\( \frac{13}{66} \)
Total balls = \( 3 + 4 + 5 = 12 \). Total ways to draw 2 balls = \( \binom{12}{2} = 66 \).
Same color:
- Red: \( \binom{3}{2} = 3 \)
- White: \( \binom{4}{2} = 6 \)
- Blue: \( \binom{5}{2} = 10 \)
Total favorable = \( 3 + 6 + 10 = 19 \).
Probability = \( \frac{19}{66} \).
Thus, the answer is \( \frac{19}{66} \).
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 :
For any natural number $k$, let $a_k = 3^k$. The smallest natural number $m$ for which \[ (a_1)^1 \times (a_2)^2 \times \dots \times (a_{20})^{20} \;<\; a_{21} \times a_{22} \times \dots \times a_{20+m} \] is: