Binomial Distribution Practice: Part A Solution & Part B Explanation

[planauts]

Homework Statement

http://puu.sh/dOcM
Answer:
http://puu.sh/dOcZ

Homework Equations

The Attempt at a Solution

I got Part A.
For part A, this is what I did:

I did Egg A: X ~ (6,(1/6)) P(X = 1) and did something similar for Egg B. I then multiplied both to get the answer for Part A.
http://puu.sh/dOfV

For Part B, I'm a bit confused. I tried doing cases. As in:
(broken eggs): AA , BB , AB (same as BA)
Then I did:
AA: X ~ (6,(1/6)) P(X = 2)
BB: Y ~ (6,(1/10)) P(X = 2)
AB: X ~ (6,(1/6)) P(X = 2) * Y ~ (6,(1/10)) P(X = 2)

My work:

Spoiler

http://puu.sh/dOh8

And then, I tried multiplying and adding the three values. But I didn't get the correct answer.

Could someone help me out please? Thanks

 

[Ray Vickson]

If two type A are broken then we must have 0 type B broken in order to have two broken altogether.

RGV

 

[planauts]

Wow, thanks. I tried it out. It works :) thanks.