<p>The probability that a positive two-digit number selected at random has its tens digit at least 3 more than its units digit is</p>
Step-by-Step Solution
Key Concept: Count two-digit numbers where tens digit t satisfies t - u \geq 3. Enumerate by t from 3 to 9.
<p>Total two-digit positive integers: 90 (from 10 to 99).</p><p>Tens digit $t$, units digit $u$, condition $t - u \geq 3$:</p><ul><li>$t=3$: $u=0$ → 1 number</li><li>$t=4$: $u \in \{0,1\}$ → 2 numbers</li><li>$t=5$: 3 numbers; $t=6$: 4; $t=7$: 5; $t=8$: 6; $t=9$: 7</li></ul><p>Total favorable $= 1+2+3+4+5+6+7 = 28$.</p><p>$P = \dfrac{28}{90} = \dfrac{14}{45}$</p>
Correct Answer: A