MHB Probability: Calculating the Chances of Success

AI Thread Summary
The discussion revolves around calculating probabilities in various scenarios. For the first question, the probability of hitting a target for the first time on the fourth shot, given a hit probability of 0.75, is determined to be 3/256. The second scenario involves a system of four independent machines, each with a 0.03 failure probability, and the probability of the system functioning properly is calculated based on the operational requirements of the administrative and customer database machines. Lastly, the probability of a salesperson making a sale through multiple attempts is explored, with the initial probabilities of success on each visit being 0.3, 0.45, and 0.2. The conversation highlights the importance of understanding independent events and cumulative probabilities in these calculations.
Rojito
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(a) If the probability of hitting a target is ¾ (0.75), find the probability of hitting the target first on the fourth shot.

More generally, if the probability of hitting the target is P, what is the probability of hitting the target for the first time on the nth shot?

(b) A computer system in a large financial institution is based on 4 mainframe machines. Each machine has its own emergency power supply and operates independently. The machines are dedicated to the following tasks:

(i) Administration
(ii) Back-up Administration System
(iii) Customer Database
(iv) Back-up Customer Database

The probability of each machine failing is 0.03. The system operates properly if both the administrative and customer database machines can be supported. What is the probability that the system functions properly?

(c) The probability that a salesperson makes a sale to a customer on a first visit is 0.3. If no sale is made, the salesperson calls again the following week and the probability of making a sale then is 0.45. If no sale is made and the salesperson calls for a third and last time the probability of a sale is 0.2. Find the probability that the salesperson will make a sale to a particular customer.
 
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Would I be correct in assuming that the chance of a mishit is .25
\therefore 3 mishits = 1/4.1/4.1/4=1/64
The next attempt (4th shot) = 3/4.1/64=3/256
 
Hello, Rojito!

Would I be correct in assuming that the chance of a miss is 1/4 ?
Therefore, 3 misses = (1/4)(1/4)(1/4) = 1/64
The next attempt (4th shot) = (3/4)(1/64) = 3/256
Correct!

 
Thanks Soroban,

I was hoping for some help on the rest of the question, which I'm totally stuck on.
 
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