There are $4$ red, $5$ green, and $6$ blue balls inside a box. If $N$ number of balls are picked simultaneously, what is the smallest value of $N$ that guarantees there will be at least two balls of the same colour? One cannot see the colour of the balls until they are picked.
Step 1: Identify the categories (pigeonholes).
There are $3$ colours: Red, Green, Blue $\Rightarrow$ $3$ pigeonholes.
Step 2: Worst-case reasoning (Pigeonhole Principle).
To avoid getting two of the same colour as long as possible, pick one ball of each colour first.
After $3$ picks, it is still possible that all $3$ balls are of different colours.
Step 3: Force a repeat.
The next (4th) ball must match one of the already chosen colours, because only $3$ colours exist.
Therefore $N=4$ guarantees at least two balls of the same colour.
\[
\boxed{N_{\min}=4}
\]
If the probability distribution is given by:
| X | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|---|---|
| P(x) | 0 | k | 2k | 2k | 3k | k² | 2k² | 7k² + k |
Then find: \( P(3 < x \leq 6) \)
If \(S=\{1,2,....,50\}\), two numbers \(\alpha\) and \(\beta\) are selected at random find the probability that product is divisible by 3 :
Consider a five-digit number PQRST that has distinct digits P, Q, R, S, and T, and satisfies the following conditions:
1. \( P<Q \)
2. \( S>P>T \)
3. \( R<T \)
If integers 1 through 5 are used to construct such a number, the value of P is:



