Consider that 37 of all emergency visits at a hospital resul

Consider that 3.7% of all emergency visits at a hospital result in a leave without being seen (LWBS). Assuming that all visits are independent, what is:

a) The probability that the fith visit is the first to LWBS?

b) The probablility that either the fifth or sixth visit is the first to LWBS?

c) THe probablity that the first visit to LWBS is among the first four visits?

d) The expected number of visits until the third LWBS?

Solution

a)

geometric distribution

P = q ^x-1 * p

x= 5 the first success ( success = LWBS)

P = (1-0.037) ^5-1 * 0.037 = 0.0318

b)

P = (1-0.037) ^5-1 * 0.037 + (1-0.037)^6-1 * 0.037 = 0.06244

c)

binomial distribution

P = 4C1 * 0.037^1 * (1-0.037)^4-1 = 0.1322

d)

expected value will be 1 / 3*0.037 = 9.009 = 9.00