## 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?

1. anonymous

use combinations

2. anonymous

@pgpilot326 u mean

3. anonymous

|dw:1444502981895:dw|

4. anonymous

i know that the fomular for combination is : $\frac{ n! }{ r!(n-1) }$

5. anonymous

for the denominator$$\left(\begin{matrix}40 \\ 3\end{matrix}\right)$$ is appropriate what about the numerator?

6. anonymous

|dw:1444503437050:dw|

7. anonymous

btw$\left(\begin{matrix}n \\ r\end{matrix}\right)=\frac{ n! }{ r!\,\,\left( n-r \right)! }$

8. anonymous

no... how many red balls are there... and how many do you need? how many yellow balls and how many do you need?

9. anonymous

there like 13 red ball and 17 yellow

10. anonymous

okay and how many of each are needed?

11. anonymous

13 red we 3 and 17 yellow for 1

12. anonymous

3 red? i don't think that's right

13. anonymous

14. anonymous

correct. so how many ways to select 2 of the 13?

15. anonymous

26

16. anonymous

no, use combinations. you are sampling without replacement and order doesn't matter

17. anonymous

ok

18. anonymous

is 286 |dw:1444503856355:dw|

19. anonymous

yep and how many ways to select 1 of the yellow balls?

20. anonymous

17

21. anonymous

yep but use combinations

22. anonymous

|dw:1444503992842:dw|

23. anonymous

|dw:1444504035212:dw|

24. anonymous

now we have like 286 for red and 17 for yellow

25. anonymous

so what you have for your probability will be this (including the blue balls): $\frac{ \left(\begin{matrix}13 \\ 2\end{matrix}\right)\cdot \left(\begin{matrix}17 \\ 1\end{matrix}\right)\cdot \left(\begin{matrix}10 \\ 0\end{matrix}\right) }{ \left(\begin{matrix}40 \\ 3\end{matrix}\right) }$

26. anonymous

$\frac{ 4862 }{ 9880 }$