Q - In a room there are 7 persons.The chances that the two of them born on the same day of week.
a) 1080/7^5 b) 2160/7^5 c) 540/7^4 d) None of these
Any hint would be greatly appreciated.
Thanks,
Ashish
Hello, Ashish!
I got "None of these" until I read the problem again . . .In a room there are 7 persons.
Find the probability that two of them are born on the same day of week.
. . $\displaystyle (a)\;\frac{1080}{7^5}\qquad(b)\;\frac{2160}{7^5}\q quad(c)\;\frac{540}{7^4}\qquad(d)\;\text{None of these}$
I believe the problem means "exactly two have the same weekday of birth".
There are: .$\displaystyle {7\choose2} = 21$ ways to pick the two people.
. . Call them $\displaystyle A$ and $\displaystyle B.$
$\displaystyle A$ can have any day for his birthday: $\displaystyle \frac{7}{7} = 1$
. . $\displaystyle B$ must have the same day: .$\displaystyle \frac{1}{7}$
. . . . $\displaystyle C$ must have one of the remaining six days: .$\displaystyle \frac{6}{7}$
. . . . . . $\displaystyle D$ must have one of the remaining five days: .$\displaystyle \frac{5}{7}$
. . . . . . . . $\displaystyle E$ must have one of the remaining four days: .$\displaystyle \frac{4}{7}$
. . . . . . . . . . $\displaystyle F$ must have one of the remaining three days: .$\displaystyle \frac{3}{7}$
. . . . . . . . . . . . . $\displaystyle G$ must have one of the remaining two days: .$\displaystyle \frac{2}{7}$
Therefore, the probability is: .$\displaystyle 21\cdot1\cdot\frac{1}{7}\cdot\frac{6}{7}\cdot\frac {5}{7}\cdot\frac{4}{7}\cdot\frac{3}{7}\cdot\frac{2 }{7}$
. . which simplifies to: .$\displaystyle \frac{2160}{7^5}$ . . . answer (c)