Question

A population in a certain town is 60% female. You select a 12-person jury from this population at random.

What’s the probability that 5 or fewer of the jury members are female?: *
a. 0.0573
b. 0.1009
c. 0.1582
d. 0.3348

Answer 1

Hi, I tried to solve this but I did not get any of the answer choices.

Answer 2

binomial for this one

Answer 3

Okay, let me try and solve it with your hint. Please do not give any more help. Thanks!

Answer 4

k i will shut up

Answer 5

lol

Answer 6

Okay, binomial distribution says the probability of getting exactly n successes is (n C k)*(p^n)*(p^n-k), where p = probability of success, and n and k are parameters.

Answer 7

For this problem n =15, and k <= 5.

Answer 8

So, I believe p(k<=5) = p(k=0) + p(k=1) + p(k=2) + p(k=3) + p(k=4) + p(k=5).

Answer 9

There might be a faster way to do that...

Answer 10

Correction, binomial distribution is (n C k)*(p^k)*(p^n-k).

Answer 11

Okay, here I go.

Answer 12

p(k=0) = (12 C 0) * (.6)^0*(.4)^12 =0.00001677721

Answer 13

p(k=1) = (12 C 1) * (.6)^1*(.4)^11 = .000301989888

Answer 14

p(k=2) = (12 C 2) * (.6)^2*(.4)^10 = .0024914166

Answer 15

p(k=3) = (12 C 3) * (.6)^3*(.4)^9 = .0124570829

Answer 16

p(k=4) = (12 C 4) * (.6)^4*(.4)^8 = .0420426547

Answer 17

p(k=5) = (12 C 5) * (.6)^5*(.4)^7 = .0124570829

Answer 18

Where (n C k) means "n choose k".

Answer 19

p(k<=5) = p(k=0) + p(k=1) + p(k=2) + p(k=3) + p(k=4) + p(k=5)
= 0.00001677721 + .000301989888 + .0024914166 + .0124570829 + .0420426547 + .0124570829
= 0.06976700419

Answer 20

Okay, now I need help!

Answer 21

i can't think of a faster way

Answer 22

The answer I am getting is not one of the answer choices. Did I make a mistake in my work?

Answer 23

Correction to the binomial theorem, it should read (n C k)*(p^k)*((1-p)^(n-k))

Answer 24

\[P(X=k) = \left(\begin{matrix}n \\ k\end{matrix}\right) p^k (1-p)^{n-k}\]

Answer 25

http://www.wolframalpha.com/input/?i=.4^12%2B12*.6*.4^%2811%29%2B66%28.6%29^2*.4^10%2B220*.4^3*.6^9%2B495*.6^4*.4^8%2B792*.6^5*.4^7

Answer 26

If they can helps us then they can has a medal.

Answer 27

sorry it took me a while

Answer 28

That's alright, let me take a look at what you posted.

Answer 29

i just added them up

Answer 30

isnt there a better way than to add up all the probability like that

Answer 31

no women plus one woman plus two women plus 3 women plus 4 women plus 5 women

Answer 32

if there is, and i am not saying that there isn't, i don't know it

Answer 33

Was there an error in your input?

Answer 34

.4^12+12*.6*.4^(11)+66(.6)^2*.4^10+220*.4^3*.6^9+495*.6^4*.4^8+792*.6^5*.4^7

Answer 35

maybe it is hard to do all of these on the computer

Answer 36

Should the fourth term be 220*.6^3*.4^9?

Answer 37

if there is, i don't see it

Answer 38

oh yes cebroski that is right

Answer 39

Corrected, I think it would look like this:
.4^12+12*.6*.4^(11)+66(.6)^2*.4^10+220*.6^3*.4^9+495*.6^4*.4^8+792*.6^5*.4^7

Answer 40

oh you are right!! i mixed up the .4 and .6!!

Answer 41

http://www.wolframalpha.com/input/?i=.4^12%2B12*.6*.4^%2811%29%2B66%28.6%29^2*.4^10%2B220*.6^3*.4^9%2B495*.6^4*.4^8%2B792*.6^5*.4^7

Answer 42

as always, the answer is C

Answer 43

C for cebroski!

Answer 44

Thanks y'all for your help. My mistake was that I was not using n C k as a coefficient.

Answer 45

Now, I understand, the reason we need n C k is because there are n C k ways to choose k women for a 12 person jury.

Answer 46

I learn good.