Question

Three numbers are chosen at random from numbers 1 to 20. The probability that they are consecutive is

• A. (\frac{1}{190})
• B. (\frac{1}{120})
• C. (\frac{3}{190})
• D. [suspicious link removed]

Hint:

For (n) numbers, the number of consecutive sets of size (k) is (n - k + 1).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We need to find the probability that 3 numbers chosen from 20 form a sequence of consecutive integers.
Step 3: Detailed Explanation:
1. Total number of ways to choose 3 numbers from 20 is \( n(S) \):
\[ n(S) = {}^{20}C_3 = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = 20 \times 19 \times 3 = 1140 \]
2. Let \( E \) be the event that chosen numbers are consecutive. Possible sets are:
\( \{1, 2, 3\}, \{2, 3, 4\}, \{3, 4, 5\}, \dots, \{18, 19, 20\} \)
The first number of the set can range from 1 to \( 20 - 3 + 1 = 18 \).
So, \( n(E) = 18 \).
3. Probability \( P(E) \):
\[ P(E) = \frac{n(E)}{n(S)} = \frac{18}{1140} \]
Divide both by 6:
\[ P(E) = \frac{3}{190} \]
Step 4: Final Answer:
The probability is \( \frac{3}{190} \).