1. ## Random Needs

Professor Jackson is in charge of a program to prepare people for a high school equivalency exam. Records show that 80% of the students need work in Math, 70% need work in English, and 55% need work in both areas. Compute the probability that a student selected at random needs:

(a) Math or English
(b) Math only
(c) English only
(d) Both
(e) None

2. ## Re: Random Needs

Originally Posted by joshuaa
Professor Jackson is in charge of a program to prepare people for a high school equivalency exam. Records show that 80% of the students need work in Math, 70% need work in English, and 55% need work in both areas. Compute the probability that a student selected at random needs:

(a) Math or English
(b) Math only
(c) English only
(d) Both
(e) None
a) $\mathcal{P}(M\cup E)=\mathcal{P}(M)+\mathcal{P}(E)-\mathcal{P}(M\cap E)$

b) $\mathcal{P}(M\cap \neg E)=\mathcal{P}(M)-\mathcal{P}(M\cap E)$

$\mathcal{P}(\neg E)$ is probability of not English.

3. ## Re: Random Needs

Thanks Plato. What about (d) and (e)?

4. ## Re: Random Needs

Originally Posted by joshuaa
Thanks Plato. What about (d) and (e)?
This is your homework. You should do it for yourself.

Now if you can do parts a & b, with my hints, then you can do the others.
Because, those are answered in parts a & b.

Give it a try and show us,

5. ## Re: Random Needs

it is obvious that (d) is 0.55 and I got it from the Question itself (55% need work in both areas), then (e) must be 0.45, but I don't understand why the book's answer is 0.44. This is why I was doubting in (d) & (e)

6. ## Re: Random Needs

Originally Posted by joshuaa
it is obvious that (d) is 0.55 and I got it from the Question itself (55% need work in both areas), then (e) must be 0.45, but I don't understand why the book's answer is 0.44. This is why I was doubting in (d) & (e)
e) $1-\mathcal{P}(M\cup E)$
Can you explain why?

7. ## Re: Random Needs

If you mean to explain why the answer of (e) looks like that, it is difficult to tell you now. I have to mimic the basic rules first!

Thanks a lot for the help and I am sure the 0.44 in the book is mistyped!

8. ## Re: Random Needs

Suppose there were a total of 100 students. 55% of them, 55 students, need work in both English and Math. 80% of them, 80, need work in Math. Since 55 also need work in English, 80- 55= 25 need work only in math. 70%, 70 students, need work in English but 55 also need work in Math, so that leaves 70- 55= 15 need work only in English. That's a total of 55+ 25+ 15= 95 students who need work in Math or English or both. 100- 95= 5 students do not need either.

The percentage who need work in
(a) Math or English is 95/100= 0.95 or 95%.

(b) Math only is 25/100= 0.25 or 25%

(c) English only is 15/100= 0.15 or 15%

(d) Both is 55/100= 0.55 or 55%

(e) None is 5/100= 0.05 or 5%

9. ## Re: Random Needs

Thanks a lot HallsofIvy. Very clear.