Question

If two numbers \( a \) and \( b \) are selected from \( S = \{1, 2, 3, \dots, 100\} \), then the probability that \( |a - b| \geq 10 \) is:

• A. \( \frac{891}{1000} \)
• B. \( \frac{119}{1000} \)
• C. \( \frac{819}{1000} \)
• D. None of these

Hint:

When calculating probabilities involving absolute differences, count the number of valid pairs for each possible value and divide by the total number of possible outcomes.

Answer 1

The Correct Option is C

To solve this problem, we need to determine the probability that the absolute difference between two numbers \(a\) and \(b\) selected from set \(S = \{1, 2, 3, \dots, 100\}\) is at least 10. The absolute difference is given by \(|a - b|\).

The total number of ways to choose two numbers from set \(S\) is calculated using the combination formula:

\[\binom{100}{2} = \frac{100 \times 99}{2} = 4950\]

This gives us the total number of possible pairs \((a, b)\).

Now, we must find the number of pairs for which \(|a - b| \geq 10\).

Consider that if \(a \geq b\), then \(|a - b| \geq 10\) implies \(a - b \geq 10\), or equivalently \(a \geq b + 10\). For each possible value of \(b\), we determine the range of \(a\).

• If \(b = 1\), then \(a\) can be \{11, 12, \dots, 100\}. This gives 90 options for \(a\).
• If \(b = 2\), then \(a\) can be \{12, 13, \dots, 100\}. This gives 89 options for \(a\.
• Continue this process until \(b = 90\), where \(a\) can only be 100, giving 1 option.

This forms an arithmetic series where the first term \(a_1 = 90\) and the last term \(a_{90} = 1\). The sum of an arithmetic series can be found using the formula:

\[S = \frac{n}{2} (a_1 + a_n)\]

where \(n\) is the number of terms.

Here, \(n = 90\), \(a_1 = 90\), and \(a_n = 1\). Thus,

\[S = \frac{90}{2} (90 + 1) = 45 \times 91 = 4095\]

The number of favorable pairs \((a, b)\) such that \(|a - b| \geq 10\) is 4095.

The probability that \(|a - b| \geq 10\) is then calculated as follows:

\[P = \frac{4095}{4950} = \frac{819}{1000}\]

Therefore, the correct answer is \(\frac{819}{1000}\).