Question

A two-digit positive number is selected at random. Then the probability that its tens digit is at least four more than its unit digit, is

• A. $\frac{1}{6}$
• B. $\frac{7}{30}$
• C. $\frac{14}{45}$
• D. $\frac{8}{17}$

Answer 1

The correct option is B $\frac{7}{30}$

Total two-digit number, $n\left(S\right)=90$
Let two-digit positive number be $xy.$
Then, $x\ge y+4$
If $y=0,$ then $x=4,5,6,7,8,9$
If $y=1,$ then $x=5,6,7,8,9$
If $y=2,$ then $x=6,7,8,9$
If $y=3,$ then $x=7,8,9$
If $y=4,$ then $x=8,9$
If $y=5,$ then $x=9$
Total favourable cases, $n\left(A\right)=6+5+4+3+2+1=21$
Required probability, $P\left(A\right)=\frac{21}{90}=\frac{7}{30}$