Question
The probability that out of $$10$$ persons, all born in April, at least two have the same birthday is :
A.
$$\frac{{{}^{30}{C_{10}}}}{{{{\left( {30} \right)}^{10}}}}$$
B.
$$1 - \frac{{{}^{30}{C_{10}}}}{{30!}}$$
C.
$$\frac{{{{\left( {30} \right)}^{10}} - {}^{30}{C_{10}}}}{{{{\left( {30} \right)}^{10}}}}$$
D.
none of these
Answer :
$$\frac{{{{\left( {30} \right)}^{10}} - {}^{30}{C_{10}}}}{{{{\left( {30} \right)}^{10}}}}$$
Solution :
There are $$30$$ days in April.
$$n\left( S \right) = $$ the number of ways in which $$10$$ persons can have birthdays in the month of April
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 30 \times 30 \times .....$$ to $$10$$ times $$ = {30^{10}}$$ (because each person can have birthday in $$30$$ ways).
$$n\left( E \right) = n\left( S \right) - $$ the number of ways in which $$10$$ persons can have different birthdays
$$\eqalign{
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\left( {30} \right)^{10}} - {}^{30}{C_{10}} \cr
& \therefore \,P\left( E \right) = \frac{{{{\left( {30} \right)}^{10}} - {}^{30}{C_{10}}}}{{{{\left( {30} \right)}^{10}}}}. \cr} $$