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?

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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
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use combinations
|dw:1444502981895:dw|

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Other answers:

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

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