In 80% of cases a worker uses the bus to go to work.If he takes the bus, there is a probability of 3/4 to arrive on time.On average, 4 days out of 6 get on time at work.Today the worker did not arrive in time to work.What's the probability he taked bus?

1 Answer
Feb 17, 2018

#0.6#

Explanation:

#P["he takes bus"] = 0.8#
#P["he is on time | he takes the bus"] = 0.75#
#P["he is on time"] = 4/6 = 2/3#
#P["he takes bus | he is NOT on time"] = ?#

#P["he takes bus | he is NOT on time"]*P["he is NOT on time"] =#
#P["he takes bus AND he is NOT on time"] =#
#P["he is NOT on time | he takes bus"]*P["he takes bus"] =#
#(1-0.75)*0.8 = 0.25*0.8 = 0.2#
#=>#
#P["he takes bus | he is NOT on time"]=0.2/(P["he is NOT on time"])#
#= 0.2/(1-2/3) = 0.2/(1/3) = 0.6#