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Havard Stat 110 Lecture 3

Birthday Problem

00:49
\(K\) people, find probability that share same birthday. Exclude Feb. 29th and assume other 365 days equally likely and independence.

Obviously if \(K>365\) , the probability is \(1\).

Let \(K\leqslant 365\) , consider \(P(\text{no match})\) :

\[ P(\text{no match}) = \dfrac{365\times 364\times \cdots (365-K+1)}{365^K} \]

So \(1-P(\text{no match}) = P(\text{match})\) .

Compute this result and we can get that:

\[ P(\text{match}) = \begin{cases}50.7\% , &K=23 \\ 97\% , &K=50 \\ 99.999\% , &K=100 \end{cases} \]

Properties of Probability

22:46

  1. \(P(\overline{A}) = 1-P(A)\) .
  2. If \(A\subseteq B\) (If \(A\) occurs that \(B\) occurs), then \(P(A)\leqslant P(B)\) .
  3. \(P(A\cup B) = P(A)+P(B)-P(A\cap B)\) . (For proof, consider \(P(A\cup B) = P(A)+P(B\cap \overline{A})\)) .
  4. \[ \begin{aligned}P(A\cup B\cup C) &= P(A)+P(B)+P(C) \\ &- P(A\cap B)-P(A\cap C)-P(B\cap C)\\ &+P(A\cap B\cap C)\end{aligned} \]

de Montmort's Problem

38:48
\(n\) cards, labeled \(1,2,\cdots,n\) . Let \(A_j\) be the event "\(j\)-th card matches".
So we need to compute \(P(A_1\cup A_2\cdots \cup A_n)\) .

\(P(A_j) = \dfrac{1}{n}\) since all position equally likely for card labled \(j\) .

\(P(A_1\cap A_2) = \dfrac{(n-2)!}{n!}\) as we can consider the first and the second one are fixed.

So \(\displaystyle P\left(\bigcap_{k=1}^n A_k\right) = \dfrac{(n-k)!}{n!}\) .

Thus we can compute that:

\[ \begin{aligned} P(A_1\cup A_2\cdots \cup A_n) &= n\frac{1}{n}- \frac{n(n-1)}{2!}\frac{1}{n(n-1)}\cdots+(-1)^{n+1}\binom{n}{n}\frac{(n-n)!}{n!} \\ &= 1- \frac{1}{2!}+ \frac{1}{3!}-\cdots+(-1)^{n+1} \frac{1}{n!} \\ &\approx 1-\frac{1}{\mathrm{e}} \end{aligned} \]

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