Question

A survey of statistics students who had completed BUS 203 revealed that 70% sought additional help at the ICC Learning Lab, while the remaining students did not utilize the Learning Lab at all. At the end of the semester, the professor went back over the data to find that 90% of students passed the class given that they used the Learning Lab. The probability of passing the class and not using the Learning Lab was 12%.

1. What is the probability of not passing the course given that the student went to the Learning Lab?
2. What is the probability of using the Learning Lab and passing the course?
3. What is the probability of passing the class?
4. Find the probability that a student utilized the Learning Lab given the condition that the student passed the class:______.
5. What is the probability of not using the Learning Lab or not passing the class?

Answer 1

a) P(NP/L) = 0.1

b) P(P/L) = 0.9

c) P(P) = 0.666

d) P(L/P) = 0.945

e) P(NL U NP) = 0.37

Explanation:

given that;

P = passing the course, NP = Not passing the course, L = Using lab, NL = not using Lab

P(L) = 70% = 0.7, P(NL) = 30% = 0.3

P(P/L) = 90% = 0.9

P(P/NL) = 12% = 0.12

a)

the probability of not passing the course given that the student went to the Learning Lab

P(NP/L) = 1 - P(P/L)

P(NP/L) = 1 - 0.9

P(NP/L) = 0.1

b)

the probability of using the Learning Lab and passing the course

P(P/L) = 0.9

c)

the probability of passing the class

P(P) = P(P/L) * P(L) + P(P/NL) * P(NL)

= (0.9 * 0.7) + (0.12 * 0.3)

P(P) = 0.666

d)

the probability that a student utilized the Learning Lab given the condition that the student passed the class

P(L/P) = [P(P/L) * P(L)] / P(P)

P(L/P) = (0.9 * 0.7) / 0.666 = 0.63/0.666

P(L/P) = 0.945

e)

the probability of not using the Learning Lab or not passing the class

P(NL U NP) = P(NL) + P(NP) - P(NP n NL)

= 0.3 + 0.334 - (0.88 * 0.3 )

= 0.634 - 0.264

P(NL U NP) = 0.37