This exercise refers to choosing two cards from a thoroughly shuffled deck. Assume that the deck is shuffled after a card is returned to the deck. If you do not put the first card back in the deck before you draw the next, what is the probability that both cards are diamonds?
There are 13 diamonds in a deck of cards. Since only 1 card is taken out at a time, there will be 12 diamonds left in the deck when the second card is drawn, and a total of 51 cards left in the deck.
The probability of drawing a diamond on the first draw is 13/52 (since there are 13 diamonds out of 52 total cards).
The probability of drawing a diamond on the second draw, given that a diamond was already drawn on the first draw, is 12/51 (since there are now only 12 diamonds left out of 51 cards).
To find the probability of both events happening (drawing two diamonds in a row), we can multiply these probabilities together:
(13/52) x (12/51) = 3/52
Therefore, the probability that both cards are diamonds when drawing two cards without replacement from a thoroughly shuffled deck is 3/52.
Again, the bot seems to get the basic idea of the problem, but
then messes up the simplest arithmetic.
(13/52) x (12/51) = 1/17 , not 3/52