# Second and Final Probability question.

• Jun 15th 2009, 09:43 AM
Kasper
Second and Final Probability question.
For this question, I tried it using $P(x success)=_{n}C_{k}P^k(1-P)^{n-k}$, but It doesn't feel right. and I'm not getting the correct answer, I think I just need a push in the right direction on this one.

Quote:

The probability of exactly 4 out of 10 people having a red ticket at a charity ball is aproximately 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.)
Am I supposed to set $P(xsuccess)=0.2508$ and solve backwards? I tried plugging $0.2508$ as P in the above formula and solving, but that was more of a wild shot.

I'm not sure where to go with this one. =/ Any help would be appreciated!
• Jun 16th 2009, 11:56 PM
Probability
Hello Kasper
Quote:

Originally Posted by Kasper
For this question, I tried it using $P(x success)=_{n}C_{k}P^k(1-P)^{n-k}$, but It doesn't feel right. and I'm not getting the correct answer, I think I just need a push in the right direction on this one.

Am I supposed to set $P(xsuccess)=0.2508$ and solve backwards? I tried plugging $0.2508$ as P in the above formula and solving, but that was more of a wild shot.

I'm not sure where to go with this one. =/ Any help would be appreciated!

Your approach is the only one I can think of. But you get an equation that looks pretty well impossible to solve: $\binom{10}{4}p^4(1-p)^6 = 0.2508$.

I have got an answer, though. It's $p = 0.4$. But I cheated, and used an Excel spreadsheet!

• Jun 17th 2009, 04:20 AM
mr fantastic
Quote:

Hello KasperYour approach is the only one I can think of. But you get an equation that looks pretty well impossible to solve: $\binom{10}{4}p^4(1-p)^6 = 0.2508$.
I have got an answer, though. It's $p = 0.4$. But I cheated, and used an Excel spreadsheet!