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

1 a)) as there are total 5 right and 4 up steps ; hence total number of ways to go from A to B =9!/(4!*5!) =126

total number of ways to go from A to C as there are 2 right and 2 up steps =4!.(2!*2!) =6

and total number of ways to go from C to B as thre are 3 right and 2 up steps from C to B =5!/(2!*3!) =10

hence total number of ways that goes from A to B and pass through C =10*6 =60

therefore probabilty you pass through component C =60/126=10/21

b) number of path to reach from C to D and then D to B =(2!/(1!*1!))*(3!/1!*2!)=6

hence total number of path C to B which does not have D =10-6 =4

therefore paths from A to B which have C but do not have D =4*6=24

hence probability =24/126=4/21

c)number of paths from A to D and then goes from D to B =6!/(3!*3!)*3!/(2!*1!)=60

hence number of paths which does not go through C or D =126-(60+60-36)=42

therefore probability =42/126 =2/6=1/3

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