1. Probability and Venn diagrams.

Two questions:

i. A six-member working group to plan a student lounge is to be selected from five teachers and nine students. If the working group is randomly selected, what is the probability of the committee including at least two teachers.

ii. 5. If a survey on teenage reads of popular magazines shows that 38% subscribe to Teen Life and 47% subscribe to Cool Teen and 35% subscribe to neither magazine, what is the probability that a randomly selected teenager,

subscribes to both magazines?
subscribes to either one magazine or both?
subscribes to only one of the two magazines?
Use a Venn diagram to support your solution.

I have no idea how I'd draw this Venn D.

2. Originally Posted by acc
Two questions:

i. A six-member working group to plan a student lounge is to be selected from five teachers and nine students. If the working group is randomly selected, what is the probability of the committee including at least two teachers.

ii. 5. If a survey on teenage reads of popular magazines shows that 38% subscribe to Teen Life and 47% subscribe to Cool Teen and 35% subscribe to neither magazine, what is the probability that a randomly selected teenager,

subscribes to both magazines?
subscribes to either one magazine or both?
subscribes to only one of the two magazines?
Use a Venn diagram to support your solution.

I have no idea how I'd draw this Venn D.
I'm sorry you weren't happy with my first reply to these problems. However, it's good that you're getting closer to posting one question per thread, which is the general etiquette around here.

Originally Posted by undefined
For problem 1, I would start with a "1" and subtract the following probabilities from it

P(0 teachers and 6 students)
P(1 teacher and 5 students)

Problem 5, you could draw a Venn diagram. (Originally, you had not written the sentence "Use a Venn diagram to support your solution.") Or using formal notation,

$\displaystyle P(A \cup B) = P(A) + P(B) - P(A \cap B)$
Since you seem to be struggling, here is what problem #1 comes out to:

$\displaystyle \text{Ways to get 0 teachers and 6 students} = \binom{5}{0}\binom{9}{6} = \binom{9}{6}=84$

$\displaystyle \text{Ways to get 1 teacher and 5 students} = \binom{5}{1}\binom{9}{5} = (5)(126)=630$

$\displaystyle \text{Ways to select any six people} = \binom{14}{6}=3003$

$\displaystyle P(\text{0 teachers and 6 students}) + P(\text{1 teacher and 5 students})=\frac{84+630}{3003}$

So

$\displaystyle P(\text{at least two teachers in committee})=1-\frac{84+630}{3003}$

For problem #5, here's a diagram (not to scale)

where T = Teen Choice and C = Cool Teen.

Using informal notation, we may write

(T) + (T and C) = 0.38

(C) + (T and C) = 0.47

1 - (T) - (C) - (T and C) = 0.35

If you're comfortable with solving systems of equations, this will be pretty easy, we can rewrite

x + y = 0.38

z + y = 0.47

1 - x - y - z = 0.35

If you're not comfortable with that, no problem. Notice that

1 - x - y - z = 0.35

is the same as

1 - (x + y + z) = 0.35

We can substitute the value for x + y as follows:

1 - (0.38 + z) = 0.35

Is this making sense to you? If not, feel free to ask questions, or maybe someone else will write up a simpler explanation.

3. For the Venn D., do I need to put a rectangle around it indicating the whole set?

4. Originally Posted by acc
For the Venn D., do I need to put a rectangle around it indicating the whole set?
Sorry, I just realized I was a bit careless with naming my variables. The maths is right but the naming is confusing.

What I meant to convey is that T = ONLY Teen Choice, and C = ONLY Cool Teen. But in practice it's better to do it differently, so I'll rewrite it again below with clearer variable assignments.

To answer your question, your teacher/professor may prefer the rectangle, and it could help with visualisation, but it's not strictly necessary.

So the more conventional way to name the variables is to let T represent the entire circle. So, we get:

YELLOW AND GREEN REGIONS

T = (T and not C) + (T and C) = 0.38

BLUE AND GREEN REGIONS

C = (C and not T) + (C and T) = 0.47

WHITE REGION

1 - (T and not C) - (T and C) - (C and not T) = 0.35

5. I worked out problem 5 and I'd love for someone to look over my answers.

Teen Life = TL, Cool Teen = CT

$\displaystyle pr(TL)=0.38$
$\displaystyle pr(CT)= 0.47$
$\displaystyle pr(TL \cup CT)'=0.35$
$\displaystyle pr(TL \cup CT)= 1-pr(TL \cup CT)'=1-0.35=0.65$

A. Subscribes to both magazines

Looking for $\displaystyle pr(TL\cap CT)$

We know that:

$\displaystyle pr(TL\cup CT)=pr(TL)+pr(CT)-pr(TL\cap CT)$

$\displaystyle 0.65=038+0.47-pr(TL\cap CT)$

$\displaystyle pr(TL\cap CT)=0.2$

B. Subscribes to either one or both magazines

Looking for $\displaystyle pr(TL\cup CT\cup(pr(TL\cap CT))$

$\displaystyle pr(TL\cup CT)=pr(TL)+pr(CT)-pr(TL\cap CT)+pr(TL\cap CT)$

$\displaystyle pr(TL\cup CT)=0.38+0.47-0.2+0.2=0.85$

C. Subscribes to only one of the two magazines

Looking for $\displaystyle pr(TL|CT')+pr(CT|TL')$

$\displaystyle \frac{pr(TL\cap CT')}{pr(CT')}+\frac{pr(TC\cap TL')}{pr(TL')}$

$\displaystyle \frac{pr(TL)-pr(TL\cap CT)}{pr(CT')}+\frac{pr(CT)-pr(TL\cap CT)}{pr(TL')}$
$\displaystyle \frac{0.38-0.2}{0.53}+\frac{0.47-0.2}{0.62}$

$\displaystyle \frac{0.18}{0.53}+\frac{0.27}{0.62}=\frac{2,547}{3 ,286}\approx 0.7751$

I'd love for someone to check my work and let me know if I'm doing this correctly or not. Thanks!

6. Originally Posted by downthesun01
I worked out problem 5 and I'd love for someone to look over my answers.

Teen Life = TL, Cool Teen = CT

$\displaystyle pr(TL)=0.38$
$\displaystyle pr(CT)= 0.47$
$\displaystyle pr(TL \cup CT)'=0.35$
$\displaystyle pr(TL \cup CT)= 1-pr(TL \cup CT)'=1-0.35=0.65$

A. Subscribes to both magazines

Looking for $\displaystyle pr(TL\cap CT)$

We know that:

$\displaystyle pr(TL\cup CT)=pr(TL)+pr(CT)-pr(TL\cap CT)$

$\displaystyle 0.65=038+0.47-pr(TL\cap CT)$

$\displaystyle pr(TL\cap CT)=0.2$

B. Subscribes to either one or both magazines

Looking for $\displaystyle pr(TL\cup CT\cup(pr(TL\cap CT))$

$\displaystyle pr(TL\cup CT)=pr(TL)+pr(CT)-pr(TL\cap CT)+pr(TL\cap CT)$

$\displaystyle pr(TL\cup CT)=0.38+0.47-0.2+0.2=0.85$

C. Subscribes to only one of the two magazines

Looking for $\displaystyle pr(TL|CT')+pr(CT|TL')$

$\displaystyle \frac{pr(TL\cap CT')}{pr(CT')}+\frac{pr(TC\cap TL')}{pr(TL')}$

$\displaystyle \frac{pr(TL)-pr(TL\cap CT)}{pr(CT')}+\frac{pr(CT)-pr(TL\cap CT)}{pr(TL')}$
$\displaystyle \frac{0.38-0.2}{0.53}+\frac{0.47-0.2}{0.62}$

$\displaystyle \frac{0.18}{0.53}+\frac{0.27}{0.62}=\frac{2,547}{3 ,286}\approx 0.7751$

I'd love for someone to check my work and let me know if I'm doing this correctly or not. Thanks!
Part A looks good. Parts B and C, it seems you didn't quite understand the quantities involved. You already found the answer to Part B at the beginning; it is 0.65. Then for Part C, that's simply 0.65 - 0.20 = 0.45. It's a lot simpler than you made it out to be. You had all the answers by part A already, you just didn't know it. You might want to revisit the Venn diagram and, on paper, draw your own circles and write in numbers to see how they all fit together.