<p>My math teacher gave this question to us as a bonus question and I was stumped on how to do it. </p>
<p>Q: If 6 people are randomly chosen, what is the probability of 2 of those people having the same birthday month? </p>
<p>I know it is more than 50% as he used our class to demonstrate this problem. He also said it is a very good idea to bet with a drunk person using this question to gain profit quick. :)</p>
<p>Are you sure it’s over 50%? I can’t say that I know how to solve it, but there’s a wikipedia article about it which states that 23 or more people are required to bring the probability over 50%.</p>
<p>hmmm that is weird 50% does not sound right… what I would have done is (1/365)(1/365) 365= days in a year…</p>
<p>well im not that sure but when he did in our class of roughly 25, it only popped out once to 2 people not having the same birthday month</p>
<p>Oh, actually I misunderstood the question, sorry.</p>
<p>[Birthday</a> problem - Wikipedia, the free encyclopedia](<a href=“http://en.wikipedia.org/wiki/Birthday_paradox]Birthday”>Birthday problem - Wikipedia)</p>
<p>Hmmm… Maybe I did understand your question. Haha.</p>
<p>there are 6 people. you count each as one event, so there are 6 events.
start at 1 and count up to 6, multiplying the accumulative possibilities as you go up.</p>
<p>there are two possible situations.
There are at least 2 people with the same birth month. One person has the same birth month as another person. Call this P(A)
There are no people with the same birth month. Call this P(B)</p>
<p>These are the only two possible situations and are mutually exclusive, so the sum of the two cases’ probabilities is 1.</p>
<p>P(A) = 1 - P(B)
for example, if P(B) has a 75% chance of occurring (0.75), the chance of it not occurring is 25% (0.25). these two add up to 1 and you can subtract either one from 1 to get the other.
We want to find P(A). We can do this by finding P(B) (which may be easier to find) and subtracting it from 1.</p>
<p>so we will find the probability that NO ONE will have the same birth month. then we will subtract that from 1.</p>
<p>P(1), P(2), P(3), P(4), P(5) and P(6) are the individual events of each person being chosen one by one, and the probabilities of them not having the same birth month as the people behind them.</p>
<p>starting at P(1):</p>
<p>P(1) = 12/12 (there is a 100% chance that no one will have the same birth month as person 1)
P(2) = 11/12 (there are 11 other months to choose from if you do not want person 2 to have the same birth month of person 1)
P(3) = 10/12 (there are 10 other months to choose from if you do not want person 3 to have the same birth month as person 1 or 2)
P(4) = 9/12
P(5) = 8/12
P(6) = 7/12</p>
<p>The product of the probabilities of the individual events is the probability of the whole system.
P(1) * P(2) * P(3) * P(4) * P(5) * P(6) = probability that none of the 6 people share the same birth month = (12/12)<em>(11/12)</em>(10/12)<em>(9/12)</em>(8/12)*(7/12) = approximately 0.2228</p>
<p>this is essentially 12 times 11 times 10 times 9 times 8 times 7 divided by 12 to the 6th power. multiplying numbers as you count down by 1 to a certain point is called a factorial. if you do not know what this is you will learn it</p>
<p>1-0.2228=0.7772=probability of there being 2 people (or more) with the same birth month=~78%</p>
<p>whoa! thx for your help</p>
<p>btw i know what factorials are</p>