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Expert-verified Found in: Page 452 ### Discrete Mathematics and its Applications

Book edition 7th
Author(s) Kenneth H. Rosen
Pages 808 pages
ISBN 9780073383095 # Question 2. To determineWhat 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 ${\mathbit{n}}$ 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 $\mathbf{1}}{\mathbf{2}}$ that there are at least two people born in the same day of the week?

1. The probability is$\frac{1}{7}$.
2. The probability is $1$ minus the people not born in that month.
3. $5$ people have been chosen.
See the step by step solution

## Step 1. Given information

1. Two people chosen at random were born during the same day of the week.
2. A group of $n$ people chosen at random there are at least two born in the same day of the week.
3. The probability greater than $1}{2}$ that there are at least two people born in the same day of the week.

## Step 2. Definition and formula to be used

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

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

## Step 3. a) Probability to born same day of week

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 $\frac{6!}{7!}$

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

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

## Step 4. b) Compute probabilities

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.

## Step 5. c) Compute probabilities

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 $n⩾8$ 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 $2⩽n⩽7$

We get $n=5$

$5$people have been chosen. ### Want to see more solutions like these? 