# Math Help - binomial expansion,combinations and probability

1. ## binomial expansion,combinations and probability

Hi; I'm having trouble matching up my combinations probability and binomial expansion probability I have 16 marbles 5 red, 11 blue and I want to know the probability of getting 2 red and 2 blue when four are picked.for the binomial I have (5/16 + 11/16)^4 expanding to 1(5/16)^4 + 4(5/16)^3(11/16) + 6(5/16)^2(11/16)^2which gives 6(5/16)^2(11/16)^2 = 0.2769 Now for the combination (5c2)(11c2)/(16c4) = 0.3022Don't understand why they have different answers?Thanks.

2. ## Re: binomial expansion,combinations and probability

Hello, anthonye!

I have 16 marbles: 5 red, 11 blue.
I want to know the probability of getting 2 red and 2 blue when four are picked.

For the binomial I have: $(\tfrac{5}{16} + \tfrac{11}{16})^4 \;=\;1(\tfrac{5}{16})^4 + 4(\tfrac{5}{16})^3(\tfrac{11}{16})^1 + 6(\tfrac{5}{11})^2(\tfrac{11}{16})^2 + \cdots$

which gives: . $6(\tfrac{5}{16})^2(\tfrac{11}{16})^2 \:=\: 0.2769$ . ← This is incorrect.

Now for the combination: . $\frac{(_5C_2)(_{11}C_2)}{_{16}C_4} \:=\: 0.3022$

Don't understand why they have different answers.

Using the Binomial formula, the events are independent.
You are drawing the four marbles with replacement.

3. ## Re: binomial expansion,combinations and probability

sO how do I do it?

4. ## Re: binomial expansion,combinations and probability

Originally Posted by anthonye
sO how do I do it?
You don't use the Binomial Formula, as it implies that the events are independent. As Soroban eluded, drawing marbles without replacement gives events that are NOT independent, so you simply cannot apply the Binomial Formula.

5. ## Re: binomial expansion,combinations and probability

There are 5 red and 11 blue marbles so the probability the first marble drawn is 5/16. Then there are 4 red and 11 blue marbles so the probability the second marble drawn is 4/15. Then there are 3 red and 11 blue marbles so the probability the third marble drawn is blue is 11/14. Then there are 3 red and 10 blue marbles so the probability the fourth marble drawn is blue is 10/13. The probability the four marbles drawn are "red, red, blue, blue" in that order is (5/16)(4/15)(11/14)(10/13). If you do the same thing for different orders, you get different fraction but the same numerators and denominators so the same product. There are 4!= 4(3)(2)(1)= 24 different orders.

Ok thanks.