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anonymous
 one year ago
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?
anonymous
 one year ago
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1444502981895:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i know that the fomular for combination is : \[\frac{ n! }{ r!(n1) }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for the denominator\(\left(\begin{matrix}40 \\ 3\end{matrix}\right)\) is appropriate what about the numerator?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1444503437050:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0btw\[\left(\begin{matrix}n \\ r\end{matrix}\right)=\frac{ n! }{ r!\,\,\left( nr \right)! }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no... how many red balls are there... and how many do you need? how many yellow balls and how many do you need?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0there like 13 red ball and 17 yellow

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay and how many of each are needed?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.013 red we 3 and 17 yellow for 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.03 red? i don't think that's right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0correct. so how many ways to select 2 of the 13?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no, use combinations. you are sampling without replacement and order doesn't matter

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is 286 dw:1444503856355:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yep and how many ways to select 1 of the yellow balls?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yep but use combinations

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1444503992842:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1444504035212:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now we have like 286 for red and 17 for yellow

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so 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
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ 4862 }{ 9880 }\]
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