Question

Question 2. To determine the smallest number of people you need to choose at random so that the probability that at least one of them has a birthday today exceeds$\raisebox{1ex}{$\mathbf{1}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{2}$}\right.$ .

Short answer

Answer

The smallest number of people is $253$.

Step-by-step solution

Step 1. Given information

One of them were both born today exceeds $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$.

Step 2. Definition and formula to be used

Formula used are:

$P\left(No\text{one born today) +}P\right(\text{at least one born today)}=1$

Step 3. Add the probabilities

$$\begin{gathered} P(Nooneborntoday)+P(atleastoneborntoday)=1 \\ P(atleastoneborntoday)=1-{\left(\raisebox{1ex}{$364$}\!\left/ \!\raisebox{-1ex}{$365$}\right.\right)}^n \\ 1-{\left(\raisebox{1ex}{$364$}\!\left/ \!\raisebox{-1ex}{$365$}\right.\right)}^n\ge \frac{1}{2} \\ \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.=1-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \\ 1-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\ge {\left(\raisebox{1ex}{$364$}\!\left/ \!\raisebox{-1ex}{$365$}\right.\right)}^n \\ \log \left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)=n\log \left(\raisebox{1ex}{$365$}\!\left/ \!\raisebox{-1ex}{$364$}\right.\right) \\ n⩽252.651 \\ n\cong 253 \end{gathered}$$

The smallest number of people is$253$