SOLUTION: Assume I shuffle together two standard (52 card) decks of playing cards: If I draw a card from the deck, replace this card, reshuffle, and draw a second card from the deck, what i

Algebra ->  Probability-and-statistics -> SOLUTION: Assume I shuffle together two standard (52 card) decks of playing cards: If I draw a card from the deck, replace this card, reshuffle, and draw a second card from the deck, what i      Log On


   

Question 1152860: Assume I shuffle together two standard (52 card) decks of playing cards:
If I draw a card from the deck, replace this card, reshuffle, and draw a second card from the deck, what is the probability that I will draw two cards that are either threes or twos? (Hint: There are 4 twos and 4 threes in a single standard deck.)
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The shuffling is irrelevant; and since the first card is replaced before the second card is drawn, the probability of getting a 2 or 3 is the same on both draws. 8 of the 52 cards are twos or threes:

%288%2F52%29%288%2F52%29+=+%282%2F13%29%282%2F13%29+=+4%2F169

ANSWER: P(both cards are twos or threes) = 4/169

Answer by ikleyn(52790) About Me  (Show Source):
You can put this solution on YOUR website!
.

P(both are twos) = %288%2F104%29%2A%288%2F104%29 = %284%2F52%29%2A%284%2F52%29 = %281%2F13%29%2A%281%2F13%29 = 1%2F169.


P(both are threes) = %288%2F104%29%2A%288%2F104%29 = %284%2F52%29%2A%284%2F52%29 = %281%2F13%29%2A%281%2F13%29 = 1%2F169.


P(under the question) = P(both are twos) + P(both are threes) = 1%2F169 + 1%2F169 = 2%2F169.    ANSWER