Here's the question you clicked on:
jesusfreak
the principal will randomly choose 6 students from a large school to represent the school in a newspaper photograph. the probability that a chose student is an athlete is 30%. (assume that this doesn't change) What is the probability that 4 athletes are chosen?
Is this for a statistics class, or something like algebra?
YTheManifold: that's not actually correct because you're not including the degree of freedom that any 4 can be chosen
Would you like to see what the possible answers could be?
Sorry \[ {6\choose 4}\cdot 0.3^4\cdot 0.7^2 \] of course
.3^4 * .7^2 is the p^k and p^(n-k) but you need the (6 choose 4) as well.
the formula YTheManifold just posted is correct - can you compute that, jesusfreak?
yes - this a Binomial Probability distribution
the general form is (n choose k) * p^k * p^(n-k). (typo in the previous one)
It all looks like gibberish. I have no idea what to do.
The answers it gives is 0.05, 0.06, 0.07, 0.08
OK, let's go through it step by step. Do you know how to compute .3^4?
th 6 4 part means the number combinations of 4 from 6
You forgot the decimal -- it's 0.3^4 we're computing. Then multiply that by 0.7^2.
ok it equals .003969
yes! Now you just need to multiply that by the (6 4) part, and you'll be done. That's called a "binomial". It's pronounced "6 choose 4", which means, if you have 6 things, how many ways can you choose 4 of them?
There's a formula for computing that, but a lot of people just type it into a calculator - it's 15 in this case.
The formula is n! / ( k! * (n-k)! )
Anyway so in this case just multiply 15 * .3^4 * .7^2 and that's your answer.