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Thread: probability 52 card problem

  1. #1
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    probability 52 card problem

    A standard deck of 52 cards is shuffled, and each of 3 people are asked to select a card at
    random from the deck. Each time a card is selected, it is returned to the deck and reshuffled.
    What is the probability that
    (a) all 3 people select different cards?
    (b) exactly one of the 3 people selects a heart?
    (c) the first person selects a heart or the last person selects an ace?

    heres what im tinking... let me know if this seens logical
    a) (1/52)(1/51)(1/50)

    b)(1/52) (1/52) (1/4)

    c)
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  2. #2
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    Quote Originally Posted by zackgilbey View Post
    A standard deck of 52 cards is shuffled, and each of 3 people are asked to select a card at random from the deck. Each time a card is selected, it is returned to the deck and reshuffled. What is the probability that
    (a) all 3 people select different cards?
    (b) exactly one of the 3 people selects a heart?
    (c) the first person selects a heart or the last person selects an ace?
    Because the card is returned to the deck after each draw, the events are independent.
    This means for c) $\displaystyle P\left( {H_1 \cup A_3 } \right) = P\left( {H_1 } \right) + P\left( {A_3 } \right) - P\left( {H_1 \cap A_3 } \right) = \left( {\frac{{13}}{{52}}} \right) + \left( {\frac{4}{{52}}} \right) - \left( {\frac{{13}}{{52}}} \right)\left( {\frac{4}{{52}}} \right)$
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  3. #3
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    Hello, zackgilbey!

    A standard deck of 52 cards is shuffled,
    and each of 3 people are asked to select a card at random from the deck.
    Each time a card is selected, it is returned to the deck and reshuffled.

    What is the probability that
    (a) all 3 people select different cards?

    The first person can select any card: .$\displaystyle \frac{52}{52}$

    The second must select one of the other 51 cards: .$\displaystyle \frac{51}{52}$

    The third must select one of the other 50 cards: .$\displaystyle \frac{50}{52}$

    $\displaystyle P(\text{3 different cards}) \;=\;\frac{52}{52}\cdot\frac{51}{52}\cdot\frac{50} {52} \;=\;\frac{1275}{1352}$




    (b) exactly one of the 3 people selects a heart?

    There are 3 choices for the person who selects the Heart.
    He gets a Heart with probability: .$\displaystyle \frac{13}{52} \,=\,\frac{1}{4}$

    The other two people must select two non-hearts: .$\displaystyle \frac{39}{52}\cdot\frac{39}{52} \,=\,\frac{9}{16}$

    $\displaystyle P(\text{exactly one Heart}) \;=\;3\cdot\frac{1}{4}\cdot\frac{9}{16} \;=\;\frac{27}{64}$




    (c) the first person selects a heart or the last person selects an ace?

    The first selects a Heart: .$\displaystyle \frac{13}{52}$

    The second can select any card . . . We can ignore him!

    The third selects an Ace: .$\displaystyle \frac{4}{52}$

    We also have: .$\displaystyle P(\text{1st Heart }and\text{ 3rd Ace}) \;=\;\frac{13}{52}\cdot\frac{4}{52} \;=\;\frac{1}{52}$


    Formula: .$\displaystyle P(A \vee B) \:=\:P(A) + P(B) - P(A \wedge B)$


    We have: .$\displaystyle P(\text{1st Heart }\vee\text{ 3rd Ace}) \;=\;P(\text{1st Heart}) + P(\text{3rd Ace}) - P(\text{1st Heart }\wedge\text{ 3rd Ace})$


    Therefore: .$\displaystyle P(\text{1st Heart }\vee\text{ 3rd Ace}) \;=\;\frac{13}{52} + \frac{4}{52} - \frac{1}{52} \;=\;\frac{16}{52} \;=\;\frac{4}{13}$

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