Question

Find ? in each case.

Answer 1

Answer:

a) For this case we have that n = 30 and p =1/6

$X \sim Binom(n=30, p=1/6)$

b)  For this case we have that n = 10 and p =1/100

$X \sim Binom(n=10, p=1/100)$

c)  For this case we have that n = 20 and p =0.3

$X \sim Binom(n=20, p=0.3)$      

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:  

$nCx=\frac{n!}{(n-x)! x!}$  

Assuming the following question : "For each of the following binomial random variables specify n and p"  

Solution to the problem

a. A fair die is rolled 30 times. X = number of times a 6 is rolled.

For this case we have that n = 30 and p =1/6

$X \sim Binom(n=30, p=1/6)$

b. A company puts a game card in each box of cereal and 1/100 of them are winners. You buy ten boxes of cereal, and X = number of times you win.

For this case we have that n = 10 and p =1/100

$X \sim Binom(n=10, p=1/100)$

c. Jack likes to play computer solitaire and wins about 30% of the time. X = number of games he wins out of his next 20 games.

For this case we have that n = 20 and p =0.3

$X \sim Binom(n=20, p=0.3)$