Probability calculations

Question

Probability calculations – exactly one occurrence You are looking for a job. Eventually you get six interviews. In each case you are one of four final candidates for the respective job. Let us assume you have a .25 probability of getting each job and the probability of getting each job offer is independent of getting any other job offer. a.. What is the probability that you get exactly one job offer of the six opportunities? Show this as a percentage with three decimal places

a.. What is the probability that you get exactly one job offer of the six opportunities? Show this as a percentage with three decimal places

 

Answer 1

In this situation, you are looking for the probability of one success and five failures. You need to consider the probability of each event, and the number of ways that the desired pattern can occur.

 

The overall pattern of exactly one success is (.25)(.75)^5. But... there are six ways that this can occur, once for each final interview. Thus, we have 6 x .25 x .75^5, about .356 or 35.6%.

 

Officially, this is written as the general equation nCr x p^r x q^(n-r)

where p is the chance of success (.25),

 

q = 1-p is the chance of failure (.75),

 

n is the number of trials (in this case 6),

 

r is the number of successes desired (1) and

 

nCr is the notation for the calculation of the number of combinations possible that give matching results.

 

nCr = n!/(r!(n-r)!) = 6!/(1! 5!) = 6