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Selecting Problem (probability)

kiroro22

New member
Joined
Jan 23, 2011
Messages
10
There are 12 eggs and among them 4 rotten eggs have inadvertently been placed in a tray of 12 eggs. the eggs are taken out of the tray, one at a time and without replacement, and inspected. Find the probability that

A) 2 of the first 4 eggs inspected are rotten.

If I think the probability of getting 4 rotten eggs in order then it will be (4/12)*(3/11)*(2/10)*(1/9) right?
but I really can't think to solve this problem,,it's been a while I studied it...

B)The third egg to be inspected is the first rotten one found.

so this mean the first two eggs have to be the normal ones and the third one inspected will be the first rotten one.

C)The last rotten egg is the tenth egg to be inspected.

Among 4 rotten eggs, 3 rotten eggs were found during the 9 inspection

What kind of probability thereom do I need to use to solve this kind of question?

please help me!!
 

soroban

Elite Member
Joined
Jan 28, 2005
Messages
5,588
Hello, kiroro22!

I believe that "Combinations" is the way to go for parts (A) and (C).


There are 12 eggs in a tray: 4 rotten eggs and 8 good eggs.
The eggs are taken out of the tray, one at a time and without replacement, and inspected.
Find the probability that:

(A) Two of the first four eggs inspected are rotten.

\(\displaystyle \text{There are 12 eggs, and we select 4 of them.}\)
. . \(\displaystyle \text{There are: }\:_{12}C_4 \:=\:{12\choose4} \:=\:\frac{12!}{4!\,8!} \:=\:495\text{ possible outcomes.}\)

\(\displaystyle \text{From the 4 rotten eggs, we want 2: }\:{4\choose2} \,=\,6\text{ ways.}\)
\(\displaystyle \text{From the 8 good eggs, we want 2: }\:{8\choose2} \,=\,28\text{ ways.}\)

\(\displaystyle \text{Hence, there are: }\:6\cdot28 \,=\,168\text{ ways to get 2 rotten and 2 good eggs.}\)

\(\displaystyle \text{Therefore: }\:p(\text{2 rotten, 2 good}) \:=\:\frac{168}{495} \:=\:\frac{56}{165}\)




(B)The third egg to be inspected is the first rotten one found.

So this mean the first two eggs have to be the normal ones
and the third one inspected will be the first rotten one. . Right!

\(\displaystyle P(\text{good-good-rotten}) \:=\:\frac{8}{12}\cdot\frac{7}{11}\cdot\frac{4}{10} \:=\:\frac{28}{165}\)




(C)The last rotten egg is the tenth egg to be inspected.

Among 4 rotten eggs, 3 rotten eggs were found during the 9 inspections. . Yes!

\(\displaystyle \text{There are 12 eggs and we select 9 of them.}\)
. . \(\displaystyle \text{There are: }\:{12\choose9} \,=\,220\text{ possible outcomes.}\)


\(\displaystyle \text{Among the first nine eggs, we want 3 rotten and 6 good eggs.}\)

\(\displaystyle \text{From the 4 rotten eggs, we want 3: }\:{4\choose3} \,=\,4\text{ ways.}\)
\(\displaystyle \text{From the 8 good eggs, we want 6: }\:{8\choose6} \,=\,28\text{ ways.}\)

\(\displaystyle \text{Hence, there are: }\:4\cdot28 \:=\:112\text{ ways to get 3 rotten and 6 good eggs.}\)

. . \(\displaystyle P(\text{3 rotten, 6 good}) \:=\:\frac{112}{220} \:=\:\frac{28}{55}\)


\(\displaystyle \text{At this point, there are 3 eggs left: 1 rotten and 2 good.}\)

. . \(\displaystyle P(\text{10th is rotten}) \:=\:\frac{1}{3}\)


\(\displaystyle \text{Therefore: }\:p(\text{10th egg is the last rotten egg}) \:=\:\frac{28}{55}\cdot\frac{1}{3} \:=\:\frac{28}{165}\)

 

kiroro22

New member
Joined
Jan 23, 2011
Messages
10
Thank you so much soroban!

I will study harder based on this~~

thanks thanks~~~
 
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