uri
  • uri
A bag contains 6 white and 8 black balls.If 4 balls are drawn.What is the probability that: a)All are black b)2 are white and 2 are black
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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uri
  • uri
@nincompoop
bahrom7893
  • bahrom7893
Sooo you have 14 balls. a) You have to pick all black: (8/14)*(7/13)*(6/12)*(5/11)
uri
  • uri
Got it,and after this? :)

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uri
  • uri
@bahrom7893
bahrom7893
  • bahrom7893
b) 2 are white and 2 are black: (6/14)*(5/13)*(8/12)*(7/11)
bahrom7893
  • bahrom7893
basically probability of picking a ball of a certain color: number of balls of that color / number of balls total
anonymous
  • anonymous
@bahrom7893 it never said you necessarily draw 2 white and then 2 black!
uri
  • uri
I don't get the b part.
bahrom7893
  • bahrom7893
@oldrin.bataku is right. you have to multiply my answer for b by 4!
bahrom7893
  • bahrom7893
but wait two of them are the same color, so 4!/2!
bahrom7893
  • bahrom7893
I meant to say 2 pairs are the same color
uri
  • uri
Okay now can you solve and show the first part? That is part a. wait we have to write it like this: \[\frac{ 6 }{ 14 }\frac{ 5 }{ 13 }\frac{ 8 }{ 12 }\frac{ 7 }{ 11 }\]
uri
  • uri
Multiply sign in between?
nincompoop
  • nincompoop
how come I didn't get a notification?
bahrom7893
  • bahrom7893
yea because basically we have to keep getting black balls: 8 black balls out of 14 total = 8/14 (one black ball is gone) 7 black balls out of 13 total = 7/13, (7 black balls are remaining, and there are 13 in total) etc. follow the pattern
uri
  • uri
I swear to god i don't get it.
uri
  • uri
I miss...@terenzreignz Now..
uri
  • uri
@terenzreignz
kirbykirby
  • kirbykirby
You can use the hypergeometric distribution
kirbykirby
  • kirbykirby
Let X = # of black balls 14 balls total You choose 4 of them If you want all of them to be black, then plug in x = 4 \[f(x)=\frac{\left(\begin{matrix}8 \\ x\end{matrix}\right) \left(\begin{matrix}6 \\ 4-x\end{matrix}\right)}{\left(\begin{matrix}14 \\ 4\end{matrix}\right)}\]
kirbykirby
  • kirbykirby
b) you can just apply the same logic but you consider the case for 2 black balls (put in x=2) then just multiply that with the result for white=2 ... you can re-write f(x) for white balls and compute it for that,
anonymous
  • anonymous
@kirbykirby is correct; hypergeometric distribution is for where we don't have replacement

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