anonymous
  • anonymous
Cards are drawn with replacement from a standard shuffled deck repeatedly until a black 8 appears 1) What is the probability of the game stopping on exactly the 5 th card? 2) What is the probability of the game lasting at least 17 cards draws?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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shamil98
  • shamil98
what game
anonymous
  • anonymous
is not a game is it just a geomtric density
anonymous
  • anonymous
a bit complec

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Rizags
  • Rizags
First one is a Bernoulli trial of the form:\[\large \left(\begin{matrix}5 \\ 1\end{matrix}\right)(\frac{1}{52})^1(\frac{51}{52})^4\times\frac{1}{5}\]
anonymous
  • anonymous
k
Rizags
  • Rizags
The second one is extremely tedious using bernoulli trials so there has to be a better way
anonymous
  • anonymous
i see
anonymous
  • anonymous
a) (50/52) ^4 * 2/52
anonymous
  • anonymous
where did the 4 come from ?
anonymous
  • anonymous
since we want probability of P( F F F F S ) where F = non black 8 S = black 8
anonymous
  • anonymous
F stands for failure, S for success
anonymous
  • anonymous
ok
anonymous
  • anonymous
4th is success and 1 is failure
anonymous
  • anonymous
also why is 2/52?
anonymous
  • anonymous
i just wonder
anonymous
  • anonymous
there are two black 8 cards
anonymous
  • anonymous
http://www.jfitz.com/cards/classic-playing-cards.png
anonymous
  • anonymous
i see
anonymous
  • anonymous
those 52 cards, 2 are black 8 i see
anonymous
  • anonymous
i guess we still need to figure out the second one
anonymous
  • anonymous
let X = number of cards till you drawing a black 8 you want P( X >= 17) = 1 - P( X < 17)
anonymous
  • anonymous
k
anonymous
  • anonymous
let X = number of cards till you draw a black 8 you want P( X >= 17) by complement rule: P( X >= 17) = 1 - P( X <17) = 1 - P( X<= 16)
Rizags
  • Rizags
IN that case, I think the second one would be:\[\large 1-\sum_{n=0}^{15}(\frac{50}{52})^n(\frac{2}{52})\]
anonymous
  • anonymous
does that mean n is 0 or 17?
anonymous
  • anonymous
n is like an index n=0,1,2,3...15
Rizags
  • Rizags
that fills cases from 1 to 16, inclusive. But n is just the index of summation like @jayzdd said. The Summation can be done with a graphing calculator or online.
anonymous
  • anonymous
that going to be a huge calculation
Rizags
  • Rizags
not for a computer
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=1-+sum%28n%3D0..15%29+%2850%2F52%29^n+*+2%2F52
anonymous
  • anonymous
ya i know but using wolfram is going to take me while to find the command
Rizags
  • Rizags
Here: This is your final answer http://www.wolframalpha.com/input/?i=1-%28sum+from+n%3D0+to+15%2C+%2850%2F52%29%5En%282%2F52%29%29
anonymous
  • anonymous
wolfram is user friendly, you dont need to know the command. i have used trial and error. a few times i have had to look up a command, for sophisticated problems
anonymous
  • anonymous
alright thank you for showing me the command on wolfram
anonymous
  • anonymous
i just google it for command lol
anonymous
  • anonymous
no thats not the command
anonymous
  • anonymous
we are using plain english
anonymous
  • anonymous
btw is 16 not 15
anonymous
  • anonymous
wolfram uses mathematica language, which has special syntax or it can use english if its precise enough to understand it
anonymous
  • anonymous
but anywhow thank so much for clarification
Rizags
  • Rizags
its 15 because I start at 0 for a specific reason
anonymous
  • anonymous
isn't the "lasting at least 17 cards"
Rizags
  • Rizags
at least means greater than or equal to 17, so I set up the sum to start at 0 and head up to 15, which functionally translates as 16 because of the corresponding Bernoulli Trial
anonymous
  • anonymous
that make more sense thank you

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