 of them are from the same suit, that suit being hearts. She is interested in the probability that Bob also has 4 cards that belong to a single suit, allowing for her knowledge of the cards she holds.
Please correct my answer if it is incorrect:
$$text{Probability}=P(text{Bob has all hearts} |text{Ann has all hearts})+P(text{Bob has all clubs/spades/diamonds}|text{Ann has all hearts})$$
$$=frac{binom{9}{4}}{binom{48}{4}}+frac{binom{3}{1}timesbinom{13}{4}}{binom{48}{4}}$$
$$=frac{757}{64860}$$
I am aware that this is a conditional probability, but I was unsure how to apply Bayes' Theorem in this scenario.
probability combinations
probability combinations
asked 21 mins ago
An Invisible Carrot
698
asked 21 mins ago
An Invisible Carrot
698
asked 21 mins ago
An Invisible Carrot
698
asked 21 mins ago
An Invisible Carrot
698
asked 21 mins ago
An Invisible Carrot
698
698
correct except $^{48}C_4$ in denominator
– idea
6 mins ago
add a comment |
correct except $^{48}C_4$ in denominator
– idea
6 mins ago
correct except $^{48}C_4$ in denominator
– idea
6 mins ago
correct except $^{48}C_4$ in denominator
– idea
6 mins ago
add a comment |
1 Answer
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$P(A)$ = P(Anne has all hearts) =$$frac{^{13}C_4}{^{52}C_4}$$
$P(B)$ = P(Bod has same suite) =$$frac{^{13}C_4cdot 4}{^{52}C_4}$$
$$P(B/A)=frac{P( A ∩ B)}{P(A)}$$
Using:
P(A ∩ B)=$$frac{^{13}C_4 cdot 3+ ^{9}C_4}{^{52}C_4}$$
You get: $$P(B/A)=frac {^{13}C_4 cdot 3+ ^{9}C_4}{^{13}C_4}$$
answered 7 mins ago
idea
2,0112923
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
$P(A)$ = P(Anne has all hearts) =$$frac{^{13}C_4}{^{52}C_4}$$
$P(B)$ = P(Bod has same suite) =$$frac{^{13}C_4cdot 4}{^{52}C_4}$$
$$P(B/A)=frac{P( A ∩ B)}{P(A)}$$
Using:
P(A ∩ B)=$$frac{^{13}C_4 cdot 3+ ^{9}C_4}{^{52}C_4}$$
You get: $$P(B/A)=frac {^{13}C_4 cdot 3+ ^{9}C_4}{^{13}C_4}$$
answered 7 mins ago
idea
2,0112923
add a comment |
up vote
0
down vote
$P(A)$ = P(Anne has all hearts) =$$frac{^{13}C_4}{^{52}C_4}$$
$P(B)$ = P(Bod has same suite) =$$frac{^{13}C_4cdot 4}{^{52}C_4}$$
$$P(B/A)=frac{P( A ∩ B)}{P(A)}$$
Using:
P(A ∩ B)=$$frac{^{13}C_4 cdot 3+ ^{9}C_4}{^{52}C_4}$$
You get: $$P(B/A)=frac {^{13}C_4 cdot 3+ ^{9}C_4}{^{13}C_4}$$
answered 7 mins ago
idea
2,0112923
add a comment |
up vote
0
down vote
up vote
0
down vote
$P(A)$ = P(Anne has all hearts) =$$frac{^{13}C_4}{^{52}C_4}$$
$P(B)$ = P(Bod has same suite) =$$frac{^{13}C_4cdot 4}{^{52}C_4}$$
$$P(B/A)=frac{P( A ∩ B)}{P(A)}$$
Using:
P(A ∩ B)=$$frac{^{13}C_4 cdot 3+ ^{9}C_4}{^{52}C_4}$$
You get: $$P(B/A)=frac {^{13}C_4 cdot 3+ ^{9}C_4}{^{13}C_4}$$
answered 7 mins ago
idea
2,0112923
$P(A)$ = P(Anne has all hearts) =$$frac{^{13}C_4}{^{52}C_4}$$
$P(B)$ = P(Bod has same suite) =$$frac{^{13}C_4cdot 4}{^{52}C_4}$$
$$P(B/A)=frac{P( A ∩ B)}{P(A)}$$
Using:
P(A ∩ B)=$$frac{^{13}C_4 cdot 3+ ^{9}C_4}{^{52}C_4}$$
You get: $$P(B/A)=frac {^{13}C_4 cdot 3+ ^{9}C_4}{^{13}C_4}$$
answered 7 mins ago
idea
2,0112923
answered 7 mins ago
idea
2,0112923
answered 7 mins ago
idea
2,0112923
answered 7 mins ago
idea
2,0112923
2,0112923
add a comment |
add a comment |
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correct except $^{48}C_4$ in denominator
– idea
6 mins ago