Question 258726
It hasn't occurred to the other tutor that although any one number is just
as likely to be chosen as any other, it is not as likely that one will be
chosen between two numbers close together as it is that one will be chosen
between two numbers far apart.
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Suppose that 10 numbers are select at random from the interval [0, 1]. Find the probability that the first 5 are less than 1/4 and at least one of the numbers is less than 1/10.

<pre><font size = 4 color = "indigo"><b>
Let's draw the interval [0, 1] and mark the point at {{{0}}} A,
the point at  {{{1/10}}} B, the point at {{{1/4}}} C, and the
point at {{{1}}} D

{{{drawing(600,70,-1,21,-3,3,
green(line(0,0,2,0)), red(line(2,0,5,0)),line(5,0,20,0),
 line(2,-.3,2,.3),
locate(-.1,1.7,A),locate(2-.1,1.7,B), locate(5-.1,1.7,C),
locate(20-.1,1.7,D),
line(5,-.3,5,.3), line(0,-.3,0,.3), line(20,-.3,20,.3),
locate(-.1,-.4,0), locate(2-.2,0,1/10), locate(5-.2,0,1/4),
locate(20-.1,-.4,1)
 )}}}  

The probability of choosing a number between A and B is {{{1/10}}}.
The probability of choosing a number between A and C is {{{1/4}}}.
The probability of choosing a number between B and C is {{{1/4-1/10=3/20}}}
The probability of choosing a number between B and D is {{{1-1/10=9/10}}}

The desired probability =

P(the first 5 are between A and C) MINUS 

P(the first 5 are between B and C AND the last 5 are between B and D)

=

{{{(1/4)^5 - (3/20)^5*(9/10)^5}}}

= .0009317221656
Edwin</pre>