Question

see below

Answer 1

Have you learned how to find binomial probabilities?

It looks like \[\Large (nCr) p^k q^{n-k}\]or http://www.mathwords.com/b/binomial_probability_formula.htm

Answer 2

like im confused.. which to put for which... nine =n? then 3 brown= p?

Answer 3

so isit (9/3)(0.21^3)(0.79^6)?

Answer 4

If you meant 9C3, that looks correct.

Answer 5

sorry what is the C?

Answer 6

Choose. 9 choose 3.

Answer 7

so my above formula should be correct?

Answer 8

Yes, as long as you mean 9 choose 3 (combinations).

Answer 9

okay so the answer would be:0.00675?

Answer 10

No, that's too small.. show your work.

Answer 11

(9/3)(0.21^3)(0.79^6)

Answer 12

Again, it's not 9 divided by 3, it's 9 choose 3, do you know how to do combinations? http://www.mathwords.com/c/combination_formula.htm

Answer 13

okay let me try to figure it out

Answer 14

http://www.mathwords.com/b/b_assets/binomial%20coefficient%20formula.gif

Answer 15

\[\Large 9C3 = \frac{ 9! }{ 3! (9-3)! }\]

Answer 16

okay so because its exactly so there is only one of the 9c3..? like i dont have to do 9c2+ 9c1?

Answer 17

ans=0.1891?

Answer 18

Correct. If at was "at most 3" that means you'd have to do four calculations, for 0, 1, 2 and 3, like so
(9C0)(0.21^0)(0.79^9) + (9C1)(0.21^1)(0.79^8) + (9C2)(0.21^2)(0.79^7) + (9C3)(0.21^3)(0.79^6) =

And correct!

Answer 19

thanks.. how about for this
(b) If you randomly select 15 m&m's, what is the probability that less than three of them are green?

Answer 20

less than 3 so does it mean i will have to take 15c3 & 15c2, 15c1?

Answer 21

Less than 3 means it has to be LESS than 3, so you can't include 3. Remember you can choose zero... You'll need a 15C0, 15C1, 15C2.

Answer 22

okay thanks.. :)

Answer 23

(c) If you randomly select 175 m&m's, what is the approximate probability that at least 18% of them are orange?

how about this?

Answer 24

For that, I think you need to use the Normal approximation to the binomial probability...