I was given the following problem on a math test, which I got wrong:
There is a group of 10 high school students, composed of 3 seniors, 4 juniors, and 3 underclassmen.
Given that a group of 5 of them is selected randomly, what is the chance that the group is composed of 2 juniors, 2 seniors, and 1 underclassman.
Here's how I solved it, and it makes sense to me.
I started out with the chance of getting 1 senior, which is 3/10. Then I multiplied that by the probability of getting a second senior (2/9). The reason the numerator and the denominator both decreased is because you've taken 1 senior, and therefore 1 person, out of the selection pool. Then I multiplied that by the chances of getting a junior, then another, which are 4/8 and 3/7 respectively. Finally, the chance of getting an underclassman is 3/6 at the end, completing the equation.
Shouldn't the final probability be (3/10)*(2/9)*(4/8)*(3/7)*(3/6)=1/140, which accounts for 2 seniors, 2 juniors, and 1 underclassman?
I'd appreciate any advice! Thanks!
There is a group of 10 high school students, composed of 3 seniors, 4 juniors, and 3 underclassmen.
Given that a group of 5 of them is selected randomly, what is the chance that the group is composed of 2 juniors, 2 seniors, and 1 underclassman.
Here's how I solved it, and it makes sense to me.
I started out with the chance of getting 1 senior, which is 3/10. Then I multiplied that by the probability of getting a second senior (2/9). The reason the numerator and the denominator both decreased is because you've taken 1 senior, and therefore 1 person, out of the selection pool. Then I multiplied that by the chances of getting a junior, then another, which are 4/8 and 3/7 respectively. Finally, the chance of getting an underclassman is 3/6 at the end, completing the equation.
Shouldn't the final probability be (3/10)*(2/9)*(4/8)*(3/7)*(3/6)=1/140, which accounts for 2 seniors, 2 juniors, and 1 underclassman?
I'd appreciate any advice! Thanks!
