Three different numbers are selected at random from the first 15 natural numbers. What is the probability that the product of two of the numbers is equal to third number?
- A. \frac{1}{91}
- B. \frac{2}{455}
- C. \frac{1}{65} ✓
- D. \frac{6}{455}
Correct Answer: C. \frac{1}{65}
Explanation
Total number of ways to select 3 numbers from 15 is \binom{15}{3} = 455. We need subsets \{a, b, c\} where ab = c. Assuming a \lt b, the possible sets are \{2, 3, 6\}, \{2, 4, 8\}, \{2, 5, 10\}, \{2, 6, 12\}, \{2, 7, 14\}, \{3, 4, 12\}, and \{3, 5, 15\}. There are 7 favorable outcomes. The probability is \frac{7}{455} = \frac{1}{65}.
Related questions on Statistics & Probability
- Let x be the mean of squares of first n natural numbers and y be the square of mean of first n natural numbers. If $\frac{x}{y}=\fra...
- What is the probability of getting a composite number in the list of natural numbers from 1 to 50?
- Two numbers x and y are chosen at random from a set of first 10 natural numbers. What is the probability that (x+y) is divisible by 4?
- A number x is chosen at random from first n natural numbers. What is the probability that the number chosen satisfies $x+\frac{1}{x} \gt...
- Three fair dice are tossed once. What is the probability that they show different numbers that are in AP?