To solve this problem, we need to determine the probability that the product of two randomly selected distinct numbers from the set \(\{1, 2, 3, \ldots, 50\}\) is divisible by 3. Let's break it down step by step:
The total number of ways to select two distinct numbers from 50 numbers is given by the combination formula: \(C(50, 2) = \frac{50 \times 49}{2} = 1225\)
The numbers divisible by 3 up to 50 are \(\{3, 6, 9, \ldots, 48\}\). This forms an arithmetic sequence with:
Using the formula for the number of terms in an arithmetic sequence: \(n = \frac{48 - 3}{3} + 1 = 16\) Thus, there are 16 numbers that are divisible by 3.
The total number of numbers not divisible by 3 from 1 to 50 is 50 - 16 = 34. The number of ways to choose two numbers from these 34 numbers is: \(C(34, 2) = \frac{34 \times 33}{2} = 561\)
\[ P(\text{Product not divisible by 3}) = \frac{561}{1225} \]
Since the complement of the event (product not divisible by 3) has a probability of \(\frac{561}{1225}\), the probability that the product is divisible by 3 is: \(1 - \frac{561}{1225} = \frac{664}{1225}\)
After reviewing the options, the probability that the product of two randomly chosen numbers is divisible by 3 is \(\frac{664}{1225}\). Thus, the correct answer given the correct option should actually be corrected in the source or verified for further detail.
P and Q play chess frequently against each other. Of these matches, P has won 80% of the matches, drawn 15% of the matches, and lost 5% of the matches.
If they play 3 more matches, what is the probability of P winning exactly 2 of these 3 matches?
If A and B are two events such that \( P(A \cap B) = 0.1 \), and \( P(A|B) \) and \( P(B|A) \) are the roots of the equation \( 12x^2 - 7x + 1 = 0 \), then the value of \(\frac{P(A \cup B)}{P(A \cap B)}\)