Below is an overhead view of a manufacturing warehouse. The lines represent aisl

Question

Below is an overhead view of a manufacturing warehouse. The lines represent aisles within the warehouse. You have been sent to retrieve 2 components within the warehouse, but are unsure of their locations. You want to spend the least amount of time in the warehouse as possible, therefore you want to avoid searching the entire warehouse. You have decided to enter the warehouse at point A and exit the warehouse at point B. At ever intersection you will flip a coin. If the coin flip is heads you move upward, and if the coin flip is tails you move right. Your hope is that you walk pass (and pick up) both components you were sent to find, which are located at points C and D. Assumptions: If you walk past a component that you need you pick it up. If you are unable to move upward, you move right instead. If you are unable to move right you move upward instead. What is the probability that you exit the warehouse with the component located at point C? What is the probability that you exit the warehouse with the component located at point C but do not have the component located at point D? What is the probability that you exit the warehouse empty handed (and get fired)?

Explanation / Answer

Solution

For ease in explaining and presentation, let F represent upward movement and R represent movement to the right. Since on flipping a coin, probability of getting a head = probability of getting a head = ½, probability of F and R are equal and hence at every intersection the probability of moving in either direction is ½.

Part (a) : There are 6 possible ways of going from A to C, as shown below: F-F-R-R; F-R-R-F; F-R-F-R; R-F-F-R; R-F-R-F; R-R-F-F.

And each possibility has a probability of (½)4.

Hence, P(exit with component at C) = 6/16 = 3/8 ANSWER

Part (b) : There are 20 possible ways of going from A to D. This can be either enumerated as above or one can use the result that ‘in (n x m) grid, the number of possible ways of reaching a corner from the opposite corner, under the restrictions as given in our problem, is {(n + m)!}/{(n!)(m!)}.

For each possibility, there are 6 intersections and so probability at each intersection is (½)6 = 1/64.

Thus, P(exit with component at D) = 20/64 = 5/16 => P(exit without component at D) = 1- 5/16 = 11/16.

Now, P(exit with component at C, but not the component at D) = P(exit with component at C) x P(exit without component at D) = (3/8)(11/16)

= 33/128 ANSWER   

Part (c)    

P(exit empty-handed) = {1- P(exit with component at C)}x{1 - P(exit with component at D)}

= (5/8)(11/16)

= 55/128 ANSWER.

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