First slide

Introduction to probability

Question

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 traveling 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)

Moderate

A

1/2
B

10/21
C

1/4
D

11/21

Solution

$\begin{matrix} n (S) = \frac{9!}{4! . 5!} = 126 \\ n (A) = O to F and F to P \\ = \frac{5!}{2! \times 3!} \times \frac{4!}{2! \times 2!} = 60 \end{matrix}$
$\Rightarrow P (A) = \frac{60}{126} = \frac{10}{21}$

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