Calculating the probability of one successful outcome of an event when repeated up to 8 times?

[###Tim###]

Hi, could anyone help?

Calculating the probability of one successful outcome of an event when repeated up to 8 times? (probability is 5% of this independent event)
once successful repeats won't be made

So far I have got
Sucess P(A) = 0.05 failure P(B) = 0.95
number of attempts (K)= 8

so guess that total probability = success on 1st atempt+ success on2nd .... +success on 8th attempt
I would work that out as the total of succeeding 1st + failing 1st and succeeding 2nd etc

1st = 0.05
+2nd(0.95power1 x 0.05) = 0.0475
+3rd (0.95power2 x 0.05) = 0.045125
+4th (0.95power3 x 0.05) = 0.04286875
+5th (0.95power4 x 0.05) = 0.0407253125
+6th (0.95power5 x 0.05) = 0.038689046875
+7th (0.95power6 x 0.05) = 0.03675459453125
+8th (0.95power7 x 0.05) = 0.0349168648046875

The total of all this is the probability of one successful outcome of an event when repeated up to 8 times?
I am ignoring the instances where total failure on all 8 attempts or more than one success occurs.

If I just add all these events I get 0.7640795687109375 so 76% chance is that right? I dont think so, doesn't sound right. I know I can't just add 5% 8 times that would equal 40% and would mean 20 attempts would guarantee success (100% )while the greater the number of attempts the probability will tend towards 100% but will never reach that. So I know I am missing a step/method.

please help, many thanks in advance. Tim

 

[HallsofIvy]

When you say "repeated up to eight times" do you mean that you stop as soon as you have a success?

If so, since the probability of success on any one trial is 0.05 and the probability of failure is 0.95, yes, what you are doing is correct. It looks to me like you have simply added wrong!

To do this more generally, take the abstract case with probability of success "p" so that the probability of failure is q= 1- p.

The probability of success on the first trial is p.
In order to have your first success on the second trial you must fail on the first trial so the probability of this is qp.
Similarly, in order to have your first success on the third trial you must fail on the first two. The probability is q^2p.
For first success on fourth, fifth, sixth, seventh, and eighth trials, the probabilities are
q^3p, q^4p, q^5p, q^6p, and q^7p. The probability of "one success in up to 8 trials" is the sum of those, p+ qp+ q^2p+ q^3p+ q^4p+ q^5p+ q^6p+ q^7p= p(1+ q+ q^2+ q^3+ q^4+ q^5+ q^6+ q^7). Do you see that this sum is a geometric sum? That sum can also be written as $\displaystyle \frac{1- q^8}{1- q}$.

With q= 0.95, that is $\displaystyle \frac{1- 0.95^8}{0.05}= \frac{1- 0.663}{0.05}= \frac{0.335}{0.05}= 6.73$. Multiplying by p= 0.05, the probability of one success in up to eight trials is 0.3365.

 

[###Tim###]

Thanks for your help, I just managed to solve it another way very simular thou
1-0.95power8 = 0.33657
which just expresses all the possible outcomes except the chance of total failure (8).
I really appreciate the help, confirmation and more detailed understanding I got.
Thanks, I got the answer I need.