Help with probability?

Question

On a test 4 students earned As, 2 students earned Bs, 2 students earned Cs, 1 student earned a D, and no students earned an F. If the teacher randomly selected three tests from the pile, what is the probability that she would select 1 A test, 1 B test, and 1 C test?
If you can help me how to set the problem up I would really appreciate it!

Answer 1

OK in the teacher's pile, there are 4 As, 2 Bs, 2Cs, 1D. Total of 9.

Let's work out the prob that she chooses an A, B and C in that order.

Initially there are 9 tests, 4 of them are As.
So Prob of first one being an A = 4/9

Now there are 8 tests remaining, 2 of them are Bs.
So prob of next one being a B = 2/8

Now, 7 tests remain, 2 are Cs
prob of next one being a C = 2/7

So the prob of all three of these happening is the product of all these probabilities
= 4/9* 2/8 * 2/7 = 2/63

But wait. the original quesiton did not say anything about the order. So we need to now find how many ways are there of arranging the A, B and C.
The answer is 3! because there are 3 of them, for a total of 6 possible arrangements, each with probability 2/63.

So the final answer = 6* 2/63 = 4/21

Answer 2

As I see it, you have 9 student papers. You take 3 of them, one at a time ,but do NOT replace them after you look at each paper. There are 6 ways to do this, in the order ABC, ACB, BCA, BAC, CAB and CBA. The probability of each combination is ABC: 4/9 * 2/8 * 2/7 = 2/63
ACB: the same
BCA: 2/9 * 2/8 * 4/7 the same
Actually all 6 would have the same prob= 2/63.
The sum of all 6 = 4/21 or slightly less than 1/5th.

Answer 3

Probability of selecting an A is 4/9, selecting a B is 2/9 and selecting a C is 2/9. (Assuming all three papers are selected together and not in an order). The total probability shd be..
4/9 * 2/9 * 2/9.

Answer 4

Answer=
Total = 4+2+2+1
= 9

A test =4/9 x 100 =
B test= 2/9 x 100 =
C test = 2/9 x 100 =


Just work out the answers from that - i'm aout 98% certain taht that is the correct way to do it

Answer 5

4A's
2B's
2C's
1D

there are 9 test total. Select 3 out of 9, the possible combination is ( 9x8x7)/3! = 84

find the favor combination of 1A, 1B and 1C
there 4 ways she can choose A, 2 ways she can choose B's and 2 way she can choose C's

4 x 2 x 2 = 16

probability: 16/84 = 4/21

Answer 6

P{1A,1B,1C}= [(4c1) (2c1) (2c1)] / (9c3) = 4/21

I hope you are familar with the (n choose k) format.

(nCk) = n!/[k! (n-k)!]

Out of the 4 A's, you choose 1, out of the 2 B's, you choose 1, out of the 2 C's, you choose one. And out of the 10 total grades, you choose 3. Thats why you divide by that. Think of it as if all of the grades are in a bucket and you are picking them out randomly.

The other people who say to muliply fractions are wrong, except for the guy who mentioned that they dont have to be in a specific order. If you muliply fractions, that means the first grade you pick MUST be an A, but in reality, the first grade you pick can be an A, B, or C. That is why this is the right way.

Answer 7

IF THE TESTS ARE PUT BACK AFTER EACH PICK:
A-4, B-2, C-2, D-1
Total: 9
P(A,B,C) = 4/9 x 2/9 x 2/9 = 16/729

IF THE TESTS ARE NOT PUT BACK AFTER EACH PICK:
A-4, B-2, C-2, D-1
Total: 9
P(A,B,C) = 4/9 x 2/8 x 2/7 = 2/63