Binomial Distribution

  • Thread starter planauts
  • Start date
  • #1
86
0

Homework Statement



http://puu.sh/dOcM [Broken]
Answer:
http://puu.sh/dOcZ [Broken]

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 [Broken]

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:
http://puu.sh/dOh8 [Broken]


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
 
Last edited by a moderator:

Answers and Replies

  • #2
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728
If two type A are broken then we must have 0 type B broken in order to have two broken altogether.

RGV
 
  • #3
86
0
Wow, thanks. I tried it out. It works :) thanks.
 

Related Threads on Binomial Distribution

  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
12
Views
953
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
1
Views
985
  • Last Post
Replies
1
Views
4K
  • Last Post
Replies
1
Views
796
  • Last Post
Replies
11
Views
5K
  • Last Post
Replies
1
Views
758
Top