# probability issue

• Aug 23rd 2007, 02:51 AM
apply-it
probability issue
I am pretty sure this is easy to solve and I am a bit embarrassed to post it but I dont quite get it (tried a tree diagram but it doesnt work for me or I might do it wrong?!):

Of a class of 100 students 80 attend lectures. The whole class is made to sit a multiple choice test. On the first question, any student who has attended lectures knows the answer. The other students each know the answer with probability 0.5. Any student who knows the answer gets it right; a student who does not know, picks an answer at random from the three choices offered. The teacher selects a student who has answered the first question correctly. Show that the probability that my student attended lectures is 6/7.

• Aug 23rd 2007, 04:19 AM
Soroban
Hello, apply-it!

Don't be embarrassed . . . Conditional probability is always tricky.

Quote:

Of a class of 100 students 80 attend lectures.
The whole class is made to sit a multiple-choice test.
On the first question, any student who has attended lectures knows the answer.
The other students each know the answer with probability 1/2.
Any student who knows the answer gets it right.
A student who does not know, picks an answer at random from the 3 choices offered.
The teacher selects a student who has answered the first question correctly.
Show that the probability that the student attended lectures is 6/7.

Are you familiar with Bayes' Theorem? . $P(A|B) \:=\:\frac{P(A \wedge B)}{P(B)}$

We want: . $P(\text{attend }|\text{ correct}) \;=\;\frac{P(\text{attend}\wedge\text{correct})}{P (\text{correct})}$

[1] Determine the numerator.
. . . $P(\text{attend}) \:=\:\frac{80}{100} \:=\:\frac{4}{5} \qquad P(\text{correct}) \:=\:1$
Hence: . $P(\text{attend}\wedge\text{correct}) \:=\:\frac{4}{5}\cdot1\:=\:\boxed{\frac{4}{5}}$

[2] Determine the denominator.
. . .There are three ways to get the correct answer.

(a) The student attended the class and got the right answer.
. . .We found this in part [1]: . $P(\text{attend}\wedge\text{correct}) \:=\:\frac{4}{5}$

(b) The student did not attend but knew the answer.
. . . $P(\text{not attend}) \:=\:\frac{1}{5}\qquad P(\text{correct}) \:=\:\frac{1}{2}$
. . .Hence: . $P(\text{not attend}\wedge\text{correct}) \:=\:\frac{1}{5}\cdot\frac{1}{2} \:=\:\frac{1}{10}$

(c) The student did not attend, did not know the answer, and guessed correctly.
. . $P(\text{not attend}) \:=\:\frac{1}{5} \qquad P(\text{did not know answer}) \:=\:\frac{1}{2} \qquad P(\text{guessed correctly}) \:=\:\frac{1}{3}$
. . .Hence: . $P(\text{not attend}\wedge\text{guessed correctly}) \:=\:\frac{1}{5}\cdot\frac{1}{2}\cdot\frac{1}{3}\: =\:\frac{1}{30}$

Then: . $P(\text{correct}) \:=\:\frac{4}{5} + \frac{1}{10} + \frac{1}{30} \:=\:\boxed{\frac{14}{15}}$

Therefore: . $P(\text{attend }|\text{ correct}) \;=\;\frac{\frac{4}{5}}{\frac{14}{15}} \;=\;\boxed{\frac{6}{7}}$ . . . . ta-DAA!

• Aug 23rd 2007, 05:52 AM
apply-it
thanks
Thank you so much!!! That makes so much sense!!Where exactly does Beye's formula come from then? Do u find the probability of A given B via finding the propbability of A^B as a proportion of B? Could u briefly explain this to me? Youre a star!!! Kati
• Aug 23rd 2007, 06:50 AM
Soroban
Hello, Kati!

I'll try to give you the derivation with a specific example . . .

Suppose there are 100 students in an English class: 60 females and 40 males.
20 of the females are math majors; 10 of the males are math majors.

A student is chosen at random.
What is the probability that the student is female and a math major?

There are two approaches to this problem.

[1] Of the 100 students, 20 are female math students.
Hence: . $P(\text{female}\wedge\text{math}) \:=\:\frac{20}{100} \:=\:\frac{1}{5}$ .[1]

[2] The probability that the student is female is: . $P(\text{female}) \:=\:\frac{60}{100} \:=\:\frac{3}{5}$
Then the probability that this female is a math major is: . $P(\text{math}|\text{female}) \:=\:\frac{20}{60} \:=\:\frac{1}{3}$
Hence: . $P(\text{female})\cdot P(\text{math}|\text{female}) \:=\:\frac{3}{5}\cdot\frac{1}{3} \:=\:\frac{1}{5}$ .[2]

Equating [2] and [1]: . $P(\text{female})\cdot P(\text{math}|\text{female}) \:=\:P(\text{female}\wedge\text{math})$

. . and we get: . $P(\text{math}|\text{female}) \;=\;\frac{P(\text{math}\wedge\text{female})}{P(\t ext{female})}$

In general: . $P(A|B) \;=\;\frac{P(A\wedge B)}{P(B)}$

• Aug 23rd 2007, 07:22 AM
apply-it
Thanks!! Thats great, really get it now!!! Perfect!