# Statistics

• May 31st 2007, 04:09 PM
superman
Statistics
For the life of me I cannot start this question to me this does not make sense?
In a certain school, it is know that 80% of the student’s user the internet for school projects 60%m use e-mail on regular basis, and 90% use the internet for school projects or for e-mail on a regular basis. A student from this school is selected at random. Determine the probability that the student used e-mail, given that the student used the internet for school projects

Would I go like this:
1/80 x 1/60 = 2.08 x 10^4 this does not make sense??????:o
• June 1st 2007, 01:43 AM
CaptainBlack
Quote:

Originally Posted by superman
For the life of me I cannot start this question to me this does not make sense?
In a certain school, it is know that 80% of the student’s user the internet for school projects 60%m use e-mail on regular basis, and 90% use the internet for school projects or for e-mail on a regular basis. A student from this school is selected at random. Determine the probability that the student used e-mail, given that the student used the internet for school projects

Would I go like this:
1/80 x 1/60 = 2.08 x 10^4 this does not make sense??????:o

No no sense at all.

Out of a nominal 1000 students

900 use one or the other
600 use "e"
800 use "i"

so 500 use both

So of the 800 that use "i" 500 also use "e", so the required probability is:

500/800 = 5/8

RonL
• June 1st 2007, 06:37 AM
Soroban
Hello, superman!

Quote:

In a certain school, 80% of the students use the internet,
60% use e-mail, and 90% use the internet or for e-mail.

A student from this school is selected at random.
Determine the probability that the student used e-mail,
given that the student used the internet.

Would I go like this: 1/80 x 1/60 = 2.08 x 10^4
This does not make sense . It certainly doesn't!

Since when does $80\% \,=\,\frac{1}{80}$ ???

Since you are given a Conditional Probability problem,

I assume you know this formula: . $P(A \cup B) \:=\:P(A) + P(B) - P(A \cap B)$ .[1]

. . and Bayes' Theorem: . $P(A|B) \:=\:\frac{P(A \cap B)}{P(B)}$ .[2]

We are given: . $P(I) = 0.80,\;P(E) = 0.60,\;P(I \cup E) = 0.90$

Substitute into [1]: . $P(I \cup E) \:=\:P(I) + P(E) - P(I \cap E)$

. . . .and we have: . . . . $0.90\:=\:0.80 + 0.60 - P(I \cap E)$

. . . . . . . . Hence: . $P(I \cap E) \:=\:0.50$

We are asked for: . $P(E|I)$

Bayes' Theorem: . $P(E|I) \:=\:\frac{P(E \cap I)}{P(I)} \:=\:\frac{0.50}{0.80} \:=\:\frac{5}{8}$