Question

The probability that birthday of six different persons will fall in exactly two calendar months is
A.$\dfrac{{341}}{{{{12}^5}}}$
B.${}^{12}{c_2} \times \dfrac{{{2^6} - 1}}{{{{12}^6}}}$
C.${}^{12}{c_2}\dfrac{{{2^6}}}{{{{12}^6}}}$
D.$\dfrac{1}{6}$

Answer 1

In this question first we find the total number of events. After that we will find the number of ways to choose two months from 12 months and also find in how many ways birthday of six persons fall in 2 months then in these ways there will be two ways in which birthday of all of six will fall in same month we have to discard those two cases and then will get favorable number of elementary events then apply the formula of probability and get the desired answer.

Complete step-by-step answer:
As we all know that anyone’s birthday can fall in one of the 12 months.
So, there are 12 ways to select one’s birthday month.
Here, we have 6 people so each person has 12 ways to select a birthday month.
Which implies the total number of elementary events is $12 \times 12 \times 12 \times 12 \times 12 \times 12$.
This further implies that,
Total number of elementary events is${12^6}$.
Now, we have to choose two months from 12 months.
So, any two months can be chosen in $^{12}{c_2}$ways.
We can choose one’s birthday month from 2 month in 2ways. So, the six birthdays can fall in 2 month in $2 \times 2 \times 2 \times 2 \times 2 \times 2$ways which is equal to ${2^6}$ways.
Out of this ${2^6}$ways there are two ways when all the six birthdays fall in one month. So, we have to discard that 2 ways from ${2^6}$ways.
Therefore, favorable number of elementary events $^{12}{c_2} \times \left( {{2^6} - 2} \right)$
Since, probability = $$$$\dfrac{{favorable{\text{ }}events}}{{total{\text{ }}number{\text{ }}of{\text{ }}events}}
Thus, required probability $= \dfrac{{^{12}{c_2} \times \left( {{2^6} - 2} \right)}}{{{{12}^6}}}$
We will use the formula $^n{c_r} = \dfrac{{n!}}{{r! \times \left( {n - r} \right)!}}$to solve the above expression.
$\Rightarrow \dfrac{{\dfrac{{12!}}{{2! \times \left( {12 - 2} \right)!}} \times \left( {{2^6} - 2} \right)}}{{{{12}^6}}}$
$\Rightarrow \dfrac{{\dfrac{{12!}}{{2! \times \left( {10} \right)!}} \times \left( {{2^6} - 2} \right)}}{{{{12}^6}}}$
$\Rightarrow \dfrac{{\dfrac{{12 \times 11 \times 10!}}{{2 \times \left( {10} \right)!}} \times 2\left( {{2^5} - 1} \right)}}{{{{12}^6}}}$
$\Rightarrow \dfrac{{12 \times 11 \times \left( {{2^5} - 1} \right)}}{{{{12}^6}}}$
$\Rightarrow \dfrac{{11 \times \left( {{2^5} - 1} \right)}}{{{{12}^5}}}$
On simplifying further we get,
$\dfrac{{11 \times \left( {32 - 1} \right)}}{{{{12}^5}}}$
$\Rightarrow \dfrac{{11 \times 31}}{{{{12}^5}}}$
$\Rightarrow \dfrac{{341}}{{{{12}^5}}}$
Thus, the probability that birthday of six different persons will fall in exactly two calendar months is $\dfrac{{341}}{{{{12}^5}}}$
Hence, option A. $\dfrac{{341}}{{{{12}^5}}}$is the correct answer.

Note: Probability is how likely an event is to occur or how likely it is that a proportion is true.
Probability of an event always lies between 0 to 1.
We use the concept of combination when we have to select something from a collection. Therefore, we have used the concept of combination in the above question because we have to select 2 months from 12 months.