A dreidel is a four-sided spinning top with the Hebrew letters nun, gimel, hei, and shin, one on each side. Each side is equally likely to come up in a single spin of the dreidel. Suppose you spin a dreidel three times. Calculate the probability of getting: (a) at least one nun? (b) exactly 2 nuns? (c) exactly 1 hei? (d) at most 2 gimels?

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

Let X the random variable of interest (number of times that appears an specific letters in n trials), on this case we now that:

[tex]X \sim Binom(n=3, p=0.25)[/tex]

The probability is 0.25 because for any letter assuming that the dreidel is fair, we have 1 possible outcome for each letter and 4 total possibilities , and 1/4=0.25. So then p =0.25 for any type of letter used.

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]

Part a

For this case we want this probability:

[tex] P(X \geq 1)[/tex]

If we use the complement rule we have this:

[tex] P(X \geq 1) = 1-P(X<1) = 1-P(X=0)[/tex]

And we can find the probability using the mass function:

[tex]P(X=0)=(3C0)(0.25)^0 (1-0.25)^{3-0}=0.422[/tex]

[tex] P(X \geq 1) = 1-0.422=0.578[/tex]

Part b

For this case we want this probability:

[tex] P(X =2)[/tex]

And we can find the probability using the mass function:

[tex]P(X=2)=(3C2)(0.25)^2 (1-0.25)^{3-2}=0.141[/tex]

Part c

For this case we want this probability:

[tex] P(X =1)[/tex]

And we can find the probability using the mass function:

[tex]P(X=1)=(3C1)(0.25)^1 (1-0.25)^{3-1}=0.422[/tex]

Part d

For this case we want this probability:

[tex] P(X \leq 2)= P(X=0) +P(X=1) +P(X=2)[/tex]

And we can find the individual probabilities using the mass function:

[tex]P(X=0)=(3C0)(0.25)^0 (1-0.25)^{3-0}=0.422[/tex]

[tex]P(X=1)=(3C1)(0.25)^1 (1-0.25)^{3-1}=0.422[/tex]

[tex]P(X=2)=(3C2)(0.25)^2 (1-0.25)^{3-2}=0.141[/tex]

[tex] P(X \leq 2)= P(X=0) +P(X=1) +P(X=2)=0.422+0.422+0.141=0.985[/tex]