Question

Help with probability?

Answer 1

What is the probability that the thirty-two residents of Lamont have thirty-two different birthdays? What is the probability that there is at least one birthday coincidence in Lamont?

Answer 2

Do you know what formula you are supposed to use?

Answer 3

uh no sorry, ill try looking for it though (i think it has something to do with combinations?)

Answer 4

i'm not really focused right now, so i don't think i will be of much help, i can't wrap my head around this problem, sorry, but i think i'm going to have to guess when i say this has something to do dependent and independent events, but i think that's wrong

Answer 5

would it be this?
(365_P_n)/(365^n) where P is permutation and n is the number of ppl (so 32)

Answer 6

$$ \large
\begin{align} \bar p(n) &= 1 \times \left(1-\frac{1}{365}\right) \times \left(1-\frac{2}{365}\right) \times \cdots \times \left(1-\frac{n-1}{365}\right) \\ &= { 365 \times 364 \times \cdots \times (365-n+1) \over 365^n } \\ \end{align}

$$

Answer 7

$$
{\begin{align}
\bar p(n) &= 1 \times \left(1-\frac{1}{365}\right) \times \left(1-\frac{2}{365}\right) \times \cdots \times \left(1-\frac{n-1}{365}\right) \\
&= { 365 \times 364 \times \cdots \times (365-n+1) \over 365^n } \\ \end{align}
\\~\\~\\\\
\bar p(32)= 1 \times \left(1-\frac{1}{365}\right) \times \left(1-\frac{2}{365}\right) \times \cdots \times \left(1-\frac{32-1}{365}\right) \\
={ 365 \times 364 \times \cdots \times (365-32+1) \over 365^n }

}
$$

Answer 8

thanks! so would that just be 365 permutation 32?

Answer 9

correct , for the numerator

Answer 10

Alright, thank you!

Answer 11

And for the second part should be completable by following this formula right? http://www.regentsprep.org/Regents/math/ALGtrig/ATS7/BLesso4.gif

Answer 12

$$
{\begin{align}
\bar p(n) &= 1 \times \left(1-\frac{1}{365}\right) \times \left(1-\frac{2}{365}\right) \times \cdots \times \left(1-\frac{n-1}{365}\right) \\
&= { 365 \times 364 \times \cdots \times (365-n+1) \over 365^n } =\frac{_{365}P_n}{365^n}\\ \end{align}
\\~\\~\\\\
\bar p(32)= 1 \times \left(1-\frac{1}{365}\right) \times \left(1-\frac{2}{365}\right) \times \cdots \times \left(1-\frac{32-1}{365}\right) \\
={ 365 \times 364 \times \cdots \times (365-32+1) \over 365^{32} } =\frac{_{365}~P~_{32}}{365^{32}}

}
$$

Answer 13

P( not all 32 birthdays are different) = 1 - P( all 32 are different)

Answer 14

oh lolol i just realized sorry that was a really stupid question, ty for the help haha

Answer 15

no problem :)