The rule for binomial distribution is this:
Pr(X=x)=nCx * p^x * q^(n-x)
Pr(X=x) is 'the probability that X is x' and in this case our X (random variable) is the number of extroverted professors out of 6 professors that are randomly chosen.
n is the sample space we're dealing with (I'm not sure sample space is the right word) but it's basically the number of things you are trialling to achieve the random variable) so in this case it's 6.
p is the probability of success, and q is the probability of failure. in these questions, success and failure refer to the probability of achieving your required outcome vs not achieving it. so in this case, the probability of success (p) is 0.45 (45% of professors are extroverted).
The probability of failure is always the probability of not succeeding. this means that q is (1-p) because it's what is left over if you havent succeeded (you can't fail and succeed at the same time)
so in our case, q=1-p=1-0.45=0.55 or 55% of professors are not extroverted. this makes sense, if 45% are extroverted then that leaves 55% to be not extroverted.
Also, nCr or in this case nCx is a 'combination' and is read 'n choose x'. it can also be written as n over an x in a bracket. the formula for nCr is n!/[r!(n-r)!] and it basically gives you the 'number of ways to sort n things into r places if you don't care about the individual order. (if you do care about order, eg you care whether professors 1 and 2 are extroverted or introverted, which we don't in methods as far as i know, you want nPr or permutations. i recommend looking up some probability khanacademy videos if you aren't following this combination/permutation stuff)
oh yeah and little x, that's the number of successes you want
Okay, onto the first question.
What is the probability thta all six are extroverts?
we want 6 successes and 0 failures.
again, the formula is Pr(X=x)=nCx * p^x * q^(n-x)
x is 6 because we want 6 successes, n is 6 because we're choosing 6 professors at random (note this makes n-x=0 which is because we want 0 failures) this makes the formula
Pr(X=6)=6C6 * p^6 * q^0
p is the probability of success, 0.45 and q is 1-p = 0.55
pr(X=6)=6C6 *0.45^6 *0.55^0
now, 6C6 (6 choose 6) is the number of ways to sort 6 things into 6 places. there's only one way, if you dont care about order, which we dont. you could use the formula, but it would give you 1. normally it wont be 1 but at the extremes (when you're looking for all successes or all failures) it is 1.
that makes the equation this:
Pr(X=6)=1 * 0.45^6 * 1 (a^0 = 1)
=0.45^6
=0.008304...........
I AM UNSURE HOW MANY DECIMAL PLACES TO ROUND THIS TO IF YOU ARE NOT TOLD TO ROUND IT TO A CERTAIN NUMBER SOMEBODY PLEASE CONFIRM?
okay, next one:
What is the probability that none of your professors is an extrovert?
0 successes, 6 failures
Pr(X=0)=6C0 * p^0 * q^(6-0)
=1 * 1 * 0.55^6
=0.55^6
=0.027681...........
this should make sense, it's almost exactly the same as the one before it in terms of working out. if you are going to use the rule for nCr to try to work out 6C0, it's important to note that 0!=1
third one is a little different
What is the probability that at least two of your professors are extroverts?
what is the probability that the number of professors who are extroverts is GREATER THAN OR EQUAL TO 2
this means what is the probability that x=2, 3, 4, 5 or 6
too easy, Pr(X=2) + Pr(x=3) + .... + Pr(X=6)
Yeah I dont want to work all that out either.
a shortcut is to work out the probability of there NOT being 2 or more extroverts, then subtracting that from 1 (to figure out the probability of there not being less than 2 extroverts which is the same as the probability of there being 2 or more extroverts)
so we can instead work it out as follows:
Pr(X>=2)=1-[Pr(X=0)+Pr(X=1)]
^greater than or equal to
We already know Pr(X=0) so let's find Pr(X=1)
1 success, that means 5 failures
Pr(X=1)=6C1*p^1*q^5
6C1 is 6, try the formula if you dont believe me, or think; how may ways can you sort 1 thing into 6 places? 6 ways. one for each place.
Pr(X=1)=6*0.45^1*0.55^5
=0.135887....
SO Pr(X>=2)=1-0.027681-0.135887
=0.836432.......
FINALLY, the last one
In a group of six professors selected at random, what is the expected number of extroverts? What is the standard deviation of the distribution?
okay, each time you chose a professor at random, you've got a 45% chance of getting and extrovert. does it kind of make sense that each professor you choose is going to average out to be 0.45 of an extrovert? the mean/the greek letter mu (mew)/the expected number of favourable outcomes for a binomial distribution is np (the number of trials times the probability of success)
for this one, that is 6*0.45=2.7
that means if you run the experiment infinite times you will average out to get 2.7 extroverts per group of 6 professors.
as for the standard deviation, there's a formula that gives us the variance of a binomial distribution and that is v (the actual symbol used is sigma squared) = npq
npq=number of trials * probability fo success * probability of failure
=6*0.45*0.55=1.485 or 297/200
that's not what the question aksed for, though. you need the standard deviation (sigma)
s=squareroot(v)
=squareroot(npq)
=sqrt(1.485)
=1.21861...... again, not sure about the decimals
Hopefully my answers are right... and hopefully that made some sense. again, hit up khanacademy for some good combinatorics (nCr and nPr) explanations and I think sal even has a section on binomial theorems so that might help too
edit: just noticed this was in the mathematics board rather than the methods board, so I guess it could be a further question or something? in that case, ignore what I said about permutations not happening because I have no idea if further covers those or not
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