1. ## Clarification?

Hi there, I was just wanting to seek some clarity/reassurance what towards a question. I was hoping someone in the know was able to please have a look over for me because I don't really feel it's right, but am having trouble getting my head round it any other way...

media research reported that 77% of New Zealanders regularly eat fresh fruit and vege and that 31% consider a low fat diet as a way of life. Suppose that 35% of the New Zealanders who regularly eat fruit and vege consider a low fat diet a way of life. If a New Zealander is selected randomly determine the following probabilities:

a) The NZer does not consider a low fat diet as a way of life given that he or she does regularly eat fresh fruit and vege. I've calculated this as 35% of 77% = 26.95%

b) The NZer does consider a low fat way diet as a way of life given that he or she does not regularly eat fresh fruit and vege. 31% of 26% = 8.06%

c) The NZer does not regularly eat fresh fruit and vegetables if it is known that he or she does consider a low fat diet a way of life. 31% of 23%(the % of NZers who don't reg eat fruit and vege) =7.13%

d) The New Zealander does not regularly eat fresh fruit and vege if it is known that he or she does not consider a low fat diet a way of life. 33% of 31% = 10.23%

2. ## Re: Clarification?

Ok, let's add some notation. Let $F$ mean that a NZer eats fruits and vegetables regularly, and $G$ mean that they consider a low-fat diet as a way of life. Add an $n$ in front to negate the letter. You are given $P(F) = 0.77$, $P(G) = 0.31$, and $P(G|F) = 0.35$. You should know that $P(A|B) = P(A\cap B)/P(B)$.

Part (a) is asking for $P(nG|F) = 1-P(G|F) = 1-0.35 = 0.65$

Part (b) is asking for $P(nG|nF)$. To figure this out, you need to figure out $P(nG \cap nF)$. From $P(nG|F) = 0.65 = P(nG \cap F)/P(F) = P(nG \cap F)/0.77$, you can solve for $P(nG \cap F) = 0.65\cdot 0.77 = 0.5005$. Now, $P(nG \cap F) + P(nG \cap nF) = P(nG)$. Plugging in the knowns, you have $P(nG \cap nF) = P(nG) - P(nG \cap F) = 1-P(G) - P(nG \cap F) = 1-0.31 - 0.5005 = 0.1895$. Finally, $P(nG|nF) = P(nG \cap nF)/P(nF) = 0.1895/(1-P(F)) = 0.1895/0.23 \approx 0.8239$

Part (c) is asking for $P(nF | G) = P(nF \cap G) / P(G)$. You are given that $P(G|F) = P(G \cap F) / P(F)$, so $P(F \cap G) = 0.35\cdot 0.77 = 0.2695$. Now, $P(F \cap G) + P(nF \cap G) = P(G) = 0.31$. So, $P(nF \cap G) = 0.31 - 0.2695 = 0.0405$. Finally, $P(nF | G) = 0.0405/0.31 \approx 0.1306$

Part (d) is asking for $P(nF | nG) = P(nF \cap nG) / P(nG)$. In part (b), we found $P(nF \cap nG) = 0.1895$, so this is $P(nF | nG) = 0.1895 / (1-P(G)) = 0.1895 / 0.69 \approx .2746$

3. ## Re: Clarification?

Rather than use formulas, I would do it this way:
Imagine 10000 New Zealanders. 77%, so 7700, "regularly eat fresh fruit and vege" and 31%, so 3100, "consider a low fat diet as a way of life".

"Suppose that 35% of the New Zealanders who regularly eat fruit and vege consider a low fat diet a way of life."
Okay that is .35(7700)= 2695. Then there are 7700- 2695= 5005 who "regularly eat fruit an vege" but do not "consider a low fat diet a way of life" and there are 3100- 2595= 505 who "consider a low fat diet a way of life" but do not "regularly eat fruit and vege". That means there are 5005+ 2695+ 505= 8205 who "regularly eat fruit and vege" or "consider a low fat diet a way of life" or both. That leaves 10000- 8205= 1795 who do neither-that is, do NOT "regularly eat fruit and vege" and do NOT consider a low fat diet a way of life".

Now:
If a New Zealander is selected randomly determine the following probabilities:

a) The NZer does not consider a low fat diet as a way of life given that he or she does regularly eat fresh fruit and vege. I've calculated this as 35% of 77% = 26.95%
No, that is backwards. You are told that "35% of the New Zealanders who regularly eat fruit and vege also consider a low fat diet a way of life". That leaves 100- 35= 65% of those who "who regularly eat fruit and vege" who do NOT "consider a low fat diet a way of life.
You could also do this using the numbers I calculated above: "Then there are 7700- 2695= 5005 who "regularly eat fruit an vege" but do not "consider a low fat diet a way of life" so the percentage is $\frac{5005}{7700}$.

b) The NZer does consider a low fat way diet as a way of life given that he or she does not regularly eat fresh fruit and vege. 31% of 26% = 8.06%
"there are 3100- 2595= 505 who "consider a low fat diet a way of life" but do not "regularly eat fruit and vege" so the percentage is 505/3100.

[quote] c) The NZer does not regularly eat fresh fruit and vegetables if it is known that he or she does consider a low fat diet a way of life. 31% of 23%(the % of NZers who don't reg eat fruit and vege) =7.13%[quote]
'there are 3100- 2595= 505 who "consider a low fat diet a way of life" but do not "regularly eat fruit and vege"' so that is 505/3100.

d) The New Zealander does not regularly eat fresh fruit and vege if it is known that he or she does not consider a low fat diet a way of life. 33% of 31% = 10.23%
Of the 10000- 3100= 6900 who do not "consider a low fat diet a way of life", 1795 also do NOT "regularly eat fruit and vege" so that is 1795/6900