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Math Help - Answer Check - Probability

  1. #1
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    Answer Check - Probability

    Q. Two cards are drawn from a deck of 52 playing cards. Find the probability of each of the following events occurring:

    a) Both cards are clubs (3 marks)

    N(S) = (52,2)
    = 1326

    C (13,2) = 78
    P(E) = 78/1326
    =39/663
    =13/221
    Therefore the probability that both cards are clubs is 13/221

    b) Both cards are red (3 marks)

    (36, 2) = 630
    P (E) = 630/1326
    =315/663
    =105/221
    Therefore, the probability that both are red is 105/221.

    c) Both cards are queens (3 marks)

    (4,2) = 6
    P(E) = 6/1326
    = 1/221
    Therefore, the probability that both cards are queens is 1/221

    d) Both cards are red queens (3 marks)

    (2,2) = 1
    P(E)= 1/1326
    Therefore, the probability that both are red queens is 1/221.

    e) Both cards are queens or both cards are red.

    (32,2) + (4,2)
    = 78 + 6
    = 84
    P(E) = 84/1326
    = 42/663
    = 14/221
    Therefore the probability that both cards are queens or both are red is 14/221.

    Please check and let me know if I am correct. THANK YOU
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  2. #2
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    Quote Originally Posted by cnmath16 View Post
    Q. Two cards are drawn from a deck of 52 playing cards. Find the probability of each of the following events occurring:

    a) Both cards are clubs (3 marks)

    N(S) = (52,2)
    = 1326

    C (13,2) = 78
    P(E) = 78/1326
    =39/663
    =13/221
    Therefore the probability that both cards are clubs is 13/221

    You have a 1 in 4 chance of picking a club to start with. Once you have picked a club, then there are 51 cards left in the pack, 12 of which are clubs. So you have a 12/51 chance. The odds of both occuring is  \frac{1}{4} \times \frac{12}{51} .

    b) Both cards are red (3 marks)

    (36, 2) = 630
    P (E) = 630/1326
    =315/663
    =105/221
    Therefore, the probability that both are red is 105/221.

    1/2 chance of picking a red card. Then there will be 25 red cards in a pack if 51. So chance of picking another one is 25/51. Total probability of both occuring:  \frac{1}{2} \times \frac{25}{51}


    c) Both cards are queens (3 marks)

    (4,2) = 6
    P(E) = 6/1326
    = 1/221
    Therefore, the probability that both cards are queens is 1/221

    4 Queens. So 4/52 chance at first. Then 3/51. Total probability \frac{4}{52} \times \frac{3}{51}. You are correct here.

    [b]
    d) Both cards are red queens (3 marks)

    (2,2) = 1
    P(E)= 1/1326
    Therefore, the probability that both are red queens is 1/221.

    2 red queens. Chance of the first being a red queen is 2/52. Chance of 2nd being a red queen is 1/51. Total probability 2/52 x 1/51

    e) Both cards are queens or both cards are red.

    (32,2) + (4,2)
    = 78 + 6
    = 84
    P(E) = 84/1326
    = 42/663
    = 14/221
    Therefore the probability that both cards are queens or both are red is 14/221.

    Just add b) and c)

    Please check and let me know if I am correct. THANK YOU
    Yes.
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  3. #3
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    Quote Originally Posted by cnmath16 View Post
    Q. Two cards are drawn from a deck of 52 playing cards. Find the probability of each of the following events occurring:

    a) Both cards are clubs (3 marks)

    N(S) = (52,2)
    = 1326

    C (13,2) = 78
    P(E) = 78/1326
    =39/663
    =13/221
    Therefore the probability that both cards are clubs is 13/221

    b) Both cards are red (3 marks)

    (36, 2) = 630
    P (E) = 630/1326
    =315/663
    =105/221
    Therefore, the probability that both are red is 105/221.

    c) Both cards are queens (3 marks)

    (4,2) = 6
    P(E) = 6/1326
    = 1/221
    Therefore, the probability that both cards are queens is 1/221

    d) Both cards are red queens (3 marks)

    (2,2) = 1
    P(E)= 1/1326
    Therefore, the probability that both are red queens is 1/221.

    e) Both cards are queens or both cards are red.

    (32,2) + (4,2)
    = 78 + 6
    = 84
    P(E) = 84/1326
    = 42/663
    = 14/221
    Therefore the probability that both cards are queens or both are red is 14/221.

    Please check and let me know if I am correct. THANK YOU
    You and Mush are both correct for (a), (c) and (d) (different methods but both methods are valid and give the same correct answer).

    -----------------------------------------------------------------------------

    You're wrong for (b). Doing it your way, it should be (26, 2) = 325. NOT (36, 2). Then you'll get the same correct answer as Mush.

    ---------------------------------------------------------------------------

    For (e) you have to be careful not to double count since two red Queens also gets counted as two red cards ....

    So it's Pr(2 red cards) + Pr(2 Queens) - Pr(2 red Queens). That is, add (b) and (c) and subtract (d).

    Both you and Mush have double counted.
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  4. #4
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    Hello, cnmath16!

    Two cards are drawn from a deck of 52 playing cards.
    Find the probability of each of the following events occurring:


    a) Both cards are Clubs (3 marks)

    N(S) = (52,2) = 1326

    C (13,2) = 78

    P(E) = 78/1326 = 13/221
    = 1/17 . . . Yes!


    b) Both cards are Red (3 marks)

    (36, 2) = 630 . . . no

    We want: . C(26,2) \:=\: 325

    Therefore: . P(\text{both Red}) \:=\:\frac{325}{1326} \:=\:\frac{25}{51}




    c) Both cards are Queens (3 marks)

    (4,2) = 6

    P(E) = 6/1326 = 1/221
    . . . Correct!



    d) Both cards are Red Queens (3 marks)

    (2,2) = 1

    P(E)= 1/1326
    . . . Yes!


    e) Both cards are Queens or both cards are Red.

    There are C(4,2) = 6 ways to get two Queens.
    There are C(26,2) = 325 ways to get two Red cards.
    . . But there is 1 way to get two Red Queens.

    Hence, there are: . 6 + 325 - 1 \:=\:330 ways to get two Queens or two Reds.

    . . P(E) \:=\:\frac{330}{1326} \:=\:\frac{55}{221}

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