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Mathematics 6 Online
OpenStudy (anonymous):

A boy takes 12 ink cartridges down from a shelf. Three of them do not have any ink. When selecting four out of the 12, what is the pobability of getting one without ink.

OpenStudy (kropot72):

\[P(1)=\frac{\left(\begin{matrix}3 \\ 1\end{matrix}\right)\left(\begin{matrix}9 \\ 3\end{matrix}\right)}{\left(\begin{matrix}12 \\ 4\end{matrix}\right)}\]

OpenStudy (anonymous):

Thank you for showing me how to figure it out .

OpenStudy (dan815):

i will teach youu

OpenStudy (dan815):

now the question says 3 of them do not have ink so how many have ink?

OpenStudy (kropot72):

You're welcome :) Do you know how to calculate the binomial coefficients?

OpenStudy (anonymous):

No

OpenStudy (dan815):

hey you answer me I Wwanna teach :)

OpenStudy (kropot72):

\[\left(\begin{matrix}n \\ r\end{matrix}\right)=\frac{n!}{r!(n-r)!}\]

OpenStudy (anonymous):

got it, thank you. Now maybe I'll be able to finsh my homework. Have a great weekend and thank you again.

OpenStudy (kropot72):

You're welcome :)

OpenStudy (dan815):

kropot!!

OpenStudy (dan815):

what are all those notations

OpenStudy (kropot72):

The solution can also be expressed as \[P(1)=\frac{3C1\times 9C3}{12C4}\]

OpenStudy (dan815):

i dont understand those Cs nad Ps is my calcuation the same thing as that?

OpenStudy (dan815):

how are u thinking about this question ?

OpenStudy (dan815):

4*(3/12*9/11*8/10*7/9 i mean that

OpenStudy (dan815):

4*(3/12*9/11*8/10*7/9) = 51% ish

OpenStudy (kropot72):

@dan815 If you calculation is \[4(\frac{4}{12}\times \frac{9}{11}\times \frac{8}{10}\times \frac{7}{9})=0.67879\] then it is not correct.

OpenStudy (dan815):

4/12 to 3/12

OpenStudy (dan815):

cuz 4 + 9 = 13, 1 more than there shud be xD

OpenStudy (kropot72):

@dan815 Your amended calculation gives the correct answer.

OpenStudy (dan815):

can you tell me how you get the same answer with Cs and Ps though this is how i did it ___,___,___,___ 4 spots and probability of getting noncolor = 3/12 and the other spots are 9/11,8/10,7/9 so the noncolor can be in 4 spots so 4*(3/12*9/11*8/10*7/9) = solution is that basically what ur doing with C and P formula

OpenStudy (kropot72):

I have used the hypergeometric distribution that treats sampling without replacement. It appears that you are basically using the same.

OpenStudy (dan815):

r u some genius??

OpenStudy (kropot72):

Not really. Just a student :)

OpenStudy (dan815):

so then how did u solve what i needed like 3 lines off work in 1 division sign

OpenStudy (kropot72):

There was much the same work in calculating the binomial coefficients and then multiplying and dividing them as in your calculation. I have left the detailed calculations to the questioner who says they understand what to do.

OpenStudy (dan815):

i think im staring to understand what u did so basically you said for a situation where there are 4 spots 3 without color go to 1 spot and 9 with color go to 3 spots and then divided by all the possibile situations with those 4 spots are 12 into 4 spots

OpenStudy (dan815):

so p(2) = (3C2)*(9C2)/(12C4)

OpenStudy (kropot72):

@dan815 Your equation for p(2) is correct.

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