1
$\begingroup$

This is for the probability that in a group of n people at least two have the same birthday, n = 3.

Hi, So for those who are familiar with the birthday paradox could you check my work:).

P(E) = [Event]/[SampleSpace]

Sample space = 365^3

Event = (365*364*363

P(E) = 0.991796

1-P(E) = 8.204 x 10^-3

This is where i am a bit lost, i dont know how to prove they are equal, any help is appreciated :).

$\endgroup$
1
$\begingroup$

Notice that $P(E)$ is the probability that in a group of 3 people, at least 2 people share the same birthday. What you are calculating is $P(\overline E)$, the probability that each person has a different birthday.

So you've calculated $1 - P(\overline E) = 0.008204$. Now you have to use counting principles to find $P(E)$.

$$ P(E) = [\mathrm{Event}]/[\mathrm{SampleSpace}] $$

Now let's find the number of events in which at least 2 people share the same birthday. Consider person 1, who has $365$ possible birthdays. Now if person 1 only shares a birthday with person 2, then person 2 can only have $1$ possible birthday—the same day as person 1! Since person 1 only shares a birthday with person 2, then person 3 must be born on a different day, any of the $364$ other days.

The same logic applies if person 1 only shares a birthday with person 3. There are also only $365$ events in which all three people have the same birthday, one event for each day of the year.

Finally, if person 1 does not share a birthday with either person 2 nor person 3, then person 2 and person 3 must share a birthday on a different day than person 1's birthday. So person 2 has $364$ different possible days to have a birthday, and person 3 has only $1$ possible day, the same day as person 2.

When we add up all these events, we get:

$365 \times 1 \times 364 + 365 \times 364 \times 1 + 365 + 365 \times 364 \times 1 = 1093 \times 365$

And, $\frac{1093 \times 365}{365^3} = 0.008204$, so we are done.

$\endgroup$
  • $\begingroup$ You found out $1 - P(\overline E) = 0.008204$. Now $1 - P(\overline E) = E$, so why do we need to find $E$ again in a roundabout way ? $\endgroup$ – true blue anil Oct 21 '15 at 6:58
  • $\begingroup$ The link says, "Calculate $P(E)$ for $n = 3$ using counting principles, and confirm that it is the same as $1 - P(\overline E)$." $\endgroup$ – eyqs Oct 21 '15 at 7:32
  • 1
    $\begingroup$ Ok, sorry, I didn't see the link. $\endgroup$ – true blue anil Oct 21 '15 at 7:49
0
$\begingroup$

You should define what $E$ is.

Any way, P(at least 2 have a common birthday) = $1 - 0.991796 = 0.008204$

which in scientific notation, becomes $8.204\times 10^{-3}$

$\endgroup$
  • $\begingroup$ You have already reached (365-n+1) = (365-3+1) = 363 in writing 365*364*363 $\endgroup$ – true blue anil Oct 21 '15 at 4:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.