Question

Can anyone answer this (revised) question:

What is the probability of drawing 3 aces

a) with replacement (and re-shuffling)?

b) without replacement?

Answer 1

With replacement: every time the probability is 4/52=1/13, so three times in a row has probability 1/13³=1/2197.

Answer 2

so then how do I find the without replacement?

Answer 3

With replacement: every time the probability is 4/52=1/13, so three times in a row has probability 1/13³
. and for without replacement directly apply the probability theorems

Answer 4

Without replacement:

You can look at this in two ways:
a. drawing them on by one
b. drawing all three at the same time

a. the first ace has a probability of 4/52 (=1/3).
second: 3/51, third: 2/51, so three in a row: (4*3*2)/(52*51*50)=1/5525

b. (you need to know about permutations and combinations)
the probability is :
\[p(3 \hspace{1 mm} aces)=\frac{ no\hspace{1 mm}of\hspace{1 mm}combinations\hspace{1 mm} of\hspace{1 mm} 3 \hspace{1 mm}out\hspace{1 mm} of \hspace{1 mm}4 }{ total\hspace{1 mm} no\hspace{1 mm} of\hspace{1 mm} combinations\hspace{1 mm} of\hspace{1 mm} 3 \hspace{1 mm}out\hspace{1 mm} of\hspace{1 mm} 52 } \]
\[=\frac{ \left(\begin{matrix}4 \\ 3\end{matrix}\right) }{ \left(\begin{matrix}52 \\ 3\end{matrix}\right) }=\frac{ \frac{ 4! }{ 3!1! } }{ \frac{ 52! }{ 49!3! } }=\frac{ \frac{ 4\cdot3\cdot2\cdot1 }{ 3\cdot2\cdot1\cdot1 } }{ \frac{ 52\cdot51\cdot50 }{ 3\cdot2\cdot1 } }=\frac{ 4\cdot3\cdot2 }{ 52\cdot51\cdot50 }=...=\frac{ 1 }{ 5525 }\]

Answer 5

Luckily, both methods yield the same result ;)

Answer 6

lol

Answer 7

thank you both.. I was pretty lost. Stats just not my cup of tea! Not today anyhow.

Answer 8

YW!