Probability of Guessing game outcome

[Destroxia]

Homework Statement

Mr. Keller filled out a bracket for the NCAA national tournament, based on his knowledge of college basketball, he has a .61 probability of guessing anyone game correctly.

What is the probability Mr. Keller will pick all 32 of the first round games correctly?

Homework Equations

Binomial distribution?

The Attempt at a Solution

This class is a calculator course, so I don't know any of the algebraic theory, but I try to plug it into my binomialpdf on my calculator and it's not coming out correctly, how would I calculate this?

 

[SammyS]

Homework Statement

Mr. Keller filled out a bracket for the NCAA national tournament, based on his knowledge of college basketball, he has a .61 probability of guessing anyone game correctly.

What is the probability Mr. Keller will pick all 32 of the first round games correctly?

Homework Equations

Binomial distribution?

The Attempt at a Solution

This class is a calculator course, so I don't know any of the algebraic theory, but I try to plug it into my binomialpdf on my calculator and it's not coming out correctly, how would I calculate this?

What buttons on your calculator have you been taught to press?

 

[RUber]

The binomial distribution looks something like:
if x is the number of correct guesses and p is the probability of a correct guess, then the probability of x correct guesses out of n tries P(x) can be written:
*edited, thank you to Ray for pointing it out*
$P(x) =\left( \begin{array}{c} n \\ x \end{array}\right) p^x(1-p)^{n-x}$
Your input for the calculator might be something like shown here where you input [n= number of trials, p=probability of correct, x = #correct].
If the output is anything close to correct, it would be the same as if you calculated the formula for P(x).
In this case, it should give something near $10^{-7}$.

 

[Ray Vickson]

The binomial distribution looks something like:
if x is the number of correct guesses and p is the probability of a correct guess, then the probability of x correct guesses out of n tries P(x) can be written:
$P(x) = p^x(1-p)^{n-x}$
Your input for the calculator might be something like shown here where you input [n= number of trials, p=probability of correct, x = #correct].
If the output is anything close to correct, it would be the same as if you calculated the formula for P(x).
In this case, it should give something near $10^{-7}$.

The formula above is wrong; it should be
$$P(x) = {n \choose x} p^x \, (1-p)^{n-x},$$
where ${n \choose x}$ is the binomial coefficient "n choose x".