need help understanding and answering the 4 questions.
got 100fg ft if that helps?

Nn fg just pm me and I can help explain and solve these for you.

Venn diagrams are quite simple. Each area (any white space enclosed by lines) represents some sort of a population.
a.) You want the probability that you are outside the "Men" circle OR inside the "Engineering" circle. In this case we know the outside of both circles (0.37), so we can just add that to the engineering circle (0.59) and we get 0.96. Another way to think about it is that the people we DON'T want are the non-engineering males. Those are on the diagram in the left moon-shaped piece that says '0.04'. So we want everyone else, that is 1- 0.04 = 0.96

b.) This is just one minus the probability of being a man. 1 - 0.6 = 0.4

c.) This is ambiguous. What they are saying is that if they only pick from engineering students, what's the probability of picking a woman. In that case this venn diagram no longer applies and the answer doesn't really matter for (d.). So it's most likely a very poorly worded question and what they really want to ask is "What's the probability of picking an engineering student who is a woman". This is all engineering students that don't overlap with "Men". That is, they belong to the group "engineers" but they don't belong to the group "men". From the diagram we see it's 0.03
If they actually ARE asking what they are asking, then we need to take the fraction of female engineers and divide that by the fraction of students that are engineers. We are basically normalizing our 0.59 fraction of engineering students to 1. So 0.03 / 0.59 = 0.051. Most likely this isn't what they want, so whoever wrote that question should get a talking to.

d.) Events are independent when the probability of both occurring together (a student is a female AND an engineer) is the same as the probabilities of each event multiplied. We found the probability of both events together in (c) and it was 0.03. The probability of a student being female is 0.4 (from b.) and the probability of a student being an engineer is given as 0.59. Multiply: 0.59 * 0.4 = 0.236. This isn't the same as 0.03, so the events are NOT independent.

nn fg

This post was edited by russian on Oct 26 2015 04:16pm

Your C is wrong. 3% of the total school population is a woman engineer, but (3/59) = 5.08% of the engineering population is a woman.

Yeah, I edited to add that part, but it's most likely not what they want.

Agreed. I always despised when you'd get questions like this on standardized tests or some buillshit so you couldn't ask the teacher if they fucked up or not..

Answer to c) is indeed 3/59.

So you can easily conclude for d) : since 3/59 is not 40%, then the events are not independant.

The way russian solved it is just another way (clearly equivalent). But nothing wrong in the text...

you guys have all been very very helpful! thank you so much. and if anyone wants the fg for helping just let me know, or I got a few more problems i wouldnt mind a sharp eye looking over and can send fg your way aswell!




here is my work:

A) 1360 / 4000 = .34
The percentage of orders from catalog users is 34% of total orders.

B ) 1826 + 33 / 4000 = .465
The percentage of orders from email or are major purchases are about 46.5% of total orders

C) The events are dependent. (this was a guess, wasn't sure how to determine on this one)

I understand the concept of dependent and independent but not sure how to determine with more complex examples as opposed to A B C > A pulled out w/o replacement > B C are left to pick.

This post was edited by Trig on Oct 26 2015 09:15pm

#3


this is my work:

3.
A) .54 + .24 +.06 = .84 or 84% of acceptable bids.
B ) .06 * .04 / .06 = .04 or 4% chance the NA bid came from the group ‘occasional’
C) .06 * .06 / .04 = .09 or 9% chance the NA bid came from the group ‘1st time’

am I making the proper calculations to determine the chance of probability for b and c?

This post was edited by Trig on Oct 26 2015 09:24pm

B would be .06 / (.06 + .06 + .04) = 37.5%
The numerator is the percentage of occasional NA, and the denominator is the total percentage of NA.

Your c is wrong as well, it would be done the same way as I just showed except the numerator will be different.

This post was edited by RzChaos on Oct 27 2015 06:43am