Probability
Geometric Probability
Grade 12
Question:
<p>Two friends visit a restaurant randomly during 5 pm to 6 pm. Among the two, whoever comes first waits for 15 min and then leaves. The probability that they meet is:</p>
<p>(a) \(\frac{1}{4}\)</p>
<p>(b) \(\frac{1}{16}\)</p>
<p>(c) \(\frac{7}{16}\)</p>
<p>(d) \(\frac{9}{16}\)</p>
Step-by-Step Solution
Key Concept: This is a geometric probability problem. They meet if the absolute difference in arrival times is at most 15 minutes. Use the coordinate plane with arrival times as axes.
<p>Let arrival times of two friends be \(x\) and \(y\) (in minutes after 5 pm), where \(0 \leq x, y \leq 60\).</p><p>They meet if \(|x - y| \leq 15\).</p><p>Total area = \(60 \times 60 = 3600\)</p><p>Favorable region: \(|x - y| \leq 15\)</p><p>Area of favorable region = Total area - Area of two corner triangles</p><p>= \(3600 - \frac{1}{2}(45)^2 - \frac{1}{2}(45)^2 = 3600 - 2025 = 1575\)</p><p>Probability = \(\frac{1575}{3600} = \frac{7}{16}\)</p>
Correct Answer: C