Question

A standard deck of 52 cards is shuffled and the cards are dealt face up one at a time until an ace appears
a. what is the probability that the 1st ace appears on the third card
Show that the probability of getting the first ace on or before the ninth card is greater than 50%

Answer 1

okay so do you like do 4C1 51 C0 /52 C1

Answer 2

If I remember correctly- it's been a little while since i did probablitiy-

Answer 3

We have to divide the deck up into groups of 13, each with 4 identical numbers. Then we do 12C1 4C1 12C1 4C1 13C1 4C1

Answer 4

Maybe? I'm not sure, since I mostly did poker hands with this sort of thing.

Answer 5

oh thanks

Answer 6

lemme try it again

Answer 7

We have to divide the deck up into groups of 13, each with 4 identical numbers. Then we do 12C1 4C1 12C1 4C1 13C1 4C1
Why \(^{13}C_1\)?

Answer 8

My attempted:
For the first and second draw, we are taking two non-ace cards. So: \[
^{52-4}P_2 = ^{48}P_2
\]
For the third draw, we are taking any ace card. So \[
4
\]
The total ways we could have done this would be: \[
^{52}P_3
\]So our probability would be : \[
\frac{4\times ^{48}P_2}{^{52}P_3}
\]

Answer 9

Remember that \(^nP_1 = ^nC_1 = n\)

Answer 10

So you could say that drawing an ace is the same as \(^4P_1\) or \(^4C_1\)

Answer 11

In this case, \(52\) is number of cards total, or \(13\times 4\). So do draw a non-ace you could say it is \(12\times 4=48\) or you can say \(13\times 4 - 4 = 52-4 = 48\).