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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

learner
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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:

    $$P(\text{A or B}) = P(A) + P(B) - P(\text{A and B})$$

     – PrincessEev Nov 20 '18 at 03:32
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    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: https://math.stackexchange.com/questions/1469242/how-can-i-convert-this-percentage-into-odds – PrincessEev Nov 20 '18 at 03:36

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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.

THIS_GAMES_DOODOO
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