Mary Ann C. answered • 05/06/20
College Professor, Certified Teacher, and Professional Tutor
We are asked to find the probability that Jim takes a taxi to work given certain over events or conditions are met, so we have a conditional proabability. The notation we use for conditional probability is P(A|B) which can be interpreted, the proabability of event A given event B has already happened.
P(A|B) = P(A and B)/P(B) or P(A|B) = P(A∩B)/P(B)
Let A be the event Jim takes a taxi to work at least three days.
This is a binomial distribution: p=.3 (the probability he takes a taxi) q=.7 (the probability he doesn't take a taxi) and n=6 (the number of days he works in one week). We can use either a table or the binomila formula to calculate binomial probabilities.
i. What is the probability that Jim takes a taxi to work at least three days given he takes a taxi at most four days?
Recall event A is "At least three days" and let B be the event that Jim takes a taxi "at most 4 days".
So the probability of A and B (the intersection of these events) is the event that he takes a taxi to work either 3 or 4 day.
P(A∩B) = P(x=3) + P(x=4) = .324 = .060 = .387
P(B) = P(x≤4) = P(x=0) + P(x=1) + P(x=2) +P(x=3) + P(x=4) = .118 + .303 + .324 + .185 + .060 = .990
P(A|B) = P(A∩B) = P(B) = .387 / .990 = .391
ii. Calculate the probability that Jim takes a taxi to work at least three days given he takes a taxi more than one day. Our new condition is 'more than one day'. So we will redefine event B. B is the event "more than one". The intersection of A and B is x = 3, 4 ,5 or 6. (Note, here the event ''at least three and 'more than 1' is the same as the event 'at least 3'. So A∩B=A )
P(A∩B) = P(x≥3) = P(x=3) + P(x=4) + P(x=5) + P(x=6) = .185 + .060 + .010 + .001 = .256
P(B) = P(x>1) = 1- P(x=0) = 1 - .118 = .882
P(A|B) = P(A∩B) / P(B) = .256 / .882 = .290
