Question

David's chance of making a free throw is 80%. In a game, he takes 10 free throws. What is the probability he makes 8 or fewer free throws out of the 10? Please leave your answer in combinations, products powers and sums. Please try to make your expression reasonably succinct.

Answer 1

Answer:

$P(X\leq 8)= 1- P(X>8) = 1-P(X\geq 9)= 1-[P(X=9)+P(X=10)]$

And we can find the individual probabilities like this:

$P(X=9)=(10C9)(0.8)^{9} (1-0.8)^{10-9}$

$P(X=10)=(10C10)(0.8)^{10} (1-0.8)^{10-10}$

And replacing we got:

$P(X\leq 8)=1-[10(0.8)^1 (1-0.8)^{1} + (0.8)^{10}]$

Step-by-step explanation:

Previous concepts

A Bernoulli trial is "a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted". And this experiment is a particular case of the binomial experiment.

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:  

$P(X)=(nCx)(p)^x (1-p)^{n-x}$  

Where (nCx) means combinatory and it's given by this formula:  

$nCx=\frac{n!}{(n-x)! x!}$  

The complement rule is a theorem that provides a connection between the probability of an event and the probability of the complement of the event. Lat A the event of interest and A' the complement. The rule is defined by: $P(A)+P(A') =1$

Solution to the problem

For this case we want this probability:

$P(X\leq 8)$

And for this case we can use the comlement rule and we got:

$P(X\leq 8)= 1- P(X>8) = 1-P(X\geq 9)= 1-[P(X=9)+P(X=10)]$

And we can find the individual probabilities like this:

$P(X=9)=(10C9)(0.8)^9 (1-0.8)^{10-9}$

$P(X=10)=(10C10)(0.8)^{10} (1-0.8)^{10-10}$

And replacing we got:

$P(X\leq 8)=1-[10(0.8)^9 (1-0.8)^{1} + (0.8)^{10}]$