Probability: Calculating the Chances of Success

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SUMMARY

The probability of hitting a target for the first time on the fourth shot, given a hit probability of 0.75, is calculated as 3/256. In a system with four independent mainframe machines, each with a failure probability of 0.03, the system functions properly if both the administrative and customer database machines are operational. The probability of this occurring is derived from the independent failure rates. Additionally, the probability of a salesperson making a sale after three attempts, with probabilities of 0.3, 0.45, and 0.2 for each attempt respectively, is also discussed, leading to a comprehensive understanding of sequential probability calculations.

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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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