Question

The probability that a positive two digit number selected at random has its tens digit at least three more than its unit digit is __________.

• A. $14/45$
• B. $7/45$
• C. $36/45$
• D. $1/6$

Answer 1

The correct option is A $14/45$

$\to$ For each unit digits, writing down the possible

tenis digits:-

$0-3,4----9=7$ $\Rightarrow$ probability $=$ Foreable cases

$1-4,---,9=6$ Total cases

$2-5,---,9=5$ $\frac{\frac{7(7+1)}{2}}{90}=\frac{7\times 8}{2\times 90}=\frac{14}{45}$

$3----=1$

$4----=3$

$5----=3$ $\Rightarrow \left(A\right)$

$6----=1$

$7----=0$

$8,9----=0$