STATISTICS HELP! Faith is a 95% free throw shooter. At practice, each player shoots 20 free throws. Let X = the number of free throws Faith makes out of 20 shots.

Question

Faith is a 95% free throw shooter. At practice, each player shoots 20 free throws. Let X =

the number of free throws Faith makes out of 20 shots.

Which of the following explains why this scenario represents a binomial setting.

A) success = made free throw, failure = missed free throw

B) each free throw is independent

C) n = 20

D) p = .95

E) all of the above

Answer 1

With the binomial theorem used in probability and statistics, each event (in this case -- a free throw) is an independent event and can be one of two possible outcomes (success or failure). With the probability of success defined as 95% we can let this represent p. This means the probability of failure is (1 - p) or 5%.

Based on this information the probability of X free throws of 20 successfully completed would be...

1) the number of combinations of making X free throws given 20 (_xC_r) [ the number of possible combinations of independent events)

2) the probability of success

3) the probability of failure

P(X,20) = ∑_i=1^X_xC₂₀ pⁱ(1-p)^20-i

So with this information (using a max of 1 free throw) ...

we see that the probability of shooting a single (1) free throw of 1 shooting is 0.95 as expected...

( ₁C₁(0.95)¹(0.05)^1-1 ) = 1 * 0.95 * 1 = 0.95

and the probability of failure is 0.05, as expected.

1 - ( ₁C₁(0.95)¹(0.05)^1-1 ) = 1 - (1 * 0.95 * 1) = 0.05

Now lets look at a max of 2 free throws to get some confidence in applying it to 20 free throws...

With 2 free throws the possibilities would be (S,S), (S,F), (F,S), (F,F) for a total of four combinations. If we want the number of combinations with at least 1 success than would be 3 of the four...

So our sum as written above is the probability of success both free throws in addition to the probability of success of one of the two free throws...

( ₁C₂(0.95)¹(0.05)^2-1 ) + ( ₂C₂(0.95)²(0.05)^2-2 ) = ( 2 * 0.95 * 0.05 ) + ( 1 * (0.95)² * 1 ) = 0.095 + 0.9025 = 0.9925.

So the high success rate tends to allow for a high rate based on the number of successful free throws...

So in this case of X out of 20, the sums runs over X and the result is the probability of making X free throws in various "orders"