Question

Suppose that you are performing the probability experiment of rolling one fair six-sided die. Let F be the event of rolling a four or a five. You are interested in how many times you need to roll the die in order to obtain the first four or five as the outcome. • p = probability of success (event F occurs) • q = probability of failure (event F does not occur)

a. Write the description of the random variable X.

b. What are the values that X can take on?

c. Find the values of p and q.

d. Find the probability that the first occurrence of event F (rolling a four or five) is on the second trial.

Short answer

a. The random variable X is the number of times we need to roll the die in order to obtain the face four or five.

b. X can take on the values as $X=1,2,3,.....$

c. The values of localid="1651910562367" $p=\frac{1}{3}$and localid="1651910594587" $q=\frac{2}{3}$

d. The probability that the first occurrence of event F is on the second trial is$0.22$

Step-by-step solution

Content Introduction

In a Bernoulli trial, the likelihood of the number of successive failures before a success is obtained is represented by a geometric distribution, which is a sort of discrete probability distribution. A Bernoulli trial is a test that can only have one of two outcomes: success or failure.

Explanation (part a)

Random variable in simple terms generally refers to variables whose values are unknown, therefore, in this case X is the number of times we need to roll the die in order to obtain the face four or five.

Explanation (part b)

Make the list of values that you want to use X may take on.

As we can see there is an upper bound for the situation at hand so,

$X=1,2,3,..........$

Explanation (part c)

We know,

p = probability of success (event F occurs)

q = probability of failure (event F does not occur)

Probability of getting$4$or$5$is,

$\begin{array}{r}P\left(4or5\right)=P\left(4\right)+P\left(5\right)\\ P\left(4\text{or}5\right)=\frac{1}{6}+\frac{1}{6}=\frac{1}{3}\end{array}$

The values are, role="math" localid="1651910849783" $p=\frac{1}{3}$

role="math" localid="1651910864409" $$\begin{gathered} q=1-p=1-\frac{1}{3} \\ \Rightarrow q=\frac{2}{3} \end{gathered}$$

Explanation (part d)

The first trial needs to be a failure which means it is two-thirds and the second trial needs to be a success which is one-third.

Let us assume $A$be the first trial and$B$be the second trial

Therefore, the probability that the first occurrence of event F is on the second trial is localid="1651910901930" $$\begin{gathered} P(A\cap B)=P\left(A\right)P\left(B\right) \\ P(A\cap B)=\frac{1}{3}\times \frac{2}{3} \\ P(A\cap B)=\frac{2}{9} \\ P(A\cap B)=0.22 \end{gathered}$$