[Math] Probability

I'm usually pretty good in math, but i'm having trouble with probability. I've asked my teacher for some help, and I just can't seem to get it.
So, I have a sample of a question we have to do. Can anyone walk me through it?

If no repetition is allowed and using the digits 0,1,2,3,4,5 how many three digit odd numbers can be formed?

Any takers?

That's not really probability, but okay.

The total is 60 possible 3 digit odd numbers with no repeats (assuming you can have 013, for instance).


First of all, there are a total of 6 numbers that can be used, and they cannot be repeated. In a basic sense, that means that there are 6 possible variations for each of the three digits in your number. When you use one digit, that means that there are five possible combinations for the next, and then four for the third.

But! There is the stipulation "odd numbers" in that question, it's best to look at that first. There are 3 odd numbers in the set, so the 1s place will always have one of those three.

For the other two, it's the same as the basic explanation. Since there are no repeats, and one of your digits was used in the 10s place, that means there are 5 that can be used in the next, and 4 after that.


...But where does 60 come from? Simple.

These values that I explained get multiplied together. 3 * 5 * 4 = 60.

Said another way, for each of the 3 odd numbers, there are 5 other numbers that can be put in the next place, and for each of those 5, there are 4 more that can be used. That does take in to account that you could have 123 and 213.

Thanks alot. You really helped me figure out how to do that.

I have two other questions that i've been desperately trying to figure out over the past 3 days. It would be greatly appriciated if you, or anybody else could show me how to do it.

1) In a certain school, 15% of the students failed math. 5% failed math and english, and 33% faild math or english. What is the probability that a student failed english?


and

2) To develop the mathematics department of a new branch of a university, the newly elected head of the department must choose 2 full professors from among 6 applicants, 2 associate professers from 10, and 6 assistant proffessors from 16 applicants. In how many different ways can she make this choice?

1) Given the ratio of students who failed Math or English you need to subtract the ratio of students who failed Math and not English to determine the number that failed English in general. To determine the size of this group (M not E), find the difference between the ratio of students failing math in general and those failing Math and English. Your final determination is 33%-(15%-5%) = 23%.
The easiest way to see this for yourself may be to attempt to populate a Venn diagram with 33 units. If you want to check your work:


18% fail English Only
10% fail Math Only
5 % fail both
23% fail English in total
15% fail Math in total
33% fail one or the other

2) This is a basic combinatorics calculation. Put the numbers into the equations you already know:
p = (6 choose 2) * (10 choose 2) * (16 choose 6) = 5,405,400.

But 33 - 15- 5 = 13. How did you get 23?

I didn't say 33 - 15- 5.
It's 33-(15-5) = 33 -15+5 = 33 - 10 =23; watch dem parentheses...

Oh, right. Well, thanks for the help. Hopfully I don't fail the test on monday.