probability?

Question

There are 50 students with different grades. If two students are selected at random, what is the probability that the first student will have a B average and the second student will hav a C acerage?

Need help! Thanks

Answer 1

It would be a 49 to 1 chance.
Here's how:
You have 50 students, you pick one (which is a "50 to 1" chance because you have 50 students) with a B average. You pick out of the remaining 49 students the one with the C average (49 to 1) because you do not know how many students hae a C average to begin with. If you did know who had a C average, you would't be picking "at random".

Answer 2

Assumming there are 5 different grades: A, B, C, D, and F.
Also assumming 10 out of 50 students get each grade. So 10 students get A, 10 get B, and so on.
Pr(first student is a B student AND second student is a C student) = Pr(first student is a B student) * Pr(second student is a C student) = (10/50) * (10/49) = 2/49 = 0.0408 or about 4%

Answer 3

I'm assuming that there are five grades, ABCDF, and ten students got each.

There is a 10 in 50 chance the first student got a B. If this is true, then there is 10 in 49 (one B student removed from the pool) the second student had a C.

(10/50)(10/49)=100/2450=10/245

If the distribution of grades is different, replace the first 10 with the number of Bs, and the second with the number of Cs.

Answer 4

You were probably given specific grades or a distribution that shows the percentage of students who got each grade. You don't give that information here, but you'll need to know it to do the problem. To solve the problem, you'd take
(# who got B's) / 50 X (# who got C's) / 49

That's because "and" means "times" in probability. You take the probability the first got an A times the probability the second got a C.

Answer 5

A 5% chance.