fernando's soccer team played a good season. Finishing with 16 wins and 4 losses. Ties are not allowed. If winning or losing a game was equally likely, what is the probability that fernando's team ended up with this record?

My Answer:
C(20,16)(.5)^16(.5)^4
=.00462
=.46%

However, the book says that it is .83%. could someone please tell me what I did wrong. Thanks

The way the question is stated, your answer is correct.

I had the same result.

To solve this problem, you need to consider the number of possible arrangements of wins and losses that result in a 16-4 record, as well as the probability of each arrangement occurring.

In this case, you want to find the number of ways to arrange 16 wins and 4 losses in a season of 20 games. This can be calculated using the formula for combinations:

C(20, 16) = 20! / (16! * (20-16)!) = 4845

So there are 4,845 possible arrangements of wins and losses that would result in a 16-4 record.

Next, you need to determine the probability of each arrangement occurring. Since winning and losing a game are equally likely, the probability of winning a single game is 0.5 and the probability of losing a single game is also 0.5. Therefore, to determine the probability of a specific arrangement, you need to multiply the probabilities together.

For the arrangement with 16 wins and 4 losses, the probability is (0.5)^16 * (0.5)^4 = 0.5^(16+4) = 0.5^20.

Finally, to calculate the overall probability of ending up with this record, you need to multiply the number of arrangements by the probability of each arrangement occurring:

Probability = C(20, 16) * (0.5)^20 = 4845 * 0.5^20 = 0.0083 = 0.83%

So the correct answer is indeed 0.83%, not 0.46%. It seems you made an error in your calculations, possibly in the calculation of C(20, 16) or in raising the probabilities to the correct exponent.