Find the largest value of $m$ Assume that property P is possessed by at least two of $a_1,b_1,c_1$ ,by at least two of $a_2,b_2,c_2,\cdots$, by at least two of $a_5,b_5,c_5$. Find the largest value of $m$ for which it can be said that at least $m$ of the $a's$, or  $m$ of the $b's$, or  $m$ of the $c's$ have the property P. 
I have tried in the following way.
If the property $P$ is possessed by at least one of a1,b1,c1 ,by at least one of $a_2,b_2,c_2,\cdots$ by at least one of $a_5,b_5,c_5$ then then at least $5$ elements satisfy the property.By pigeon hole principle at least $2$ of the $a$ or $b$ or $c$ satisfy the property $P$. So largest value of m in this case is $2$. Can this argument be extended to the property "at least two" instead of "at least one" ?
 A: Let $A$ be the number of indices among $\{1,2,3,4,5\}$ for which $a_i$ has property $P$. Similarly, let $B$ (respectively, $C$), denote the number of indices among $\{1,2,3,4,5\}$ for which $b_i$ (respectively, $c_i$), has property $P$. I claim that in every case
$$\max\{A,B,C\}\geq 4$$
so that the answer to your question is $4$. (It is clearly not $5$). The proof is brute-force. 
If $A=0$, then clearly $B=C=5$. 
If $A=1$, then
in four cases both $b_i$ and $c_i$ must have property $P$, and in one additional case one of them must appear, so $\max\{B,C\}=5$. 
If $A=2$, then in three cases $a_i,b_i$ must have property $P$, and in two cases at least one of them occurs, so $\max\{B,C\}\geq 4$.
If $A=3$, then in $2$ cases $b_i,c_i$ must have property $P$. In each case of the other three cases, where $a_i$ already has property $P$, at least one of $b_i,c_i$ must have property $P$, so one of them must have it in two of the remaining three cases, therefore, again, $\max\{B,C\}\geq 4$.
If $A\geq 4$, then clearly $\max\{A,B,C\}\geq 4$.
Edit:
As is often the case, after a solution has been presented, there is a much easier way to do it. As Jaap Scherphuis pointed out, the sum $A+B+C$ is clearly at least $10$, so that the average is at least $10/3$, from which it follows that the maximum is at least $4$. Nonetheless, I am keeping the book-keeping solution above, because I like it. 
