Question

The probability that among $7$ persons, no $2$ were born on the same day of a week is
A.$\dfrac{2}{7}$
B.$\dfrac{{7!}}{7}$
C.$\dfrac{{7!}}{{{7^7}}}$
D.$\dfrac{2}{{{7^7}}}$

Answer 1

Hint: Probability of any given event is equal to the ratio of the favourable outcomes with the total number of all the outcomes. Probability is the state of being probable and the extent to which something is likely to happen in the particular favourable situations. Here, we will find the probability by using the formula, $P(A) = \dfrac{{n(A)}}{{n(S)}}$

Complete step-by-step answer:
Given that the total number of persons $ = 7$
The sample space, $n(S)$ is the number of the possible ways $7$ persons have birthday in a week is ${7^7}$
Therefore, $n(S) = {7^7}{\text{ }}......{\text{(1)}}$
Let, A be an event that all have birthdays in the different days of the week.
Therefore, $n(A)$ is equal to the total number of ways of selecting $7$different days for $7$ persons.
$\therefore n(A) = {}^7P{}_7$
Simplify the right hand side of the equation –
$\eqalign{
  & \therefore n(A) = \dfrac{{7!}}{{(7 - 7)!}} \cr
  & \therefore n(A) = \dfrac{{7!}}{{(0)!}} \cr} $
 $\eqalign{
  & As,{\text{ 0! = 1}} \cr
  & \therefore {\text{n(A) = 7!}}\,{\text{ }}.......{\text{(2)}} \cr} $
Now, the required probability that no $2$ were born on the same day of a week is –
$\Rightarrow P(A) = \dfrac{{n(A)}}{{n(S)}}$
Place values of equation $(2){\text{ and (1) in the above equation}}$
$\Rightarrow P(A) = \dfrac{{7!}}{{{7^7}}}$
Therefore, the required answer - The probability that among $7$ persons, no $2$ were born on the same day of a week is $\dfrac{{7!}}{{{7^7}}}$

So, the correct answer is “Option C”.

Note: For this type of probability problems, just follow the general formula for probability and basic simplification properties for the fractions. Always remember that the probability of any event lies between zero and one. $1$. It can never be negative or the number greater than one. The probability of an impossible event is always zero.