[luvele73]

Bob believes he must train a master pizza maker rather than rely on his dad. He notes that the probability of a trainee actually becoming a master is only 1 in 6. The costs of trainees and their training are:
1= $30,000
2= $ 55,000
3=$75000
4= $85000
5=$95000

a. As the advisor what would you tell Bob is the likelihood of getting one master out of a group of 3 trainees?

b. How large a group of trainees would you advise him to use? If the training ratio of success to failure was 1/3 would that make a substantial difference in your advice?

:eek: I am confused. If it had said 1in 5, it would make sense. If the probability is 1 in 6 , where do I start?Please help

[jiten55]

1. "He notes that the probability of a trainee actually becoming a master is only 1 in 6.".

You take 3 trainees. (A, B, C)

a. Chances that none becomes maser is: (5/6)^3

Chances that AT LEAST ONE becomes master = 1 - (5/6)^3

b. Chances that exactly one out A, B, C becomes master

= 3 times (1/6)(5/6)^2 (Because each A, B, C could be the One)


2. Regarding the COSTS, there is CONFUSION in the question. Maybe you should study the question again. As the question is framed, COSTS make NO DIFFERENCE to Probabilities.

Costs become a consideration when we know what is a master's worth to the business, or how much profit a master is going to bring.



3. It will obviously make a significant difference if it was 1 out of 3 rather than out of 6.

It is plain common sense. But you can verify Mathematically.

[jiten55]

COSTS:

Assuming his aim is to minimise Expected Loss:


COST CONSIDERATION: MAYBE WE CAN LOOK AT IT LIKE THIS:

1. Lets suppose he tries 3 employees.

Cost: 75000

Probability of Loss = (5/6)^3 = 0.578703704 (Not even one master)

Expected Loss = $75,000 times 0.578703704 = $43403



2. Lets suppose he tries 4 employees.

Cost: 1= $85,000

Probability of Loss = (5/6)^4 = 0.482253086 (Not even one master)


Expected Loss = $85,000 times 0.482253086 = $40991



4. Lets suppose he tries 5 employees.

Cost: $95000

Probability of Loss = (5/6)^5 = 0.401877572 (Not even one master)


Expected Loss = $95000 times 0.401877572 = $38178



HENCE (Mathematical) Expectation of Loss is least when he takes on 5 employees.