Ok. On this planet, you are either a rogger or a bogger. there is an 8% chance of you being a bogger. 7% of rogger have ears. 23% of boggers have ears. Given that you have an ear, what are the chances that you are a bogger????
2007-07-15 12:07:13 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Yes, this problem is solvable. If you draw a simple Venn diagram --- starting with 92% roggers and 8% boggers --- you can figure out what percentage of total ear-possessors is boggers.
I'm pretty sure the correct answer is: if you have an ear, there is a 22.222% chance that you are a bogger.
2007-07-15 12:20:54 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Given 8% are boggers, then 92% are roggers.
7% of roggers have ears, 23% of boggers have ears.
Total chances of roggers having ears is 6.44%
Total chances of boggers having ears is 1.84%
Chances of having an ear & being a bogger 3.5%
2007-07-15 19:22:03 · answer #2 · answered by Robert S 7 · 0⤊ 1⤋
Per 100 total roggers and boggers:
8 boggers for every 92 roggers
1.84 boggers have ears; 6.44 roggers have ears.
So the chances of being a rogger if you have an ear is 6.44%
2007-07-15 19:29:32 · answer #3 · answered by ironduke8159 7 · 0⤊ 1⤋
Yes there is an 8% chance
2007-07-15 19:20:08 · answer #4 · answered by The Universe is an atom 1 · 0⤊ 1⤋
Use Bayes Theorem
E: has an ear
R: rogger
B: bogger
P(B|E) = P(E|B)P(B) / [P(E|B)P(B) + P(E|R)P(R)]
= (0.23 * 0.08) / (0.23*0.08 + 0.07*0.92)
= 0.0184 / (0.0184 + 0.0644)
= 0.0184 / 0.0828
= 0.2222
So there is a 22.22% chance that if you have an ear, you are a bogger.
2007-07-15 19:16:31 · answer #5 · answered by whitesox09 7 · 0⤊ 0⤋
P(Bogger&Ear)
=P(Ear|Bogger)P(Bogger)
=.23*.08=.0184
P(Rogger&Ear)
=P(Ear|Rogger)P(Rogger)
=.07*(1-.08)=
.0644
P(Ear)
=P(Bogger&Ear)
+P(Rogger&Ear)
=.0184+.0644=.0828
P(Bogger|Ear)
=P(Bogger&Ear)/P(Ear)
=.0184/.0828
=.2222...
=2/9
2007-07-15 19:26:38 · answer #6 · answered by Anonymous · 0⤊ 0⤋