# Thread: Tricky question (tree diagram)

1. ## Tricky question (tree diagram)

A randomised response technique was used to investigate rates of illegal fishing of red abalone in Northern California. The study involved interviews with people at various sites. Once verbal consent to participant was obtained, each respondent was given a coin and an envelope containing two cards.
On one card was written the question In the past year have you ever taken abalone under the minimum legal size limit? while on the other card was written the question Did you get heads on the coin toss? Without the interviewer watching, each respondent first tossed the coin, noting the outcome. They then chose a card at random and answered the question on it to the interviewer. Suppose that out of 279 respondents, 102 answered 'Yes'.

The estimated proportion of people who have actually taken abalone under the minimum legal size limit in the last year is

--> I have tried to do this so many time with the help of tree diagrams and have asked other people but no one seems to get it right.
Do you guys have any idea how to tackle this, cudos if u can!!!

2. ## Re: Tricky question (tree diagram)

You have four outcomes:
1. They chose the coin card and they flipped heads (wrote 'Yes')
2. They chose the coin card and they flipped tails (wrote 'No')
3. They chose the other card and they wrote 'Yes'
4. They chose the other card and they wrote 'No'

So, let's break it down into cases. Suppose 0 people picked the coin card. That has probability $\dfrac{1}{2^{279}}$. That means that the proportion of people who illegally took abalone was $\dfrac{102}{279}$.
Suppose exactly 1 person picked the coin card. The odds of that happening is $\dbinom{279}{1}\dfrac{1}{2^{279}}$. Then, there were either 102 yeses for illegal abalone or 101 with a 50/50 chance of each. So, it would be:
$\dbinom{279}{1}\dfrac{1}{2^{279}}\left(\dfrac{1}{ 2}\dfrac{102}{278} + \dfrac{1}{2}\dfrac{101}{278} \right)$
Next, suppose exactly 2 people picked the coin card. That happens with probability $\dbinom{279}{2}\dfrac{1}{2^{279}}$. Then there were 102, 101, or 100 yeses for illegal abalone out of 277. Probability for this scenario is:
$\dbinom{279}{2}\dfrac{1}{2^{279}}\left( \dbinom{2}{2}\dfrac{1}{2^2}\dfrac{102}{277} + \dbinom{2}{1}\dfrac{1}{2^2}\dfrac{101}{277} + \dbinom{2}{0}\dfrac{1}{2^2}\dfrac{100}{277} \right)$

This pattern continues up to 102 people chose the coin card. When exactly $n$ people chose the coin card and $0 \le n \le 102$, you have the following probability:
$\displaystyle \dbinom{279}{n}\dfrac{1}{2^{279}}\sum_{k=0}^n \dbinom{n}{k}\dfrac{1}{2^n}\dfrac{102-k}{279-n} = \dbinom{279}{n} \cdot \dfrac{1}{2^{279}} \cdot \dfrac{n-204}{2(n-279)}$

From $n=103$ to $n=177$ people choosing the coin card, you have the following probability:
$\displaystyle \dbinom{279}{n}\dfrac{1}{2^{279}}\sum_{k=0}^{102} \dbinom{n}{k} \dfrac{1}{2^n}\dfrac{102-k}{279-n}$

And lastly, from $n=178$ through $n=279$ people choosing the coin card. This is the most difficult one to figure out. It is less than:
$\displaystyle \dbinom{279}{n}\dfrac{1}{2^{279}}$

So, the total expected proportion from $n=178$ through $n=279$ is less than
$\displaystyle \sum_{n=178}^{279}\dbinom{279}{n} \dfrac{1}{2^{279}} \approx 3\times 10^{-6}$
http://www.wolframalpha.com/input/?i...,178,279%7D%5D

So, the total expected proportion of people who have actually taken abalone under the minimum legal size limit would be:
$\displaystyle \sum_{n=0}^{102}\left[ \dbinom{279}{n}\dfrac{1}{2^{279}}\dfrac{n-204}{2(n-279)} \right] + \sum_{n=103}^{177} \left[ \dbinom{279}{n}\dfrac{1}{2^{279}}\sum_{k=0}^{102} \dbinom{n}{k} \dfrac{1}{2^n}\dfrac{102-k}{279-n} \right] + \epsilon \approx \dfrac{23}{100}$ where $\epsilon < 3\times 10^{-6}$

This can be calculated with a numerical processing system. The first and last summations will likely have a significance of less than $10^{-5}$. So, you really only need to worry about the middle summation.
http://www.wolframalpha.com/input/?i...,103,177%7D%5D

3. ## Re: Tricky question (tree diagram)

I figured out what $\epsilon$ equals in my post above.

$\displaystyle \epsilon = \sum_{n=178}^{278} \left[ \dbinom{279}{n}\dfrac{1}{2^{279} }\sum_{k=n-177}^{102} \dbinom{n}{k} \dfrac{1}{2^n} \dfrac{102-k}{279-n} \right] \approx 8.744 \times 10^{-42}$

So, extremely insignificant. Note that in the extremely small chance that every respondent chose the the card for coin flip, we cannot make any determination about the proportion of people who illegally took home abalone.

Anyway, I stick by my answer of about $\dfrac{23}{100}$

4. ## Re: Tricky question (tree diagram)

Hello
Is it possible if u can show it has a tree diagram?

5. ## Re: Tricky question (tree diagram)

Originally Posted by ana1237
Hello
Is it possible if u can show it has a tree diagram?
I never made tree diagrams before. I just know how to solve the problem. What does a tree diagram look like? Do you have an example?

7. ## Re: Tricky question (tree diagram)

Would be so appreciative if you could show it like this, I attached a webpage about tree diagram

8. ## Re: Tricky question (tree diagram)

Oh, then no. This tree would have more leaves than there are atoms in the universe. While such a tree exists in theory, there is no way to physically draw it.