Combinatorics / Probability ( checking my answer)

[Matejxx1]

Homework Statement

The chance that an event A will happen is always equal to 0,3. We stop the experiment once event A occurs What is the probability that :
a) we have to attempt the even at least 4 times
b)we then do the event a fourth time
I am asking this because the professor did not tell us the answers for the problems because he wanted us to solve this ourselfs

Homework Equations

A=event
A'= complemetary event

The Attempt at a Solution

a)
ok so here I was thinking that if the event has to occur at least 4 times that means that the event A fails to occur the first 3 times. therefore I tried writting it like this
P(A)+P(A')=1
P(A)=1-P(A')
P(A)=1-(0,7*0,7*0,7) fails the first 3 times
P(A)=0,657=66%
What I was trying to do here was to find what are the chances that the experiment will fail the first 3 times, and then I just substracted that from 1 (1 meaing that the event is 100% sure to happen)
b)we then do the event a fourth time
I was thinking that this means , that the event succeded the 3 time and then for some reason we repeat the event a 4 time
I tried solving this one like this :
P(A'A'AA+A'A'AA')...... A= A has happened A'= A has no happened
the first part demonstrates that the 2 the times the event fails and then the 3 &4 times is succeds , but because we don't know if the 4 th time the event succeds we also have to take into account and check was happends of the event fails
and so if I add those two things up:
P(A)=(0,7*0,7*0,3*03+0,7*0,7*0,3*0,7)
P(A)=0,*7²*0,3(0,3+0,7)
P(A)=0,7²*0,3(1)
P(A)=0,15=15%
thanks

 

[Matejxx1]

UPDATE: I have tried thinking a little more about a)
if I wanted to do complemetary events I should have checked the chances to see if the event succeds the 1st ,2nd or 3th time
so I think it should have looked something like this now
P(A)=1-(A')
P(A)=1-(0,3+0,7*0,3+0,7*0,7*0,3)
P(A)=1-0,657
P(A)=0,343=34,3%

 

[HallsofIvy]

Yes, the simplest thing to do is to calculate the probability this terminates in the first second or third event, then subtract that from 1.

The probability you get "A" in the first event is 0,3. To get it in the second event you must first NOT get A in the first event, which has probability 0.7, then get in in the second. The probability of that is (0.7)(0.3)= 0.21. To get it in the third event you must not get A in the first or second events but get it in the third event. The probability of that is (0.7)(0.7)(0.3)= .0.147. Add those, then subtract from 1 get precisely the answer you give- although I would not call that "P(A)".

The second question is "b)we then do the event a fourth time". I don't know what that means. If it is the probability we do have to do this a fourth time, whether it is successful or not, is just the probability we do NOT get "A" on the first, second, or third trials which is (0.7)(0.7)(0.7)= 0.343, surprisingly the same answer as before!

 

[Matejxx1]

thanks as you said the first answer was corret, however my profesor interperated the b) part differently and told me this :" it means that we do the event the fourth times because the first 3 times have failed. But we also don't know if the even will occur that 4th some so we may also need to repeat a 5th or a 6th . . . n time until the event happens." and the way that he told me to solve it was exactly the same as yours
Thanks

 

[haruspex]

The wording "we do the event a fourth time" is most confusing. The experiment consists of a number of trials. The only 'event' described is event A, and when that occurs the experiment stops. So should b) be asking for the probability that we do a fourth trial? As it stands, it could be construed as asking for the probability that the event occurs on the fourth trial.