Solve the given problem involving probability without replacement

AI Thread Summary
The discussion revolves around solving a probability problem involving drawing balls without replacement, with participants sharing their methods and seeking a more systematic approach. One user successfully calculated the probability of drawing a green ball through trial and error, while others suggested using combinations or sequences for a more concrete solution. The conversation also touches on the scoring system for the problem, with a consensus that 2 marks are appropriate for a complete solution, and fewer marks for partial solutions. Various formulas for calculating probabilities were shared, highlighting the complexity of the problem. Overall, the participants are focused on refining their understanding of probability calculations in this context.
chwala
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Homework Statement
This is apast paper question. Allow me to post it as it is.
Relevant Equations
Probability
1728080305462.png


Okay, i was able to solve it by trial and error, i am seeking for a more concrete approach. Can combination work here? or a more solid approach using sequences? or probability itself?

My trial and error,
##P_{green} = \dfrac{9}{12}×\dfrac{8}{11} ×\dfrac{7}{10}×\dfrac{6}{9}×\dfrac{3}{8} = \dfrac{21}{220}##

##n=5##.

and why 2 marks? or there is a shorter method.
 
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chwala said:
and why 2 marks?
And how should we know how many marks there are supposed to be? Is 2 not a perfect score for this question?
 
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chwala said:
a more concrete approach
Excel:
1728096424538.png


The calculated value in C6 equals the target value in D6 when n=5.
 

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  • 1728096320696.png
    1728096320696.png
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Hill said:
Excel:
View attachment 351877

The calculated value in C6 equals the target value in D6 when n=5.
ok @Hill, I can see that you're still using the same approach as I am, but you're doing it differently. For instance, following your method,

##P(2) = \dfrac{9}{12}×\dfrac{3}{11}=0.204525##

and so on.

Cheers man!
 
chwala said:
ok @Hill, I can see that you're still using the same approach as I am, but you're doing it differently. For instance, following your method,

##P(2) = \dfrac{9}{12}×\dfrac{3}{11}=0.204525##

and so on.

Cheers man!
These formulas are behind my numbers:
1728108388293.png


I.e., ##P(n)=\frac {3}{13-n}+\frac{10-n}{13-n}*P(n-1)##

In my approach, ##P(2) = 0.454545##, but rather ##P(2)-P(1)=0.204545##.
 
Last edited:
## \begin{align}
P(n)&=\frac{9!}{(10-n)!}\cdot \frac{(12-n)!}{12!}\cdot3\nonumber\\
&=\frac{(12-n) \cdot (11-n)}{12\cdot 11\cdot 10}\cdot 3\nonumber\\
&=\frac{n^2-23n+132}{440}\nonumber
\end{align} ##

where ## P(n) ## is the probability that Marie picks a green ball on pick ## n ##.
 
I forgot to say that in my posts 3 and 5, ##P(n)## is the probability of picking a green ball on picks ##1## - ##n##. Thus, ##P(n)-P(n-1)## is the probability of picking a green ball on pick ##n##.
 
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chwala said:
Homework Statement: This is apast paper question. Allow me to post it as it is.
Relevant Equations: Probability

View attachment 351870

Okay, i was able to solve it by trial and error, i am seeking for a more concrete approach. Can combination work here? or a more solid approach using sequences? or probability itself?

My trial and error,
##P_{green} = \dfrac{9}{12}×\dfrac{8}{11} ×\dfrac{7}{10}×\dfrac{6}{9}×\dfrac{3}{8} = \dfrac{21}{220}##

##n=5##.

and why 2 marks? or there is a shorter method.
I would do it this way.
 
We can work backwards from \frac{21}{220} until we get something we can recognise as P(n) for some n. We have a factor of 7 and a factor of 3 in the numerator, so we need at least factors of 9 and 8, which we get by multiplying by 9/9 = 1 and 8/8 = 1. We don't hae a factor of 12 in the denominator, but we can get that by mupltiying by 6/6 = 1 and combining the 6 with a factor of 2. Hence
\begin{split}<br /> \frac{21}{220} &amp;= \frac{7 \times 3}{2 \times 10 \times 11} \\<br /> &amp;= \frac{9}{9} \times \frac88 \times \frac{6}{6} \times \frac{7 \times 3}{2 \times 10 \times 11} \\<br /> &amp;= \frac{9}{6 \times 2} \times \frac{8}{11} \times \frac{7}{10} \times \frac{6}{9} \times \frac{3}{8} \\<br /> &amp;= P(5).\end{split}
 
Last edited:
  • #10
phinds said:
And how should we know how many marks there are supposed to be? Is 2 not a perfect score for this question?
I agree that 2 marks is sufficient. Question requires logical thinking.
 
  • #11
chwala said:
I agree that 2 marks is sufficient. Question requires logical thinking.
As I understand it, there can be three cases.

The first case is a complete solution. That means there is a correct trial for n=5 (the fifth ball is green). In this case a student scores the mark 2.
The second case is an incomplete solution. That means there is a correct trial for n=4 (the fourth ball is green) or n=3 (the third ball is green) or n=2 (the second ball is green). In this case a student scores the mark 1.
The third case is an incomplete solution too. But in this case that means there is only 1 correct trial for n=1 (the first ball is green) or there is not a correct trial. In this case a student does not score a mark.
 
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