Question

Question 2. To determine

1. What is the probability that two people chosen at random were born during the same day of the week?
2. 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?
3. 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?

Short answer

Answer

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.

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 $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$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

$$\begin{gathered} p=7^n \\ p\left(A\right)=1-p\left(A^c\right) \end{gathered}$$

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

$$\begin{gathered} p=\frac{6!}{7!} \\ \Rightarrow \frac{1}{7} \end{gathered}$$

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.