Question

Question 2. To determine

1. What is the probability that two people chosen at random were born during the same month of the year?
2. What is the probability that in a group of$\mathit{n}$people chosen at random, there are at least two born in the same month of the year?
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 month of the year?

Short answer

Answer

1. The probability is$\frac{1}{12}$.
2. The probability is$1$minus the people not born in that month.
3. people have been chosen.

Step-by-step solution

Step 2. Definition and formula to be used

Probability deals with the random event.

Step 3. a) Compute the probability

The total number of months in a year is $12$. for the first person to be born on any one of the month is$1$

For the second person to be born on the same month is of the probability$\frac{11}{12}$

$$\begin{gathered} p=\frac{11!}{12!} \\ =\frac{1}{12} \end{gathered}$$

The probability is$\frac{1}{12}$.

Step 4. b) Compute probabilities

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

$1-\left(\frac{12}{12}\right)\left(\frac{11}{12}\right)\left(\frac{10}{12}\right)\left(\frac{9}{12}\right)\left(\frac{8}{12}\right)=61.8$

where$n=5$

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

Step 5. c) Compute probabilities

When $23$people are gathered, there is more chance that not that $2$if them have the same birthday. $23$people have a slightly overrole="math" localid="1668588532105" $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ probability of two of them sharing the same birthday.