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johnson32

  • 3 years ago

Can anyone answer this (revised) question: What is the probability of drawing 3 aces a) with replacement (and re-shuffling)? b) without replacement?

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

  2. johnson32
    • 3 years ago
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    so then how do I find the without replacement?

  3. harshit1202
    • 3 years ago
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    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

  4. ZeHanz
    • 3 years ago
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    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 : \[p(3 \hspace{1 mm} aces)=\frac{ no\hspace{1 mm}of\hspace{1 mm}combinations\hspace{1 mm} of\hspace{1 mm} 3 \hspace{1 mm}out\hspace{1 mm} of \hspace{1 mm}4 }{ total\hspace{1 mm} no\hspace{1 mm} of\hspace{1 mm} combinations\hspace{1 mm} of\hspace{1 mm} 3 \hspace{1 mm}out\hspace{1 mm} of\hspace{1 mm} 52 } \] \[=\frac{ \left(\begin{matrix}4 \\ 3\end{matrix}\right) }{ \left(\begin{matrix}52 \\ 3\end{matrix}\right) }=\frac{ \frac{ 4! }{ 3!1! } }{ \frac{ 52! }{ 49!3! } }=\frac{ \frac{ 4\cdot3\cdot2\cdot1 }{ 3\cdot2\cdot1\cdot1 } }{ \frac{ 52\cdot51\cdot50 }{ 3\cdot2\cdot1 } }=\frac{ 4\cdot3\cdot2 }{ 52\cdot51\cdot50 }=...=\frac{ 1 }{ 5525 }\]

  5. ZeHanz
    • 3 years ago
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    Luckily, both methods yield the same result ;)

  6. harshit1202
    • 3 years ago
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    lol

  7. johnson32
    • 3 years ago
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    thank you both.. I was pretty lost. Stats just not my cup of tea! Not today anyhow.

  8. ZeHanz
    • 3 years ago
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    YW!

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