Question

The unit interval (0,1) is divided into 5 equal parts. Then 5 points are chosen at random from the unit interval. What is the probability that rach selected point lies in a different part?

Answer 1

Since each part is divided into same length, the probability of picking up a given length is $\frac{1}{5}$.

Now, the probability of choosing five points such that that each selected point lies in a different part will have a Multinomial Distribution Property as:

$\frac{5!}{1!\times 1!\times 1!\times 1!\times 1!}\times \frac{1}{5}\times \frac{1}{5}\times \frac{1}{5}\times \frac{1}{5}\times \frac{1}{5}=\frac{24}{625}$

Thus, the probability that each selected point lies in a different part is $\frac{24}{625}$

Answer 2

The unit interval (0, 1) is divided into 5 equal parts

1/5 = .2

However, in order to select that each of the parts is selected in the random order then there are 5! Ways to select the number where there are 5 choices

Hence, Probability = 5/5!

= 5/ 5 * 4 * 3 * 2 * 1

= 1/24