anonymous
  • anonymous
A box has 40 balls, of which 13 are red, 10 are blue, and 17 are yellow, as before. Balls are extracted at random but this time not returned to the box. What is the probability that after performing this test three times exactly two are red and one is yellow?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
use combinations
anonymous
  • anonymous
@pgpilot326 u mean
anonymous
  • anonymous
|dw:1444502981895:dw|

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More answers

anonymous
  • anonymous
i know that the fomular for combination is : \[\frac{ n! }{ r!(n-1) }\]
anonymous
  • anonymous
for the denominator\(\left(\begin{matrix}40 \\ 3\end{matrix}\right)\) is appropriate what about the numerator?
anonymous
  • anonymous
|dw:1444503437050:dw|
anonymous
  • anonymous
btw\[\left(\begin{matrix}n \\ r\end{matrix}\right)=\frac{ n! }{ r!\,\,\left( n-r \right)! }\]
anonymous
  • anonymous
no... how many red balls are there... and how many do you need? how many yellow balls and how many do you need?
anonymous
  • anonymous
there like 13 red ball and 17 yellow
anonymous
  • anonymous
okay and how many of each are needed?
anonymous
  • anonymous
13 red we 3 and 17 yellow for 1
anonymous
  • anonymous
3 red? i don't think that's right
anonymous
  • anonymous
misread is 2 red
anonymous
  • anonymous
correct. so how many ways to select 2 of the 13?
anonymous
  • anonymous
26
anonymous
  • anonymous
no, use combinations. you are sampling without replacement and order doesn't matter
anonymous
  • anonymous
ok
anonymous
  • anonymous
is 286 |dw:1444503856355:dw|
anonymous
  • anonymous
yep and how many ways to select 1 of the yellow balls?
anonymous
  • anonymous
17
anonymous
  • anonymous
yep but use combinations
anonymous
  • anonymous
|dw:1444503992842:dw|
anonymous
  • anonymous
|dw:1444504035212:dw|
anonymous
  • anonymous
now we have like 286 for red and 17 for yellow
anonymous
  • anonymous
so what you have for your probability will be this (including the blue balls): \[\frac{ \left(\begin{matrix}13 \\ 2\end{matrix}\right)\cdot \left(\begin{matrix}17 \\ 1\end{matrix}\right)\cdot \left(\begin{matrix}10 \\ 0\end{matrix}\right) }{ \left(\begin{matrix}40 \\ 3\end{matrix}\right) }\]
anonymous
  • anonymous
\[\frac{ 4862 }{ 9880 }\]

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