00:01
Okay, so before we go on to this problem, i'm going to simplify this a little bit.
00:08
So remember that in a 52 card deck, there is going to be 13 of each suit.
00:15
And since all of these problems are only dealing with the suits, we can simplify this problem by considering a four -card deck consisting of the ace of each suit, spade, club, hard, diamond.
00:32
And the probability is going to be the same because we had the condition that each card gets replaced into the deck before the next one is picked.
00:42
And of course, we're assuming that we are shuffling the deck each time.
00:51
So we can...
00:53
So for all these problems, instead of dealing with 13 over 52 or 39 over 52, we're going to deal with instead one -fourth.
01:04
So the probability that all three cards are hearts is that is basically the probability where you pick the ace of hearts in the first row, in the first straw, shuffle it, get the ace of hearts again, and then shuffle, and then get the ace of hearts again.
01:27
So that's going to be equal to one fourth, which is the third power, which is going to be 1 over 64, or if you're going to be 1 over 64, or if you're you want the decimal, 0 .15625.
01:53
Part b is asking for exactly two of the cards are space, spades, sorry.
02:00
So the probability that you draw three times with replacing and reshuffling, and two of them end up being the space of spades...