Consider 30 MACM students chosen at random. What are the chances that at least 2 of them have the same birthday?
Suppose 30 students signed their name on a calendar. They sign their name in the box that corresponds to their
birthday. How many different calendars might be possible?
Each student has 365 possibilities (a permutation with replacement)
By Rule of Product, get 36530 different calendars
In general, the permutation of r of n distinguishable objects with replacement can be done in nrpossible ways.
How many calendars possible which have at least one box with 2 signatures?
Solution: Count backwards
# of calendars with at least one box with 2 signatures = total # of calendars - # of calendars with all different boxes
= 36530 - (365!/335!)
If all calendars are equally likely, then the fraction of the total number of calendars is called the probability
For 30 students, P = (36530 - (365!/335!))/36530 = 1 - 365!/(335!36530) = 70.6%
Definition: A probability is a number between 0 and 1 inclusive, that represents the likelihood that an event will occur in
the future. 1 = a certainty of success; 0 = a certain failure.
P(event) = (# of ways for event to occur)/(# of possible outcomes)
If all outcomes are equally likely, we can use counting to compare the probability:
P (rolling a 1 or 2) = (# of successful outcomes)/(# of outcomes)
E.g.: If we roll a die once, what is the probability of rolling a 1 or a 2?
E.g. Suppose a fair coin was tossed 4 times. What's the probability that heads came up exactly twice?
# of possible outcomes = 24(permutation with replacement)
# of successful outcomes (exactly 2 heads) = "4 choose 2" = 6
So, P (exactly 2 heads) = ("4 choose 2")/24= 6/16 = 3/8 = 0.375
If you roll a 1, game is over and you lose
On any other result, I roll my die. If you have a larger number than I do, then you win. Otherwise you lose.
E.g. A game is played with 2 dice: one for you and one for me. You roll first.
What's the probability that you will win?
Even though there are 31 outcomes, not all events are equally likely. To fix this, I will roll my die even if you roll a 1
==> 36 possible outcomes
January 14, 2016
Lecture Notes Page 1