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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.

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.

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,

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

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" $$\begin{gathered} q=1-p=1鈭�\frac{1}{3} \\ 鈬�q=\frac{2}{3} \end{gathered}$$

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" $$\begin{gathered} P(A鈭�B)=P\left(A\right)P\left(B\right) \\ P(A鈭�B)=\frac{1}{3}脳\frac{2}{3} \\ P(A鈭�B)=\frac{2}{9} \\ P(A鈭�B)=0.22 \end{gathered}$$