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Thread: Deck of cards with no replacement

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    Question Deck of cards with no replacement

    Answer requires justification

    A sequence of 3 cards is drawn from a standard deck of cards (with no replacement).

    (a) How many of such sequences will have at least one king?
    (b) How many of such sequences will have at least one king or one queen (or both)?
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    Re: Deck of cards with no replacement

    Quote Originally Posted by inayat View Post
    Answer requires justification

    A sequence of 3 cards is drawn from a standard deck of cards (with no replacement).

    (a) How many of such sequences will have at least one king?
    (b) How many of such sequences will have at least one king or one queen (or both)?
    Let $\mathcal{P}^n_j$ denote the number of permutations of $n$ objects taken $j$ at a time.

    (a) $\mathcal{P}^{52}_3-\mathcal{P}^{48}_3$
    Now you job is to reply with a explanation why that answer is correct.
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    Re: Deck of cards with no replacement

    First, calculate the probability of NO king. Given that there are 4 kings in 52 cards, so 48 non-kings, the probability the first card is not a king is 48/52= 12/13, the probability the second card is not a king is 47/51, and the probability the last card is not a king is 46/50= 23/25. The probability of no king is (12/13)(47/51)(23/25). The probability of "at least one king" is 1 minus that.

    There are 52 cards in a deck and 52- 8= 44 are neither king nor queen. The probability the first card drawn is neither a king nor a queen is 44/52= 11/13. There are then 51 cards left, 43 neither king nor queen. The probability the second card is neither a king nor a queen is 43/51. There are then 50 cards left, 42 neither king nor queen. The probability the last card is neither a king nor a queen is 42/50= 21/25. The probability there is no king nor queen is (11/13)(43/51)(21/25). The probability there is at least one king or one queen is 1 minus that.
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    Re: Deck of cards with no replacement

    Quote Originally Posted by HallsofIvy View Post
    First, calculate the probability of NO king. Given that there are 4 kings in 52 cards, so 48 non-kings, the probability the first card is not a king is 48/52= 12/13, the probability the second card is not a king is 47/51, and the probability the last card is not a king is 46/50= 23/25. The probability of no king is (12/13)(47/51)(23/25). The probability of "at least one king" is 1 minus that.

    There are 52 cards in a deck and 52- 8= 44 are neither king nor queen. The probability the first card drawn is neither a king nor a queen is 44/52= 11/13. There are then 51 cards left, 43 neither king nor queen. The probability the second card is neither a king nor a queen is 43/51. There are then 50 cards left, 42 neither king nor queen. The probability the last card is neither a king nor a queen is 42/50= 21/25. The probability there is no king nor queen is (11/13)(43/51)(21/25). The probability there is at least one king or one queen is 1 minus that.
    Please note that the question asks nothing about probability.
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