Question

Find the probability that the birthdays of 6 different persons will fall in exactly two calendar months.

Answer 1

Hint: Find S = the no. of choices for 6 people to have birthdays in the 12 months. Find A = the no. ways in which 2 months can be chosen from 12 months. Find B = the no. of ways in which the 6 birthdays fall in exactly 2 months. Then find the probability using $P = \dfrac{{A \times B}}{S}$


Complete step by step solution:
We have 6 different people.
We have to choose 2 months and find the number of possibilities that the birthdays of these 6 people fall exactly in these two months.
The formula for probability of an event A is:
$P(A) = \dfrac{{n(A)}}{{n(S)}}$
Where $n(S)$ is the no. of possible events in the sample space
$n(A)$ is the number of possibilities for the event A.
As there are 12 months in all, the number of ways in which 6 people can choose from 12 months ${ = ^{12}}{C_6}$
We can choose any two months from 12 months in $^{12}{C_2}$ ways.
Now, each person has 2 months as choices and there are 6 people.
Therefore, the number of ways in which this choice can be made $ = {(2 \times 2 \times 2 \times 2 \times 2 \times 2)_{6times}} = {2^6}$
Now, we need these 6 people to not have birthdays falling in the same month.
This is possible in only 1 way.
Therefore, the number of ways in which the birthdays of 6 people can fall in exactly two months$ = {2^6} - 1$
Thus, the probability that the birthdays of 6 different persons will fall in exactly two calendar months $ = \dfrac{{^{12}{C_2} \times ({2^6} - 1)}}{{^{12}{C_6}}}$


Note: Students tend to get confused between the formula for permutation and combination.
The formula for combination is:$^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$ and that of permutation is $^n{P_r} = \dfrac{{n!}}{{(n - r)!}}$
Therefore, the best way to remember is by memorizing the relation between these two formulae which is given by $^n{C_r} = \dfrac{{^n{P_r}}}{{r!}}$. Dividing by $r!$ ensures that the repetitions are not counted.