Question

two numbers between $0$ and $1$ on a number line are to be chosen at random. what is the probability that the second number chosen will exceed the first number chosen by a distance greater than $\frac 14$ unit on the number line? express your answer as a common fraction.

Answer 1

The probability that the second number chosen will exceed the first number chosen by a distance greater than 1/4 unit on the number line is 1/2.

Step-by-step explanation:

To calculate the probability that the second number chosen will exceed the first number chosen by a distance greater than 1/4 unit on the number line, we can use geometric probability. Imagine the first number as a point on the number line. The probability that the second number chosen will be greater than the first number by a distance greater than 1/4 unit is equal to the length of the region on the number line that satisfies this condition divided by the total length of the interval from 0 to 1.

The length of the region on the number line where the second number exceeds the first number by a distance greater than 1/4 unit is (1 - 1/4) - (0 + 1/4) = 3/4 - 1/4 = 1/2 unit. The total length of the interval from 0 to 1 is 1 unit. Therefore, the probability is (1/2) / 1 = 1/2.

Answer 2

Therefore, the probability that the second number chosen will exceed the first number chosen by a distance greater than
`$(1)/(4)$` unit on the number line is
`${(3)/(4)}$`.

Let's consider the range between
`$0$` and
`$1$` on the number line. We can divide this range into four equal parts:
`$[0, (1)/(4)]$`,
`$((1)/(4), (1)/(2)]$`,
`$((1)/(2), (3)/(4)]$`, and
`$((3)/(4), 1]$`.

In order for the second number to exceed the first number by a distance greater than
`$(1)/(4)$` unit, we need to choose a second number that falls into the intervals
`$((1)/(4), (1)/(2)]$`,
`$((1)/(2), (3)/(4)]$`, or
`$((3)/(4), 1]$`.

Let's analyze each interval separately:

- Interval
`$((1)/(4), (1)/(2)]$`: The length of this interval is
`$(1)/(2)-(1)/(4) = (1)/(4)$` unit.

- Interval
`$((1)/(2), (3)/(4)]$`: The length of this interval is
`$(3)/(4)-(1)/(2) = (1)/(4)$` unit.

- Interval
`$((3)/(4), 1]$`: The length of this interval is
`$1-(3)/(4) = (1)/(4)$` unit.

Since all three intervals have the same length of
`$(1)/(4)$` unit, the probability of choosing a second number within one of these intervals is
`$(1)/(4) + (1)/(4) + (1)/(4) = (3)/(4)$`.