Mr. A lives at origin on the Cartesian plane and has his office at (4, 5). His friend lives at (2, 3) on the same plane. Mr. 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 Mr. A passed his friends house is (shortest path for any event must be considered)

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Solution

The correct option is B
$n (S) = \frac{9!}{4! 5!} = 126$
n(A)=0 to F and F to P
= $\frac{5!}{2! 3!} \times \frac{4!}{2! 2!} = 60 \Rightarrow P (A) = \frac{60}{126} = \frac{10}{21}$

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The correct option is B n(S)=9!4!5!=126 n(A)=0 to F and F to P =5!2!3!×4!2!2!=60⇒P(A)=60126=1021

The correct option is B
$n (S) = \frac{9!}{4! 5!} = 126$
n(A)=0 to F and F to P
= $\frac{5!}{2! 3!} \times \frac{4!}{2! 2!} = 60 \Rightarrow P (A) = \frac{60}{126} = \frac{10}{21}$