You are at a stall at a fair where you have to throw a ball at a target. There are two versions of the game. In the first
version, you are given three attempts, and you estimate that your probability of success on any given throw is 0.1.
In the second version, you are given five attempts, but the target is smaller, and you estimate that your probability of
success on any given throw is 0.05. The prizes for the two versions of the game are the same, and you are willing to
assume that the outcomes of your throws are independent. Which version of the game should you choose? (Hint: In
the first version of the game, the probability that you do not get the prize is the probability that you fail on all three
attempts.)

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".

Solution to the problem

Alternative 1

Let X the random variable of interest, on this case we now that:

[tex]X \sim Binom(n=3, p=0.1)[/tex]

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

[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]

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

[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]

We can find the probability of loss like this P(X=0) and if we find this probability we got this:

[tex]P(X=0)=(3C0)(0.1)^0 (1-0.1)^{3-0}=0.729[/tex]

And the probability of loss with the first wersion is 0.729

Alternative 2

Let Y the random variable of interest, on this case we now that:

[tex]Y \sim Binom(n=5, p=0.05)[/tex]

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

[tex]P(Y)=(nCy)(p)^y (1-p)^{n-y}[/tex]

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

[tex]nCy=\frac{n!}{(n-y)! y!}[/tex]

We can find the probability of loss like this P(Y=0) and if we find this probability we got this:

[tex]P(Y=0)=(5C0)(0.05)^0 (1-0.05)^{5-0}=0.774[/tex]

And the probability of loss with the first wersion is 0.774

As we can see the best alternative is the first version since the probability of loss is lower than the probability of loss on version 2.