# Thread: Trying to manipulate the equation Px=nCr(p)^x(q)^n-x, not going well.

1. ## Trying to manipulate the equation Px=nCr(p)^x(q)^n-x, not going well.

The probability of exactly 4 out of 10 people having a red ticket at a charity ball is approximately 0.2508. What is the probability, as a percentage, of a random person having a red ticket? Round your answer to the nearest percent. (Hint: red or not red)

The answer I got for this, using the equation Px=nCx(p)^x(q)^n-x, where n=10, x=4, and Px=0.2508 was a probability of approx. 0.5688 for the value of p(probability of success). I plugged my answer back into the equation where x=4 with a probability of success of approx. 0.5688 and received a Px value of approx. 14%, whereas in the question it is stated that it should be equal to 0.2508. Clearly, I have done something wrong but I can't see it. Any help would be greatly appreciated.

2. ## Re: Trying to manipulate the equation Px=nCr(p)^x(q)^n-x, not going well.

I think that you have simply made a mistake in solving the equation for p. I solved it using, in effect, a trial and error approach, the equation is

$210(1-p)^{6}p^{4}-0.2508=0$

and got a value of $p\approx0.4 .$