If \(S=\{1,2,....,50\}\), two numbers \(\alpha\) and \(\beta\) are selected at random find the probability that product is divisible by 3 :
Step 1: Understanding the Question:
We are selecting two distinct numbers from the set S = \(\{1, 2, ..., 50\}\). We need to find the probability that their product, \(\alpha\beta\), is a multiple of 3. It's often easier to calculate the probability of the complementary event.
Step 2: Complementary Event:
The complementary event is that the product \(\alpha\beta\) is NOT divisible by 3. This occurs if and only if neither \(\alpha\) nor \(\beta\) is divisible by 3.
Step 3: Total Number of Outcomes:
The total number of ways to choose two distinct numbers from 50 is given by the combination formula: \[ \text{Total Outcomes} = ^{50}C_2 = \frac{50 \times 49}{2 \times 1} = 25 \times 49 = 1225 \]
Step 4: Favorable Outcomes for the Complementary Event:
First, we count the numbers in S that are not divisible by 3.
Numbers divisible by 3 in S are \(\{3, 6, 9, ..., 48\}\). The number of such terms is \(\frac{48}{3} = 16\).
Numbers NOT divisible by 3 in S are \(50 - 16 = 34\). For the product \(\alpha\beta\) to not be divisible by 3, both \(\alpha\) and \(\beta\) must be chosen from these 34 numbers. The number of ways to choose 2 numbers from these 34 numbers is: \[ \text{Favorable Outcomes for Complement} = ^{34}C_2 = \frac{34 \times 33}{2 \times 1} = 17 \times 33 = 561 \]
Step 5: Calculating Probabilities:
The probability of the complementary event (product not divisible by 3) is: \[ P(\text{not divisible by 3}) = \frac{\text{Favorable Outcomes for Complement}}{\text{Total Outcomes}} = \frac{561}{1225} \] The probability of the desired event (product is divisible by 3) is 1 minus the probability of the complementary event: \[ P(\text{divisible by 3}) = 1 - P(\text{not divisible by 3}) = 1 - \frac{561}{1225} \] \[ P(\text{divisible by 3}) = \frac{1225 - 561}{1225} = \frac{664}{1225} \]
Step 6: Final Answer:
The probability that the product is divisible by 3 is \(\frac{664}{1225}\).
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: \( P(3 < x \leq 6) \)

