Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

johnson32 Group Title

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

  • one year ago
  • one year ago

  • This Question is Closed
  1. ZeHanz Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  2. johnson32 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    so then how do I find the without replacement?

    • one year ago
  3. harshit1202 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    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

    • one year ago
  4. ZeHanz Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  5. ZeHanz Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Luckily, both methods yield the same result ;)

    • one year ago
  6. harshit1202 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    lol

    • one year ago
  7. johnson32 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  8. ZeHanz Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    YW!

    • one year ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.