All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Probability
S.S. Minnow SOS (Posted on 2020-09-02) Difficulty: 3 of 5
The coast guard station has received a distress call from the S.S. Minnow, sinking near one of two islands in opposite directions from the station. The call was interrupted by radio failure on the Minnow before the tourboat could specify which of the two islands it was near.

The station chief knows from experience with that tour operator that there is a 20% chance the boat is near island A and an 80% chance it's near island B. The station has 13 rescue boats, and, again from experience, it is known that each rescue boat has, independently of the other search boats, a 20% probability of finding a distressed boat if indeed a distressed boat is present, effecting a rescue.

How should the 13 boats be split between the two islands to maximize the probability that the people aboard the Minnow will be rescued? What is the probability that they will in fact be rescued if that optimal strategy is followed?

Part 2:

Suppose the coast guard station has 40 boats available but each one has only a 5% probability, independently, of finding a ship in trouble (given there is one in that location to be found). And further, there's only a 10% probability the boat is near island A, 90% of being near island B.

See The Solution Submitted by Charlie    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Possibly solution | Comment 2 of 6 |
If x ships are sent to a location that has a distressed boat and the probability of each is p then the chance that at least one of them will find the boat is 1-(1-p)^x.  Splitting the ships among two locations using the numbers for part 1 gives the total probability

(1-.8^x)*.2+(1-.8^(13-x))*.8

A quick table shows the probability has a maximum of 0.8117 when x=3.  So we should send 3 ships to A and 10 to B.

Part two similarly gives

(1-.95^x)*.1+(1-.95^(40-x))*.9

Which shows a maximum of 0.7843 when x=0.  So we should send all of the ships to B.

This was a bit of a surprise.   Looking closer, though, the difference between sending 39 vs 40 ships to B is greater than the difference between sending 1 vs 0 ships to A when we consider how unlikely it is to have a distressed boat at A.  


  Posted by Jer on 2020-09-02 09:38:59
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (1)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (22)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information