Calculate the probability of and odds for being dealt a king or a heart
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your are dealt one card from a 52-card deck. Calculate the probability of and odds for being dealt a king or a heart
how to find this there are 4 kings and 13 hearts are there
so $P(k text{ or } h)=frac{4}{52}+frac{13}{52}-frac{1}{52}$ is im correct
probability
asked Nov 20 at 3:31
learner
957
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your are dealt one card from a 52-card deck. Calculate the probability of and odds for being dealt a king or a heart
how to find this there are 4 kings and 13 hearts are there
so $P(k text{ or } h)=frac{4}{52}+frac{13}{52}-frac{1}{52}$ is im correct
probability
asked Nov 20 at 3:31
learner
957
2
Yup, you're correct! Just an anecdote, in terms of a formula, you can use $$P(A cup B) = P(A) + P(B) - P(A cap B)$$ If you're not familiar with that notation, it's basically saying the following: $$P(text{A or B}) = P(A) + P(B) - P(text{A and B})$$
– Eevee Trainer
Nov 20 at 3:32
2
Though a footnote: the "odds" of something and "probability" of something are colloquially the same, but if you're finding both you want to use the formal definition of "odds," since technically they're different. (The probability is some % or fraction of something, where the odds are "5 to 1" or whatever - you might hear the latter during gambling, horse races in particular come to mind. What you have is the probability.) You can see this question on converting probability to odds: math.stackexchange.com/questions/1469242/…
– Eevee Trainer
Nov 20 at 3:36
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
your are dealt one card from a 52-card deck. Calculate the probability of and odds for being dealt a king or a heart
how to find this there are 4 kings and 13 hearts are there
so $P(k text{ or } h)=frac{4}{52}+frac{13}{52}-frac{1}{52}$ is im correct
probability
asked Nov 20 at 3:31
learner
957
your are dealt one card from a 52-card deck. Calculate the probability of and odds for being dealt a king or a heart
how to find this there are 4 kings and 13 hearts are there
so $P(k text{ or } h)=frac{4}{52}+frac{13}{52}-frac{1}{52}$ is im correct
probability
probability
asked Nov 20 at 3:31
learner
957
asked Nov 20 at 3:31
learner
957
asked Nov 20 at 3:31
learner
957
asked Nov 20 at 3:31
learner
957
asked Nov 20 at 3:31
learner
957
957
2
Yup, you're correct! Just an anecdote, in terms of a formula, you can use $$P(A cup B) = P(A) + P(B) - P(A cap B)$$ If you're not familiar with that notation, it's basically saying the following: $$P(text{A or B}) = P(A) + P(B) - P(text{A and B})$$
– Eevee Trainer
Nov 20 at 3:32
2
Though a footnote: the "odds" of something and "probability" of something are colloquially the same, but if you're finding both you want to use the formal definition of "odds," since technically they're different. (The probability is some % or fraction of something, where the odds are "5 to 1" or whatever - you might hear the latter during gambling, horse races in particular come to mind. What you have is the probability.) You can see this question on converting probability to odds: math.stackexchange.com/questions/1469242/…
– Eevee Trainer
Nov 20 at 3:36
add a comment |
2
Yup, you're correct! Just an anecdote, in terms of a formula, you can use $$P(A cup B) = P(A) + P(B) - P(A cap B)$$ If you're not familiar with that notation, it's basically saying the following: $$P(text{A or B}) = P(A) + P(B) - P(text{A and B})$$
– Eevee Trainer
Nov 20 at 3:32
2
Though a footnote: the "odds" of something and "probability" of something are colloquially the same, but if you're finding both you want to use the formal definition of "odds," since technically they're different. (The probability is some % or fraction of something, where the odds are "5 to 1" or whatever - you might hear the latter during gambling, horse races in particular come to mind. What you have is the probability.) You can see this question on converting probability to odds: math.stackexchange.com/questions/1469242/…
– Eevee Trainer
Nov 20 at 3:36
2
2
Yup, you're correct! Just an anecdote, in terms of a formula, you can use $$P(A cup B) = P(A) + P(B) - P(A cap B)$$ If you're not familiar with that notation, it's basically saying the following: $$P(text{A or B}) = P(A) + P(B) - P(text{A and B})$$
– Eevee Trainer
Nov 20 at 3:32
Yup, you're correct! Just an anecdote, in terms of a formula, you can use $$P(A cup B) = P(A) + P(B) - P(A cap B)$$ If you're not familiar with that notation, it's basically saying the following: $$P(text{A or B}) = P(A) + P(B) - P(text{A and B})$$
– Eevee Trainer
Nov 20 at 3:32
2
2
Though a footnote: the "odds" of something and "probability" of something are colloquially the same, but if you're finding both you want to use the formal definition of "odds," since technically they're different. (The probability is some % or fraction of something, where the odds are "5 to 1" or whatever - you might hear the latter during gambling, horse races in particular come to mind. What you have is the probability.) You can see this question on converting probability to odds: math.stackexchange.com/questions/1469242/…
– Eevee Trainer
Nov 20 at 3:36
Though a footnote: the "odds" of something and "probability" of something are colloquially the same, but if you're finding both you want to use the formal definition of "odds," since technically they're different. (The probability is some % or fraction of something, where the odds are "5 to 1" or whatever - you might hear the latter during gambling, horse races in particular come to mind. What you have is the probability.) You can see this question on converting probability to odds: math.stackexchange.com/questions/1469242/…
– Eevee Trainer
Nov 20 at 3:36
add a comment |
1 Answer
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So you are dealt another card from a 51 card deck due to the first card being already picked? I am going to answer for if you drew neither a king or heart and then king or heart, then finally if you drew a King which is also a heart.
P(K or H) [Where the initially drawn card isn't K or H) = 13/51 + 4/51 - 1/51 = 16/51
P(K or H) [Where the initially drawn card is only K) = 12/51 + 4/51 - 1/51 = 15/51
P(K or H) [Where the initially drawn card is only H) = 13/51 + 3/51 - 1/51 = 15/51
P(K or H) [Where the initially drawn card is both K and H) = 12/51 + 3/51 - 0 = 15/51
If I may have misunderstood your question, and you are asking the initial card from a 52 card deck, then your answer is absolutely correct.
answered Nov 20 at 3:54
THIS_GAMES_DOODOO
212
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
So you are dealt another card from a 51 card deck due to the first card being already picked? I am going to answer for if you drew neither a king or heart and then king or heart, then finally if you drew a King which is also a heart.
P(K or H) [Where the initially drawn card isn't K or H) = 13/51 + 4/51 - 1/51 = 16/51
P(K or H) [Where the initially drawn card is only K) = 12/51 + 4/51 - 1/51 = 15/51
P(K or H) [Where the initially drawn card is only H) = 13/51 + 3/51 - 1/51 = 15/51
P(K or H) [Where the initially drawn card is both K and H) = 12/51 + 3/51 - 0 = 15/51
If I may have misunderstood your question, and you are asking the initial card from a 52 card deck, then your answer is absolutely correct.
answered Nov 20 at 3:54
THIS_GAMES_DOODOO
212
add a comment |
up vote
0
down vote
So you are dealt another card from a 51 card deck due to the first card being already picked? I am going to answer for if you drew neither a king or heart and then king or heart, then finally if you drew a King which is also a heart.
P(K or H) [Where the initially drawn card isn't K or H) = 13/51 + 4/51 - 1/51 = 16/51
P(K or H) [Where the initially drawn card is only K) = 12/51 + 4/51 - 1/51 = 15/51
P(K or H) [Where the initially drawn card is only H) = 13/51 + 3/51 - 1/51 = 15/51
P(K or H) [Where the initially drawn card is both K and H) = 12/51 + 3/51 - 0 = 15/51
If I may have misunderstood your question, and you are asking the initial card from a 52 card deck, then your answer is absolutely correct.
answered Nov 20 at 3:54
THIS_GAMES_DOODOO
212
add a comment |
up vote
0
down vote
up vote
0
down vote
So you are dealt another card from a 51 card deck due to the first card being already picked? I am going to answer for if you drew neither a king or heart and then king or heart, then finally if you drew a King which is also a heart.
P(K or H) [Where the initially drawn card isn't K or H) = 13/51 + 4/51 - 1/51 = 16/51
P(K or H) [Where the initially drawn card is only K) = 12/51 + 4/51 - 1/51 = 15/51
P(K or H) [Where the initially drawn card is only H) = 13/51 + 3/51 - 1/51 = 15/51
P(K or H) [Where the initially drawn card is both K and H) = 12/51 + 3/51 - 0 = 15/51
If I may have misunderstood your question, and you are asking the initial card from a 52 card deck, then your answer is absolutely correct.
answered Nov 20 at 3:54
THIS_GAMES_DOODOO
212
So you are dealt another card from a 51 card deck due to the first card being already picked? I am going to answer for if you drew neither a king or heart and then king or heart, then finally if you drew a King which is also a heart.
P(K or H) [Where the initially drawn card isn't K or H) = 13/51 + 4/51 - 1/51 = 16/51
P(K or H) [Where the initially drawn card is only K) = 12/51 + 4/51 - 1/51 = 15/51
P(K or H) [Where the initially drawn card is only H) = 13/51 + 3/51 - 1/51 = 15/51
P(K or H) [Where the initially drawn card is both K and H) = 12/51 + 3/51 - 0 = 15/51
If I may have misunderstood your question, and you are asking the initial card from a 52 card deck, then your answer is absolutely correct.
answered Nov 20 at 3:54
THIS_GAMES_DOODOO
212
answered Nov 20 at 3:54
THIS_GAMES_DOODOO
212
answered Nov 20 at 3:54
THIS_GAMES_DOODOO
212
answered Nov 20 at 3:54
THIS_GAMES_DOODOO
212
212
add a comment |
add a comment |
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Yup, you're correct! Just an anecdote, in terms of a formula, you can use $$P(A cup B) = P(A) + P(B) - P(A cap B)$$ If you're not familiar with that notation, it's basically saying the following: $$P(text{A or B}) = P(A) + P(B) - P(text{A and B})$$
– Eevee Trainer
Nov 20 at 3:32
2
Though a footnote: the "odds" of something and "probability" of something are colloquially the same, but if you're finding both you want to use the formal definition of "odds," since technically they're different. (The probability is some % or fraction of something, where the odds are "5 to 1" or whatever - you might hear the latter during gambling, horse races in particular come to mind. What you have is the probability.) You can see this question on converting probability to odds: math.stackexchange.com/questions/1469242/…
– Eevee Trainer
Nov 20 at 3:36