| Scenario | Jar 1 | Jar 2 | Chance of Picking White |
|---|---|---|---|
| 25 white and 25 black in each | 25 white, 25 black | 25 white, 25 black | 50% |
| White in one and 99 in the other | 1 white | 99 (49 white, 50 black) | 50% |
| 50 white in one and 50 black in the other | 50 white | 50 black | 50% |
| All hundred in one | 100 (50 white, 50 black) | 0 | 50% |
To maximize the probability of picking a white marble, strategy 2 is the most effective. In this setup, you put 1 white marble in Jar 1 and the rest (49 white, 50 black) in Jar 2. When you draw from Jar 1, the probability of getting a white marble is 100%. If you draw from Jar 2, the probability is 49/99. Thus, the overall probability of picking a white marble is the average of the probabilities when drawing from each jar. Assume you always choose Jar 1 when available:
Prob(White) = Prob(Choose Jar 1) * Prob(White from Jar 1) + Prob(Choose Jar 2) * Prob(White from Jar 2) = 0.5 * 1 + 0.5 * (49/99) ≈ 0.745, which is 74.5%.
This approach yields the highest possible likelihood of picking a white marble, significantly better than the 50% baseline in other configurations.
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 :