Question

1) You take a test, and 60% pass the first time and 40% fail.
The second time around, 80% pass the test and 20% fail.
The third time, 90% pass and 10% fail.

a. What is the probability that you fail the 1st OR the 2nd time?
b. What is the probability that you fail on the 3rd time?
c. What is the probability you fail?

Answer 1

is everyone retaking the test each time or just those that failed before ?

Answer 2

actually the original question was this:

1) You take a test, and 60% pass the first time and 40% fail.
The second time around, 80% pass the test and 20% fail.
The third time, 90% pass and 10% fail.

a. What is the probability that you pass the 1st OR the 2nd time?
b. What is the probability that you pass on the 3rd time?
c. What is the probability you fail?

Answer 3

but i decided to flip the question...does it not make sense?

Answer 4

?? they are both exactly the same

Answer 5

this one its "probability of pass" the other was "probability of fail"

Answer 6

oh sorry

Answer 7

hey is this thread ok?

Answer 8

ok well i will assume you only retake if you failed before...that seems to fit what they are asking

1st time: pass 60% fail 40%
2nd time (only 40% represented) pass:40%*80% = 32% fail: 40% *20% = 8%
3rd time (only 8% represented) pass: 8%* 90% = 7.2% fail: 8% *10% = 0.8%

a) P(fail1st OR 2nd) = 40% + 20% - 8% = 52%

b)P(fail 3rd) = 10%

c) P(fail) = 0.8%

Answer 9

wait b) should be same answer as part c) ... 0.8%
probability you fail all 3 times is 0.8%

Answer 10

i just dont get it because it doesn't add up to 100 you know?

Answer 11

very well. so, can you guess what are the possible cases in here?
(refer to the original question)

Answer 12

Okay why it doesn't add upto 100---> because this is "conditional" probability. You have constraints applied.

When you are calculating the P of a person passing in, suppose the second test, you automatically assume he fails in the first. That is a must. So, one case (which is not possible according to the circumstances) is P1F2.


But suppose there would have been no constraints like if you pass the first the first test, you need not appear in the second, that is every candidate appears in all tests. Then the probability of all cases would definitely add up to 100.

Answer 13

ok so like i said i am tempted to say:
F1= .40
F2= .40 x .20= 8%
F3= .40 x .20 x. 10= .8%

How does that add up to 100?

Answer 14

the "OR" part is messing with me
sorry i believe i got part a wrong

should be 40%

Answer 15

maybe this will help