Why is my line so slow?

We often wait on checkout lines at the grocery store, and look around to see that other lines are going faster than ours. Suppose your grocery store has five lines, what is the chance of this happening?

Note: By "this", it means "some other line is faster".

4 5 \frac45 1 5 \frac15 3 4 \frac34 1 1

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Alexa Kottmeyer
May 9, 2015

Since there are 5 lines and you chose 1 of them then, assuming equal likelihood of each line being fastest, there is a 1/5 chance that you picked the fastest and therefore a 4/5 chance that you did not.

If we see individually each line has 3 possibilities either it is faster ,or it at the same speed,or it is slower than our's. The possibilities of each line being faster is 1/3. total probability will be 4×(1/3)(2/3)(2/3)(2/3) +6×(1/3)(1/3)(2/3)(2/3)+4×(1/3)(1/3)(1/3)(2/3)+1×(1/3)(1/3)(1/3)(1/3)=65/81

Shubham Agarwal - 5 years, 6 months ago

There are 5 lines all together!! All in all none of the answers are correct. Assuming 50% chance of another of the 4 lines being faster then the chance that NO other line is faster is 1/16 hence the chance that at least one is faster is 15/16.

David Rubinstein - 6 years, 1 month ago

Log in to reply

This would imply that each line has a 1/16 chance of being the fastest, meaning altogether there is a 5/16 chance that there is one fastest line, which makes no sense because at least one line must be the fastest. The fault in your logic is that if your line is compared with another, and you find your line is faster, then it is more likely that your line is faster than the "average" line. This means that if you are compared with a third line, your are 66.7% likely to be in a faster line. Mathematically, 1/2 x 2/3 x 3/4 x 4/5 = 1/5 = probability of being the fastest

This is somewhat similar to the Monty Hall problem.

Jonathan Delgadillo - 6 years, 1 month ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...