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Question

There are N blue sweets in a bag. For every 2 blue sweets in the bag there are 3 red sweets in the bag. 3 sweets are chosen at random and taken out of the bag. Given that the probability of choosing 3 blue sweets is 1/30, show that 23N^2-114N+88 equals 0.

I don't need help in actually finding the answer to this problem, I am just unsure on what method I have to use as I solved that quadratic equation equal to zero at the end and I got N is equal to 0.95652174 or 4, however I don't think that solving the quadratic at the end is perhaps the best approach to this question. So, do you have to use frequency trees? or thinking in terms of the ratios?   

Answer 1

The question doesn't ask you to solve the quadratic equation, it just tells you to show that $N$ satisfies the equation.

The total number of sweet is equal to $\frac52N.$

$$\frac{N}{\frac52 N}\frac{N-1}{\frac52N-1}\frac{N-2}{\frac52N-2} = \frac1{30}$$

Hence

$$\frac85\frac{N-1}{5N-2}\frac{N-2}{5N-4} = \frac1{30}$$

$$48(N-1)(N-2)=(5N-2)(5N-4)$$

Can you complete the rest to show that the equation above is equal to the quadratic equation in the question?

Answer 2

Solving the quadratic equation isn't the right way to answer the question you were asked. But it is a small part of one possible approach to the problem. Much more is required, though.

I know you said you don't need help answering the question, but your idea of answering it by solving the quadratic is pretty far off, so here is the full process with some details left to you:

You know that for every $2$ blue sweets, there are $3$ red sweets. So the ratio of blue sweets to red sweets is: $$\frac{\text{blue}}{\text{red}} = \frac23$$

You also know that there are $N$ blue sweets. So how many red sweets are there? Use the ratio to find out: $$\frac23 = \frac Nr$$ The $r$ above represents the number of red sweets. Solve that for $r$. Note that $r$ will be an expression in terms of $N$.

At this point you know what $r$ is in terms of $N$. The total number of sweets in the bag is $N + r$ (number of blue sweets plus number of red sweets). Therefore you know the total number of sweets in the bag in terms of $N$.

3 sweets are taken out of the bag, seemingly without replacement. We want the probability of choosing a blue sweet three times in a row. So we multiply the probabilities of choosing a blue sweet at each step.

The probability of choosing a blue sweet first is $\dfrac N{N+r}$. (Recall that you already know what $N+r$ is in terms of just $N$.)

The probability of choosing a blue sweet second is $\dfrac {N-1}{N+r-1}$. The $-1$ on top is because we already pulled a blue sweet out of the bag. The $-1$ on the bottom is because our sample space has decreased by $1$ since we already pulled one sweet out of the bag.

The probability of choosing a blue sweet third is $\dfrac{N-2}{N+r-2}$.

You're given that multiplying these together will equal $\dfrac1{30}$. So you have a big expression involving $N$ that is equal to $\dfrac1{30}$. Rearrange that expression to get $23N^2 - 114N + 88 = 0$. (Or you can solve the equation you get and verify that you get the same values of $N$ as you did when you solved the quadratic.)

Answer 3

When something is taken drawn out of a bag and without replacement, the question usually concerns the hypergeometric distribution, (unless there are a large number of objects then it can be approximated by the binomial distribution which is for drawing objects with replacement).

In this question we can say there are $N=2m$ blue sweets and $3m$ red sweets.

The wikipedia page has a nice example at the bottom. If we follow this analogously we get

The probability of drawing these objects is \begin{equation} \frac{\binom{2m}{3}\binom{3m}{0}}{\binom{5m}{3}} = \frac{1}{30} \end{equation} where the right hand side came from the question we can replace this with the equivalent expressions \begin{equation} \frac{4-12m+8m^2}{10-75m+125m^2} = \frac{1}{30} \end{equation} replace $2m=N$ and rearrange \begin{align} 4-6N+2N^2 = \frac{1}{120}(40-150N+125N^2)\\ 23N^2-114N+88=0 \end{align} we did not need to know how many sweets there were at all in the end.