## anonymous 3 years ago pete and sean decide to raise money for a charity by having a carnival in their backyard. in one of the games that they set up, the probability that a person will win is 0.4. if robyn plays that game nine times, what is the probability that she wins exactly four times? A)9C5^5 (0.4)^5 (0.4)^5 B)9C4 (0.5)^4 (0.5)^5 C) 9C4 (0.4)^4 (0.6)^5 D)9C5 (0.4)^5 (0.6)^4 and how did you get really confused on probability formula thx

1. anonymous

what does 9C5^5 mean?

2. anonymous

oh it's suppose to be 9C5 (0.4)^5 (0.4)^4 sorry

3. anonymous

the answer anyway is $(0,4)^4 * (0,6)^5 * \frac{ 9! }{ 4!5! }$ where (0.4) is the probability to win and (0.6) is 1 - 0.4 equals the probability to lose. Do you need more explanations why this is the answer?

4. anonymous

I guess 9C4 means $\left(\begin{matrix}9 \\ 4\end{matrix}\right) = \frac{ 9! }{ 4!5! }$ then answer C is equivalent to my solution

5. anonymous

why do you substract it by 1

6. anonymous

well, what does the probability of 1 mean? It means that the case you're looking at is going to happen EVERY time (100% of the tries). Let p be the probability to win, and q be the probability to lose. In this game we assume that anytime you play, you either win or lose. This OR translates into the language of probability as a '+', that means p+q = 1. Because when you look at all possible cases and add them , that is equivalent to calculating the probability of one of the possible cases happening - which happens 100% of the time, hence the probability is 1. In this example, you the probability to win (p) and the probability to lose (q) are all the possible cases, hence p + q = 1 ==> q = 1-p

7. anonymous

okay what do the exponents represents

8. anonymous

They're kind of the translation of 'AND' into the language of probability. For example, if you tried once to win the game you had a probability of 0.4. If you tried twice, the probability to win twice would be (0.4) * (0.4) = (0.4)^2. The probability to play 9 times and win 9 games would be (0.4)^9 Similarly, if you played 5 times and lost all 5 games the probability for that would be (0.6)^5 Can you see the scheme here?

9. anonymous

yeah! okay that pretty much sums most of my confusion thank you so much!