Question

‘A’ lives at origin on the cartesian plane and his office at (4, 5). His friend lives at (2, 3) on the same space. ‘A’ can go to his office travelling one block at a time either in the +y or +x direction. If all possible paths are equally likely, then the probability that ‘A’ passed his friend’s house is

• A. $\frac{1}{2}$
• B. $\frac{10}{21}$
• C. $\frac{1}{4}$
• D. $\frac{11}{21}$

Answer 1

The correct option is B

$\frac{10}{21}$

$n\left(s\right)=\frac{9!}{4!5!}=126$
n(A)=0 to F and F to p
$=\frac{5!}{2!3!}.\frac{4!}{2!2!}=60P\left(A\right)=\frac{60}{126}=\frac{10}{21}$