What is the value of K in the equation 21K = 1?

[Jason76]

What is the value of K?

$$K(x^{6}) = 1$$

 

[MarkFL]

Suppose $x\ne0$ and you divide both sides by $x^6$...what do you get?

 

[Jason76]

Suppose $x\ne0$ and you divide both sides by $x^6$...what do you get?

$$K(x^{6}) = 1$$

$$\dfrac{K(x^{6})}{x^{6}} = \dfrac{1}{x^{6}}$$ where $$x \ne 0$$

$$K = \dfrac{1}{x^{6}}$$

Actually, here is the original problem:

A die has its six faces loaded so that P(roll is i)=K*x for x=1,2,3,4,5,6. It is rolled until an even number appears. Let X be the number of rolls needed.

Next finding K is needed. Sorry if in wrong section. This is for probability.

 

[MarkFL]

I've moved the thread, and will wait until someone more proficient at probability to chime in. :D

 

[Jason76]

I've moved the thread, and will wait until someone more proficient at probability to chime in. :D

It seems like a long time ago, the professor said something about differentiation and also used Wolfram Alpha.

 

[I like Serena]

Hi Jason,

I take it that should be P(roll is i)=K*i?
If so, K must be such that all chances sum to 1.
So:
$$K\cdot 1 + ... + K\cdot 6 = 1$$
Can we find K from that?

 

[MarkFL]

...I take it that should be P(roll is i)=K*i?...

I also thought that might be the intended problem. :D

 

[Jason76]

Contacted the professor.

$$K + 2K + 3K + 4K + 5K + 6K = 1$$

is the correct format so

$$21K = 1$$

so

$$K = \dfrac{1}{21}$$