Jen's Circus TripProbability puzzles require you to weigh all the possibilities and pick the most likely outcome.
Jennifer lives in a rural community and farms with her parents for a living. Most of the time she loves this life and enjoys farming immensely. However, sometimes she really enjoys getting a break from the hard farming life and going somewhere to just have fun. Her favourite place to do this is the circus, and this February they're in town! This February is not on a leap year, and it is unique in that it has four of each day (4 Sundays, 4 Mondays, 4 Tuesdays, etc).
Jen is all excited about the circus, and wants to go very badly. But, she still has chores to do around the farm. Every other Wednesday (the first and third week), she collects the eggs from the chickens and refills their feed bucket. On Sundays, she and her entire family go to church, as they are very religious. On the second week of the month, many of their sheep are due to have lambs, so she has to stay home and help. Fridays are taken up by grooming the horses and cows.
If she puts her savings together, she will be able to afford one regular price ticket. But then she will have no money left for candy floss. Luckily, the circus has a special on: on 5 random days of the month, they will have a discount for farmers only. This discount will allow Jen to buy a ticket as well as lots of candy floss. There's only one catch. They only announce that the special will be on the day before. There would be no planning ahead for it. This makes Jen feel blue, and she thinks that she probably won't make it on any of the discount days.
What is the probability of Jen being able to go on a discount day?
HintThe number of days that Jen is busy can be simplified into one number.
Answer15 days out of 28 Jen is busy (2 Wednesdays, 4 Fridays, 4 Sundays, and 1 week for sheep, less Sunday & Friday). The chance of Jen missing one of the discount days is 15/28 or 53.57%. The chances of Jen missing the second discount day is 14/27 or 51.85%, the 3rd 13/26 (50%), the 4th 12/25 (48%), the 5th 11/24 (45.83). The chance of her missing all 5 would be 53.57%x51.85%x50%x48%x43.83% or ((15*14*13*12*11)/(28*27*26*25*24)) = 3%.
Therefore, the probability that Jen is able to go on a discount day is about 97% (96.94% actually).
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