A bag contains the numbers 1 through 20 inclusive. If three are selected at random and placed in order from smallest to largest, find the probability that the arrangement is of three consecutive integers.
Printable View
A bag contains the numbers 1 through 20 inclusive. If three are selected at random and placed in order from smallest to largest, find the probability that the arrangement is of three consecutive integers.
Hello,Quote:
Originally Posted by punmaster
I made a diagram to analyze the described situation. Maybe you can use it for your calculations
EB
Here is another way.Quote:
Originally Posted by punmaster
---
Probability is favorable to possible.
What is favorable?
{1,2,3} All premutations of this set = 6
{2,3,4} All premutations of this set = 6
.....
{18,19,20} All premutations of this set = 6
In total we have,
18 Sets
Thus,
$\displaystyle 18\cdot 6=108$
The possible outcomes is the number of ways of selecting 3 out of twenty.
Which is,
$\displaystyle _{20}C_3=1140 $
Thus,
$\displaystyle \frac{108}{1140}$
There are 18 subsets of three consecutive integers.Quote:
Originally Posted by ThePerfectHacker
There are $\displaystyle _{20}C_3=1140 $ total subsets three integers.
Thus, $\displaystyle \frac{18}{1140} = 0.01579$
The order does not come to this.
If we do use order then we would have $\displaystyle \frac{108}{(20)(19)(18)} = 0.01579$
You are right. When I was doing this problem I first did it with order then realized that order is not important. So I re-edited the problem but forgot to remove the 3!Quote:
Originally Posted by Plato
