Half of a floor

Probability Level 3

A number $x \in \mathbb R$ is chosen from the closed interval $[2,10]$ , the probability that $\left\lfloor \frac x2 \right\rfloor$ is even can be expressed as $\frac ab$ , where $a$ and $b$ are positive co-prime integers, find $a+b$

Details and Assumptions

$\left\lfloor x\right\rfloor$ is the floor function. $\left\lfloor x\right\rfloor$ Is the largest integer not greater than $x$ . For example: $\left\lfloor 3\right\rfloor = 3$ , $\left\lfloor 5.8\right\rfloor = 5$ , $\left\lfloor \pi \right\rfloor = 3$

The answer is 3.

2 solutions

Yan Yau Cheng
Apr 14, 2014

The length of the interval $[2,10]$ is 8 units. If $4\leq x <6$ or $8\leq x <10$ then $\left\lfloor\frac x2\right\rfloor$ is an even integer. The sum of the lengths of the interval $[4,6)$ and $[8,10)$ is 4.

By Geometric Probability, the probability that $\left\lfloor \frac x2 \right\rfloor$ is even is $\frac 48 = \frac 12$ . Therefore $a=1$ and $b=2$ , and $a+b = \boxed{3}$

But wait. $x$ is even when it belongs to $[4;6)$ (in, say, $p$ cases) or $[8;10)$ (in $p$ cases) and odd when it belongs to $[2;4)$ (in $p$ cases) or $[6;8)$ (in $p$ cases) or it is equal to $10$ (in one case). Therefore, the probability that it is even is $\frac{p}{2p+1}$ , which is not exactly $\frac{1}{2}$ . If the probability is indeed $\frac{1}{2}$ , I'd say that mathematics is weird and doesn't make sense in this case...

mathh mathh - 6 years, 11 months ago

hey "x" is a number not an integer so how can you take sum of lengths of interval =4

Gautam Sharma - 6 years, 10 months ago

reply awaited

Gautam Sharma - 6 years, 10 months ago

Kïñshük Sïñgh
Jul 11, 2014

There are total 8 ranges... X/2 belongs to [2,3), [3,4)...upto [9,10].....and from these... Only 4 will give even value.... So probability will be 4/8 = 1/2 a+b = 3

Half of a floor

The answer is 3.

2 solutions

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