# Thread: hmwk problem

1. ## 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

2. 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)

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

so:

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

RonL

3. 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:

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

Now:

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

so:

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

RonL