Person and week problem.

**a69356** · April 1st 2009, 06:22 AM

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

**Soroban** · April 1st 2009, 05:41 PM

Hello, Ashish!

In a room there are 7 persons.
Find the probability that two of them are born on the same day of week.

. . $(a)\;\frac{1080}{7^5}\qquad(b)\;\frac{2160}{7^5}\q quad(c)\;\frac{540}{7^4}\qquad(d)\;\text{None of these}$

I got "None of these" until I read the problem again . . .

I believe the problem means "exactly two have the same weekday of birth".

There are: . ${7\choose2} = 21$ ways to pick the two people.
. . Call them $A$ and $B.$

$A$ can have any day for his birthday: $\frac{7}{7} = 1$
. . $B$ must have the same day: . $\frac{1}{7}$
. . . . $C$ must have one of the remaining six days: . $\frac{6}{7}$
. . . . . . $D$ must have one of the remaining five days: . $\frac{5}{7}$
. . . . . . . . $E$ must have one of the remaining four days: . $\frac{4}{7}$
. . . . . . . . . . $F$ must have one of the remaining three days: . $\frac{3}{7}$
. . . . . . . . . . . . . $G$ must have one of the remaining two days: . $\frac{2}{7}$

Therefore, the probability is: . $21\cdot1\cdot\frac{1}{7}\cdot\frac{6}{7}\cdot\frac {5}{7}\cdot\frac{4}{7}\cdot\frac{3}{7}\cdot\frac{2 }{7}$

. . which simplifies to: . $\frac{2160}{7^5}$ . . . answer (c)

**a69356** · April 1st 2009, 08:20 PM

Thanks a lot Soroban.

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