SOLUTION: Suppose that you start with a normal deck of 52 cards and remove everything except the twos, threes, and fours. So you now have a deck of twelve cards consisting of four twos, four

Algebra ->  Probability-and-statistics -> SOLUTION: Suppose that you start with a normal deck of 52 cards and remove everything except the twos, threes, and fours. So you now have a deck of twelve cards consisting of four twos, four      Log On


   



Question 1149440: Suppose that you start with a normal deck of 52 cards and remove everything except the twos, threes, and fours. So you now have a deck of twelve cards consisting of four twos, four threes, and four fours. A card is drawn at random and replaced then another card is drawn at random and replaced. Make a probability distribution for the sum of the two numbers that are drawn.
Answer by greenestamps(13200) About Me  (Show Source):
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Since each card that is drawn is replaced before drawing the next card, the probability of each number on each draw is 4/12, or 1/3. Then

P(4) = P(2 and 2) = (1/3)(1/3) = 1/9
P(5) = P(2 and 3 OR 3 and 2) = (1/3)(1/3)+(1/3)(1/3) = 2/9
P(6) = P(2 and 4 OR 3 and 3 OR 4 and 2) = (1/3)(1/3)+(1/3)(1/3)+(1/3)(1/3) = 3/9
P(7) = P(3 and 4 OR 4 and 3) = (1/3)(1/3)+(1/3)(1/3) = 2/9
P(8) = P(4 and 4) = (1/3)(1/3) = 1/9

Note that the sum of those probabilities is 9/9=1, giving us some reassurance that our method of calculating them is correct.