Twenty people at a meeting were born during the month of Sep

at least two of the people = 1- no two people share the same bday
no two people share the same bday = (1st pick a day in the 30 days) * (2rd pick another day in the left 29 days)
= (1/30) * (29/29)
so, at least two of the people share differ = 1-(1/30) * (29/29) = 29/30 = 99%

I still don't get it. I thought it would be 1-(29/30)*(28/30). Does anyone have another way of figuring this out?

hi vad3tha,

could you explain the red part please?

It should be 30^20, instead of 20^30: each out of 20 people has 30 options - 30*30*...*30 = 30^20.

It should be 30/30*29/30*28/30*27/30*26/30*...*11/30.

It is easy to solve the problem by finding the probability where each person is born on a different day and subtracting it from 1.

Let us start with a single way. The first person can be born on Sep 1 , the second on Sep 2 and so on. So the probability of 20 persons born on different days = (1/30)*(1/30) *..20 times =1/(30^20)

How many such ways are there?

(1) the 20 days can be chosen from 30 days in 30C20 ways
(2) The birthdays of the 20 persons can be arranged in 20! ways

For the probability, we have to multiply 1/(30^20) by 30C20 and 20!

So the probability that the birthdays fall on different days = 30C20 * 20! / (30^20)

The probability that at least two persons share the same birthday is 1 - (30C20 *20!) / (30^20) = 99%(approx)

\(? = 1 - P\left( {\underbrace {{\rm{all}}\,\,20\,\,{\rm{different}}\,\,{\rm{birthday}}\,\,{\rm{dates}}}_{{\rm{unfavorable}}}} \right)\)


\({\rm{Total}}:\,\,30 \cdot 30 \cdot \ldots \cdot 30 = {30^{20}}\,\,\,{\rm{equiprobable}}\,\,{\rm{possibilities}}\,\,\,\)

\({\rm{unfavorable}} = \,\,30 \cdot 29 \cdot \ldots \cdot 11\)


\(P\left( {{\rm{unfavorable}}} \right) = {{30 \cdot 29 \cdot \ldots \cdot 11} \over {{{30}^{20}}}} = 1 \cdot \underbrace {{{29} \over {30}} \cdot {{28} \over {30}} \cdot \ldots {{14} \over {30}}}_{ < < < \,\,1} \cdot \underbrace {{{13} \over {30}} \cdot {{12} \over {30}} \cdot {{11} \over {30}}}_{ \cong \,\,0.05} < < < 0.05 = 5\%\)

\(?\,\,\, > > > \,\,\,100\% - 5\% \,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left( {\rm{E}} \right)\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

The probability that at least two of the people in the room share the same birthday is equivalent to subtracting from 1 the probability that no two people in the room share the same birthday.

The first person can have a birthday on any of the 30 days of September. In order to avoid a birthday match, the second person can have a birthday on any of the remaining 29 days. Similarly, to avoid a match with either of the first two people, the third person can have a birthday on any of the remaining 28 days. And so forth, down to the twentieth person. We can then express each event as a probability by dividing by 30, the total number of days in September. The first person’s probability of not matching is 30/30 (because they can be born on any day). The second person’s probability of not matching the first person is 29/30, and the third person’s probability of not matching either of the first two is 28/30. This follows in a similar fashion to the twentieth person.

The probability that no two people in the room share the same birthday (i.e., that they all have different birthdays) is:

30/30 x 29/30 x 28/30 x … x 11/30
(30 x 29 x 28 x … x 11)/(30 x 30 x 30 x … x 30)

30P20 / 30^20 ≈ 0.0002

Therefore, the probability that at least two of the people in the room do share the same birthday is:

1 - 0.0002 = 0.9998 = 99.98%

Answer: E

How can we know that (30*29*28*...*11)/30^20 is far less than 0.05? I can only reason that its value to must be greater than (20*20*20*...*20)/30^20 which means (2/3)^20 ~ 0.66^20, but still not convincing that it must be less than 0.05