In a class of 100 students, 40 are computer science majors, 36 are mechanical engineering majors, 18 are civil engineers and the rest are general engineering majors. Assume students only have one major.
If a student is chosen at random what is the probability they are: (Round all answers to two decimal places)

a) a civil engineering major or mechanical engineering major?

b) Suppose five students from the class are chosen at random. What is the probability none are mechanical engineering majors?

Question

In a class of 100 students, 40 are computer science majors, 36 are mechanical engineering majors, 18 are civil engineers and the rest are general engineering majors. Assume students only have one major. 
If a student is chosen at random what is the probability they are: (Round all answers to two decimal places)

a) a civil engineering major or mechanical engineering major?

b) Suppose five students from the class are chosen at random. What is the probability none are mechanical engineering majors?

raffle_snaffle · Answer

@tkhunny

raffle_snaffle · Answer

For a is it P = (36+18)/100 = 0.54

tkhunny · Answer

We're just counting.

(18+36)/100  --  Why is it harder than that?

Second is a little trickier.  Shall we approximate or produce the exact value?

bahrom7893 · Answer

Probability that 5 chosen at random and non are mech eng:
1/(100-36) * 1/(99-36)...
Am I right?

bahrom7893 · Answer

Sorry it's been a while, wanted to post before seeing tk's solution

raffle_snaffle · Answer

LOL because I did this problem and I got it wrong... I added 6 the first time byt now I see why my other answer makes more sense.

raffle_snaffle · Answer

I don't know if you are right.

raffle_snaffle · Answer

I don't have the answers.

jim_thompson5910 · Answer

`36 are mechanical engineering majors`
so 100-36 = 63 are majored in something else

x = 63 C 5 = number of ways to pick 5 non-mechanical engineering students
y = 100 C 5 = number of ways to pick any 5 students

the answer will be in the form x/y

I'm using this combination formula (n C r)

$$\Large _n C_r = \frac{n!}{r!*(n-r)!}$$

jim_thompson5910 · Answer

sorry not 63, made a typo

should be 64

raffle_snaffle · Answer

Where did you get that equation? *looking in text book*

bahrom7893 · Answer

@jim_thompson5910 and sorry for turning your question into a discussion @raffle_snaffle , but how would I approach this if I didn't remember the formulas?

jim_thompson5910 · Answer

look for the combinations or permutations section

raffle_snaffle · Answer

I don't know @bahrom7893

bahrom7893 · Answer

(taking probability seminar in a week, need to review all these things myself and I have a terrible memory for formulas)

raffle_snaffle · Answer

I don't think we have covered that formula yet... Can I use Binomial distribution or Possion DIstrbution?

jim_thompson5910 · Answer

think of 5 slots (A,B,C,D,E)

if you have 100 people to choose from

100 choices for slot A
99 for slot B
98 for slot C
97 for slot D
96 for slot E

multiply those choices out 100*99*98*97*96 = 9,034,502,400

9,034,502,400 is the number of ways to pick 5 out of a pool of 100 where order matters. This is equal to 100 P 5

since order doesn't matter, you divide by 5! = 120

9,034,502,400/120 = 75,287,520


this is exactly equal to 100 C 5

jim_thompson5910 · Answer

I divided by 5! = 120 because there are 120 ways to arrange 5 items

eg: ABCDE is the same as ABCED

raffle_snaffle · Answer

Okay @jim_thompson5910

triciaal · Answer

why wouldn't you use 64?

bahrom7893 · Answer

Nvm I just saw your example

raffle_snaffle · Answer

I am going to ask my instructor next week since I have one more attempt for this problem. Thanks for the help everyone.