SOLUTION: 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. s

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Question 258726: 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.
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Found 2 solutions by Fombitz, Edwin McCravy:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
The assumptions is that all numbers are equally likely.
Choosing each number is an independent event.
P%285+numbers%3C0.25%29=%280.25%29%280.25%29%280.25%29%280.25%29%280.25%29=0.0009766
P%281+number%3C0.10%29=0.10

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
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.

Let's draw the interval [0, 1] and mark the point at 0 A,
the point at  1%2F10 B, the point at 1%2F4 C, and the
point at 1 D

  

The probability of choosing a number between A and B is 1%2F10.
The probability of choosing a number between A and C is 1%2F4.
The probability of choosing a number between B and C is 1%2F4-1%2F10=3%2F20
The probability of choosing a number between B and D is 1-1%2F10=9%2F10

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)

=

%281%2F4%29%5E5+-+%283%2F20%29%5E5%2A%289%2F10%29%5E5

= .0009317221656
Edwin