# Thread: Choose eggs/rotten batch and good batch

1. ## Choose eggs/rotten batch and good batch

Thank you for your anticipated help!

You have 12 eggs, 2 of which are rotten, and you choose 4 randomly. Afterwards, 4 more are randomly chosen from the remaining 8. Find the conditional prob. that the second batch of four were all good, given that at least one of the first four was rotten.

THANKSSSS!!!

2. Hello, bjanela!

I had to baby=talk my way through this one . . .

You have 12 eggs, 2 of which are bad, and you choose 4 randomly.
Afterwards, 4 more are randomly chosen from the remaining 8.

Find the conditional probbability that the second batch of four were all good,
given that at least one of the first four was bad.
We are concerned with two outcomes in the first draw:
. . 1. One bad egg was drawn.
. . 2. Both bad eggs were drawn.

Case 1: One bad egg was drawn.

There are 12 eggs: 2 bad $\displaystyle (B)$ and 10 good $\displaystyle (G).$
There are: .$\displaystyle {12\choose4} \:=\:495$ possible outcomes.
To get 1 B and 3 G, there are: .$\displaystyle {2\choose1}{10\choose3} \:=\: 240$ ways.
. .Hence: .$\displaystyle P(1B,\text{first draw}) \:=\:\frac{240}{495} \:=\:\frac{16}{33}$

For the second draw, there are: 1B, 7G.
There are: .$\displaystyle {8\choose4} \:=\:70$ possible outcomes.
To get 4G, there are: .$\displaystyle {7\choose4} = 35$ ways.
. . Hence: .$\displaystyle P(\text{4G, 2nd draw}) \:=\:\frac{35}{70} \:=\:\frac{1}{2}$

For Case 1: .$\displaystyle P(\text{4G, 2nd draw}) \;=\;\frac{16}{33}\cdot\frac{1}{2} \:=\:{\color{blue}\frac{8}{33}}$

Case 2: Both bad eggs were drawn.

There are: .$\displaystyle {12\choose4} = 495$ possible outcomes.

To get 2B and 2G, there are: .$\displaystyle {2\choose2}{10\choose2} = 45$ ways.
Hence: .$\displaystyle P(\text{2B, first draw}) \:=\:\frac{45}{495} \:=\:\frac{1}{11}$

For the second draw, there are: $\displaystyle 8G$

The probability of drawing 4G is: .$\displaystyle 100\% \:=\:1$

For Case 2: .$\displaystyle P(\text{4G, 2nd draw}) \:=\:\frac{1}{11}\cdot1 \:=\:{\color{blue}\frac{1}{11}}$

Therefore: .$\displaystyle P(\text{4G, 2nd draw}) \;=\;\frac{8}{33} + \frac{1}{11} \;=\;\frac{11}{33} \;=\;\frac{1}{3}$