you have a class with 40 students — 10 freshmen, 12 sophomores, and 18 juniors.
You pick two students at random, one at a time. What is the probability that both are freshmen?
Hello, delgeezee!
You have a class with 40 students: 10 freshmen, 12 sophomores, and 18 juniors.
You pick two students at random, one at a time. What is the probability that both are freshmen?
One Method
$\displaystyle P(\text{1st is Freshman}) \:=\:\frac{10}{40}$
$\displaystyle P(\text{2nd is Freshman}) \:=\:\frac{9}{39}$
$\displaystyle \text{Therefore: }\:P(\text{both Freshmen}) \:=\:\frac{10}{40}\cdot\frac{9}{39} \:=\:\frac{3}{52}$
Another Method
There are: .$\displaystyle _{40}C_2 \:=\:\frac{40!}{2!\,38!} \:=\:780$ possible outcomes.
We want 2 of the 10 Freshmen: .$\displaystyle _{10}C_2 \:=\:\frac{10!}{2!\,8!} \:=\:45$ ways.
$\displaystyle \text{Therefore: }\:P(\text{both Freshmen}) \:=\:\frac{45}{780} \:=\:\frac{3}{52}$
The way I would have done this: You start with 40 students, 10 of whom are freshmen. Assuming every student is equally likely to be chosen (unstated but typically assumed) the probability that the first student you choose is a freshman is 10/40= 1/4. If you do in fact choose a freshman, you now have 39 students, 9 of whom are freshmen. (And, since in order that we "choose two freshmen", we must have chosen a freshman the firt time, we can assume that.) The probability, then, that the second chosen is also a freshman is 9/39. The probability that the two students chosen are both freshmen is (1/4)(9/39)= 9/156.