Calculating Rocket Reliability: Probability of Survival in Launch and Failure

[matt222]

Homework Statement

rockwt has reliability of 0.81 for the first trail. if the manned rocket fails there is probability of 0.05 of a catastrophic explosion in which case the abort system can not be used. id the abort can be attempted the system has a reliability of 0.90

find 1- the probability of each possible outcome
2-the chance of the crew surviving the launch
3- the chance of the crew surviving rocket failure

Homework Equations

The Attempt at a Solution

we have two events, first event abort used call it A, and second event abort not used event B, both event sharing the same reliability of the first aunch 0.81,
1-
p=(1-0.81)=0.19 for event A
P=(1-0.81)*(1-0.90)=0.019, this will give 1-0.019=0.981 for event B
2-
Crew surviving should be in the case of A, 0.19*0.05=0.00035
for B 0.981

3-

lost here and not sure about the above work

 

[Antiphon]

Look up the conditional probability formula by Bayes.

 

[HallsofIvy]

Homework Statement

rockwt has reliability of 0.81 for the first trail. if the manned rocket fails there is probability of 0.05 of a catastrophic explosion in which case the abort system can not be used. id the abort can be attempted the system has a reliability of 0.90

find 1- the probability of each possible outcome
2-the chance of the crew surviving the launch
3- the chance of the crew surviving rocket failure

Homework Equations

The Attempt at a Solution

we have two events, first event abort used call it A, and second event abort not used event B, both event sharing the same reliability of the first aunch 0.81,
1-
p=(1-0.81)=0.19 for event A
P=(1-0.81)*(1-0.90)=0.019, this will give 1-0.019=0.981 for event B

Have you posted the full text of the question? You had only "find the probability of each possible outcome" but then went immediately to what happens if the launch is aborted. Isn't "success" an outcome?

2-
Crew surviving should be in the case of A, 0.19*0.05=0.00035
for B 0.981

Again, you are ignoring the case of a successful launch.

3-

lost here and not sure about the above work

THIS is where you should only consider the two kinds of "aborts". But you don't want two different answers- combine the probabilities you had above.

 

[matt222]

i will start solving by stepsnow:
first from 0.81 success
failure going to be 0.19,(0.05) catastrophic failure
if abort used from 0.95 with 0.9 abort reliability success we have 0.1 failure
2 failure case and 2 success

for answering first question = for each event the success going to be 0.81*0.9, for failure 0.05*0.1



for answering the 2nd question=0.81*0.9/0.81*0.9+0.05*1

for answering the 3rd question=combine the above two

is it ok

 

[D H]

No, it's not OK. You do have four possible outcomes, but your answer is completely wrong. Hint: You should end up with four mutually exclusive events that describe all possible outcomes. What should the probability sum to?

For the second question, take advantage of the way you answered the first question.

For the third question, the fact that the rocket failed is a given. You haven't shown any work on this other than some very vague words. Show some more work and we will be able to give you a bit more help.

 

[matt222]

for the first question the sum should end up to 1 so that,

event A of 0.81*0.05 will give me 0.0405 success
event B we have now 3 possibles, first one is 0.95*0.1 will give me 0.095 failure, second possible 0.05*0.19 will give me 0.0095 failure, third one going to be 0.9*.95 success will give me 0.855

sum all of the event give me 1

2nd question will be simply the success which is the even of A success and B success which is 0.8955

3rd question we have the total of success and subract it from failure so 0.8955 success -0.1045=0.791
i hope now is ok

 

[D H]

That is just wrong.

As a starter, why are you multiplying 0.81*0.05? The probability of success is 0.81, period.

 

[matt222]

event A of 0.81success sorry got mistake by multiply it with 0.05,


event B we have now 3 possibles, first one is 0.95*0.1 will give me 0.095 failure, second possible 0.05 failure, third one going to be 0.9*.95 success will give me 0.855

sum all 3 of the B event give me 1

2nd question will be simply the success which is 0.855

3rd question we have the total of success and subract it from failure so 0.855 success -0.1045=0.751

 

[D H]

Still wrong, even on part 1. Be specific and name your outcomes. Don't just rattle off a number. Remember that the sum of all of your events should add to one.

 

[matt222]

event A of 0.81success,

event B of 0.19 failure we have 3 possibles, first one is 0.05 failure, second one 0.9*.95 success will give me 0.855 when abort system is used,third one going to be 0.95*0.1 of the failure if the abort is attempted will give me 0.095
sum all 3 of the B event give me 1, the problem here i got only the 3 event in B sumed to 1 when the abort system is used what about the first one of 0.81 how to include it and where is my mistake?

 

[Apphysicist]

Why don't you start off with:

The probability of the rocket working perfectly or failing (given by your reliability). Then
If the rocket fails (0.19 times out of 1), then how likely is it to explode? Then
If the rocket fails (0.19), then how likely is it to be able to use the abort fcn and then for it to fail as well? Then
If the rocket fails (0.19), then how likely is it to be able to use the abort fcn and then for it to work properly?

Then consider which of these cases result in the survival of those inside.
Then consider which of these cases results in the survival of those inside AFTER failure.

Rather than rushing through the problem, you'd benefit by taking the time to develop each case clearly.

 

[matt222]

Give up really confused