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 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 1

.6 .8 .9
.4 .2 .1

Answer 2

Yes Im trying to use matrices and vectors to solve it,

Answer 3

i think you use conditional probabilitiy

Answer 4

Yeah I know. Im just not used to that.

Answer 5

so for what is the probability that you pass 2nd time?
it would be P (pass 2nd time)= P (fail first time) x P (passed 2nd time) but I don't why its like that

Answer 6

were you able to solve it Romero?

Answer 7

Romero is a matrix fanatic, but it would be more valuable if you could explain how to solve 3 by 3 matrices by Gaussian elimination method.

Answer 8

Isn't the first one
\[\frac{60 + 80}{2} = 70\]

Answer 9

the answer key is
1. P (pass 2nd time)= P (fail 1st time) x P (pass 2nd time) but i dont know why its like that

Answer 10

I told you I wasn't good at probability.

Answer 11

Not to worry. I can find the proper solutions unless @apoorvk already has it

Answer 12

can i give you a another one Hero

Answer 13

Let's see what @apoorvk comes up with first.

Answer 14

ok

Answer 15

how aer you doing apporvk

Answer 16

Okay.. (Pn and Fn represent Pass or fail test #n)



so, either 1st time or second time, sum of probabs --> 0.92

All possible events:
P1-->0.6
F1, P2 --> 0.4 x 0.8 = 0.32
F1,F2,P3-->0.6 x 0.2 x 0.9=0.0.108
F1, F2,F3-->0.6x0.2x0.1=0.012
SUM=1.04

So #1
passing first time means --> P1 = a probability of 0.6
passing on the second attempt means Failing on the first, 'and' then passing on the second ---->F1,P2 = 0.4 x 0.8 = 0.32
sum of probabilities of the conditions=0.92


Therefore P= 0.92/1.06

Answer 17

There is a BUG!!!!!

Answer 18

thank you so much but can you please explain the question...because when they ask what is the probability of passing the second time wouldn't it be 80% as they stated...i am confused

Answer 19

Well it seems the question has not really been quoted so well. It can only mean that if a student fails the test, he reappears again,and again, till he passes it.

Answer 20

wait so P (passing 2nd time)...what does that mean? the chances of the student passing on the second time? but it says "second t ime around, 80% pass and 20% fail" so doesnt it mean that the chances of passing second time is 80%?

Answer 21

for passing the second time , you ll have to appear the second time, and for that you must fail the first time. so, first you must fail, then pass. so P of passing in second test = (P of failing in first test)*(P of passing in the second)

Answer 22

I think I see how to explain this now.

Answer 23

Let me try

Answer 24

ok got it- you're great!
so for passing the first time it will be 60% and for pass the 3rd time it is
p(pass 3 time)= P (fail 1 time) x P (fail 2nd time) x P (pass 3rd time)

For fail is will be P (fail first ime) x (P fail second time) x (pail 3rd time)?

Answer 25

EXACTLY!!!! BRAINWAV-O!!! :))))

Answer 26

Well, if you already get it, then nvm

Answer 27

yay! i understand now...it think understanding the concept is really important

Answer 28

HUNDRED percent!!!! (no one has understood this kinda probability here in my one month on OS as fast as you!)

Answer 29

can i give you another one?

Answer 30

Yeah, it's the only thing that's important actually. and EXTREMELY so. If you know your basics and concepts right, you can solve simply any thing.

Answer 31

Yeah, but do post it as a separate question. This thread is already too long now.

Answer 32

ok perfect i'll post it now