# hmwk problem

• May 23rd 2006, 08:23 PM
swoopesjr01
hmwk problem
Students who read their statistics text have a .98 probability of passing the course. Those who don't have a .85 probability of failing. If 70% of the students do the readings, then:

a) What is the probability that a student will pass the course?

b) If a student were to pass, what would the probability have been that he or she did the readings?

a) .371
b) .398

__________________________________________________ ________
This is what I have come up with so far:

Let P = pass course and T= read text

(P|T) = 0.98 so P(P'|T) = 0.02
(P'|T') = 0.85 so (P|T') = 0.15
• May 27th 2006, 03:16 AM
CaptainBlack
Quote:

Originally Posted by swoopesjr01
Students who read their statistics text have a .98 probability of passing the course. Those who don't have a .85 probability of failing. If 70% of the students do the readings, then:

a) What is the probability that a student will pass the course?

b) If a student were to pass, what would the probability have been that he or she did the readings?

a) .371
b) .398

__________________________________________________ ________
This is what I have come up with so far:

Let P = pass course and T= read text

(P|T) = 0.98 so P(P'|T) = 0.02
(P'|T') = 0.85 so (P|T') = 0.15

a)

$
p(P)=p(P|T)p(T)+p(P|T')p(T')
$

so:

$
p(P)=0.98\times 0.70+0.15 \times 0.30=0.73
$

RonL
• May 27th 2006, 03:26 AM
CaptainBlack
Quote:

Originally Posted by swoopesjr01
Students who read their statistics text have a .98 probability of passing the course. Those who don't have a .85 probability of failing. If 70% of the students do the readings, then:

a) What is the probability that a student will pass the course?

b) If a student were to pass, what would the probability have been that he or she did the readings?

a) .371
b) .398

__________________________________________________ ________
This is what I have come up with so far:

Let P = pass course and T= read text

(P|T) = 0.98 so P(P'|T) = 0.02
(P'|T') = 0.85 so (P|T') = 0.15

b) Here we need what I think of as 1/2Bayes' Theorem:

$
p(T|P)p(P)=p(T \mbox{ and }P)
$

Now:

$
p(T \mbox{ and }P)=0.7 \times 0.98=0.686
$

so:

$
p(T|P)=p(T \mbox{ and }P)/p(P)=0.686/0.73\approx 0.940
$

RonL