StudySmarter AI is coming soon!

- :00Days
- :00Hours
- :00Mins
- 00Seconds

A new era for learning is coming soonSign up for free

Suggested languages for you:

Americas

Europe

Q18E

Expert-verifiedFound in: Page 452

Book edition
7th

Author(s)
Kenneth H. Rosen

Pages
808 pages

ISBN
9780073383095

**Question 2. To determine**

**What is the probability that two people chosen at random were born during the same day of the week?****What is the probability that in group of ${\mathit{n}}$****people chosen at random there are at least two born in the same day of the week?****How many people chosen in random are needed to make the probability greater than $\raisebox{1ex}{$\mathbf{1}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{2}$}\right.$****that there are at least two people born in the same day of the week?**

**Answer**

** **

- The probability is$\frac{1}{7}$.
- The probability is $1$ minus the people not born in that month.
- $5$ people have been chosen.

** **

- Two people chosen at random were born during the same day of the week.
- A group of $n$ people chosen at random there are at least two born in the same day of the week.
- The probability greater than $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ that there are at least two people born in the same day of the week.

** **

Given a group of n people, the sample space of days of the week in which they were born is

$p={7}^{n}\phantom{\rule{0ex}{0ex}}p\left(A\right)=1-p\left({A}^{c}\right)$

** **

The total number of days in a week is $7$.For the first person to be born on any one the day is $1$.

For the second person to be born on the same day is of the probability $\frac{6!}{7!}$

$p=\frac{6!}{7!}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{7}\phantom{\rule{0ex}{0ex}}$

The probability is $\frac{1}{7}$

** **

If there are people and two were born in the same day.

$1-\left(\frac{7}{7}\right)\left(\frac{6}{7}\right)\left(\frac{5}{7}\right)\left(\frac{4}{7}\right)\left(\frac{3}{7}\right)=0.85$

where $n=5$

The probability is $1$ minus the people not born in that month.

** **

Given a group of n people, the sample space of days of the week in which they were born is $p={7}^{n}$

Let $A$ be the event that at least two of them were born on the same day. Then $p\left(A\right)=1-p\left({A}^{c}\right)$ Now ${A}^{c}$ is the set that none of them share a birthday, so you need to pick $n$ different days out of $7$. This can be done in $\left(\begin{array}{c}7\\ n\end{array}\right)$ ways. Assuming the people are ordered, given a set of $n$ days we have to consider all permutations, hence:

$p\left({A}^{c}\right)=\frac{n!\left(\begin{array}{c}7\\ n\end{array}\right)}{{7}^{n}}$

If $n\u2a7e8$ then obviously at least two share a birthday, which agrees with our formula since $p\left({A}^{c}\right)=0$ for$n>7$ . Now use the formula above to compute the probabilities for $2\u2a7dn\u2a7d7$

We get $n=5$

$5$people have been chosen.

94% of StudySmarter users get better grades.

Sign up for free