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Posted by on Thursday, February 24, 2011 at 9:28pm.

A standard deck of cards has had all the face cards (Jacks, queens, and kings) removed so that only the ace through ten of each suit remains. A game is played in which two cards are drawn (without replacement) from this deck and a six-sided die is rolled. For the purpose of this game, an ace is considered to have a value of 1.

I know that the possible outcomes for this game is 9360.

but I get this question wrong.

b) Find the probability of each of the following events

i)one even card is drawn and an even number rolled

Second question I need help is this

A bag contains four red, three green, and five yellow marbles. Three marbles are drawn, one at time, without replacement. Determine the probablity that the order in which they are selected is

a) yellow, red, green
b)yellow, green,green

  • math - , Thursday, February 24, 2011 at 10:34pm

    1a. "Two cards are drawn" means that the order is not important.
    So the outcome for the cards is
    = 40*39/2!
    = 780
    Multiply by the 6 outcomes of the independent event of the die to give 4680.

    b. I interpret "one even card is drawn" to mean "exactly one even card is drawn" and not "at least one even card is drawn".
    Out of the 4 possible outcomes:
    Odd/Odd, even/even, even/odd, and odd/even, the probability that exactly one even card is drawn is 2/4. Similarly, the probability for an even number on the die is 3/6.
    So the joint probability can be obtained by the multiplication rule.

    Given 4 red, 3 green and 5 yellow marbles (total = 12).
    Three are drawn without replacement
    a) To be drawn in order YRG:
    To get a yellow, there are 5 out of 12
    To get a red on the next, there are 4 out of 11
    To get a green on the next, there are 3 out of 10.
    The probability is therefore:

    ii) left to you as an exercise.

  • math - , Thursday, February 24, 2011 at 10:52pm

    I still don't understand 1b.. the answer is 10/39 but i don't know...

    For 2ii) its

    Yellow: 5/12

    (5/12)*(3/11)*(2/10= 1/44

  • math - , Thursday, February 24, 2011 at 11:29pm

    2ii is correct.

    For 1B, there are two cases.

    He draws 1 even followed by 1 odd:
    1 even: 20 out of 40 cards
    1 odd : 20 out of 39 cards.
    Probability : (20/40)*(20/39)=400/1560=10/39

    Similarly, 1 odd followed by 1 even:
    1 even: 20 out of 40 cards
    1 odd : 20 out of 39 cards.
    Probability : (20/40)*(20/39)=400/1560=10/39

    Total of two cases : (10+10)/39 = 20/39

    For the die, there are 3 even out of 6, so the probability is 1/2.

    Overall probability = (20/39)*(1/2) = 10/39.

  • math - , Thursday, February 24, 2011 at 11:39pm

    Thank you! I understand now.

  • math :) - , Thursday, February 24, 2011 at 11:44pm


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