1. A standard deck of 52 cards comes with 2 jokers, which are usually removed fr
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Question
1. A standard deck of 52 cards comes with 2 jokers, which are usually removed from the deck (leaving behind the 52 other cards) before playing a card game. Suppose that you remove the hearts, clubs and spades from the deck of 52 cards, and then add in the 2 okers, before beginning this experiment. Suppose at you will draw3cards from this depleted at a time and without replacement, and count the number of jokers that you end up with. Letting "J" represent picking a card that WAS a joker, and N" present picking a card that was NOT a joker: a. Draw the complete tree diagram for this experiment, with all of the twig- probabilities included in their proper place. (10points) LabelExplanation / Answer
b) you now have 13 diamonds and 2 jokers(similar looking)...which is 15 cards. Since we are picking without replacement Then (P) = (13/15)x(12/14)x(11/13) x 3!=12/35..from 15 we can choose any one diamond from the 13 available...then from the 14 remaining(we have picked one), we can choose any of the remaining 12 diamonds, and lastly from the 13 remaining we can choose any 11 diamonds...please note they can come in any order, hence 3 cards can be arranged in 3! ways.eg i choose 3 different diamond cards A,B and C...the ways they can come in are ABC, ACB, BAC, BCA, CAB and CBA. so 6.
c) (P)= (2/15)x(13/14)*12*(13)x3! = 24/35
d) here the concept is sameexcept when we do the arrangement in the end(3! is what we did in the last 2 sums), we must remember that 2 jokers are identical, and hence we multiply by 3!/2!= 3 the 3 arrangements are JJD,DJJ and JDJ...Therefore the (P)=(2/15)x(1/14)*(13/13)x(3!/2!)= 1/35
Note: It does not matter if you pick diamonds first or joker first. The base remains the same, and thats why we are multiplying by the no of arrangements possible.
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