Question

A deck of cards (4 suits of 13 cards each) is shuffled and a gambler draws 4 cards with replacement. What is the probability of getting at most three aces?

Answer 1

Note that since the gambler draws the cards with replacement, the probability of drawing an ace on all the 4 draws is same and independent for each draw. Now, since a deck of cards has 4 aces out of a total of 52 cards, thus we get:

P(drawing an ace on a draw) = 4/52 = 1/13

Now, the probability of getting at most 3 aces on 4 draws of cards with replacement is given by:

P(at most 3 aces) = 1 - P(more than 3 aces)

= 1 - P(4 aces on 4 draws)

= 1 - P(ace on first draw)*P(ace on second draw)*P(ace on third draw)*P(ace on fourth draw)

[Since, the cards were drawn with replacement, all the four draws are independent]

= 1 - (1/13)*(1/13)*(1/13)*(1/13)

= 1 - 1/28561

= 28560/28561 [ANSWER]

= 0.999965 [ANSWER]