the probability of exactly 4 out of 10 people having a red ticket is 0.2508. what is the probability as a percentage,of a random person having a red ticket?
I'll assume the odds of 1 person having a red ticket is independent of others having a red ticket so that we can use the binomial formula to know that $\displaystyle .2508=\binom{10}{4}p^4(1-p)^6$ solve for p to get your answer...
oddly (for this type of question), you'll get two answers that are plausible, .397919 and .402083 so idk, interpret as you will, the answers are close enough where I would just say p=.4
Well lets consider this:
Say the probability of a person have a red ticket is "p"
Then what is the probability that out of a group of 10 randomly selected people, four people have red tickets?
The answer is $\displaystyle \binom{10}{4}p^4(1-p)^6$
So use this for your question in reverse.
If the question stated the the probability of 4 out of 10 is .250822656, then we would get p=.4, so the answer is definitely p=.4, the problem is that the decimal they give you isn't exact enough