Question

If $10$ things are distributed among $3$ persons, the chance of a particular person having more than $5$ of them is $\frac{1507}{19683}$.

Answer 1

Let the $3$ men be $A,B,C$ and we need to find the probability of $A$ (random) receiving more than $5$ things

Number of cases when the $A$ receive all $10$ things ${=}^{10}{C}_{10}=1$

Number of cases when the $A$ receive $9$ things and the remaining is distributed among $B$ and $C$ ${=}^{10}{C}_9\cdot 2=20$

Number of cases when the $A$ receive $8$ things and the remaining is distributed among $B$ and $C$ ${=}^{10}{C}_8\cdot 2^2=180$

Number of cases when the $A$ receive $7$ things and the remaining is distributed among $B$ and $C$ ${=}^{10}{C}_7\cdot 2^3=960$

Number of cases when the $A$ receive $6$ things and the remaining is distributed among $B$ and $C$ ${=}^{10}{C}_6\cdot 2^4=3360$

Total number of ways of distributing $10$ things among $A,B$ and $C$ $=3^{10}$

Therefore, probability of $A$ recieving more than $5$ things $=\frac{1+20+180+960+3360}{3^{10}}=\frac{1507}{19683}$