Solve the given problem involving probability without replacement

[chwala]

Homework Statement

This is apast paper question. Allow me to post it as it is.

Relevant Equations

Probability

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.

 

[phinds]

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?

 

[Hill]

a more concrete approach

Excel:

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

 

[chwala]

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!

 

[Hill]

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:

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 $P(2)-P(1)=0.204545$.

 

[Gavran]

$\begin{align} P(n)&=\frac{9!}{(10-n)!}\cdot \frac{(12-n)!}{12!}\cdot3\nonumber\\ &=\frac{(12-n) \cdot (11-n)}{12\cdot 11\cdot 10}\cdot 3\nonumber\\ &=\frac{n^2-23n+132}{440}\nonumber \end{align}$

where $P(n)$ is the probability that Marie picks a green ball on pick $n$.

 

[Hill]

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 $n$.

 

[Hornbein]

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.

 

[pasmith]

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
$$\begin{split} \frac{21}{220} &= \frac{7 \times 3}{2 \times 10 \times 11} \\ &= \frac{9}{9} \times \frac88 \times \frac{6}{6} \times \frac{7 \times 3}{2 \times 10 \times 11} \\ &= \frac{9}{6 \times 2} \times \frac{8}{11} \times \frac{7}{10} \times \frac{6}{9} \times \frac{3}{8} \\ &= P(5).\end{split}$$