The geometric distribution is related to the binomial distribution because it also consists of a set of trials where successes or failures can occur. Furthermore, each set is independent of the other, however, there is no exact number n of trials, rather the experiment is repeated until success is achieved.
All this explained means that:
P[X=1]=P[\text{success in the first session}]=p
P[X=2] = P[\text{failure in the first session and success in the second}]=
P[\text{failure in the first}]P[\text{success in the second}]=(1-p)(p)
By doing this, successively:
P[X = 3] =(1-p)(1-p)p
P[X = k] = (1-p)^{k - 1}p
The case when X = 2 or when X = 3, etc. it means that I want to be successful in trial number 2 or 3, therefore, it must have failed in the previous trials. If I want success in trial 2, it must have failed in the first trial, this is what explains the equation that was written above P[X = 2]. The same happens when X = 3, if I want success only in trial 3, it must have failed in the first trial and in the second trial. Everything will be clearer when we go to the examples.
Properties of the geometric distribution
Expected value:
E(X) = \cfrac{1}{p}
Variance:
Var(X) = \cfrac{1 - p}{p^{2}}
Cumulative distribution:
F(k) = 1 - q^{k}
If we want success in all trials, we will use the following formula:
P(k > r) = q^{r}
First example of the geometric distribution, solved
In a soccer tournament, “A Country” has a 60% probability of winning a match. “A Country” plays until lose.
- Find the probability that “A Country” plays at least 4 games.
For “A Country” play four matches, it has to win three matches, it does not matter if it loses or wins the fourth match because what is counted is that “A Country” will play four matches or more, but for it to be able to play four matches or more must have won yes or yes the previous three games. And the probability is calculated as follows:
P = P(k > 3) = (0.6)^{3} = 0.216
So the probability of “A Country” play at least four games is 0.216.
- Find the probability that “A Country” wins the tournament if there are 64 teams registered.
To win the tournament, you must win six matches. It is easy to know that you “A Country” has to win six games because there are 64 teams, so the first game is played when there are 64, the second game when 32 remain, the third when 16 remain, the fourth when 8 remain, the fifth when 4 remain and the sixth game is played when only 2 teams remain, therefore, the sixth game must also be won to win the tournament.
As the probability that “A Country” wins a match is 60%, the probability that it wins the tournament is calculated as follows:
P=(0.6)^{6} = 0.0467
Finally, the probability that “A Country” wins the tournament is 0.0467.
Second solved example of geometric distribution
When filming a television commercial, the probability is 0.30 that the “Actor” will say his lines correctly in any session.
- What is the probability that the “Actor” will say his lines correctly for the first time in the sixth session?
For the “Actor” get a sixth session, he had to have made a mistake in the previous 5 sessions. First you have to calculate the probability that he will NOT say his lines correctly, that is calculated by subtracting the probability that he will correctly say his lines from the integer:
1 - 0.3 = 0.7
Well, the probability that you don’t say your lines correctly is 0.7. It remains to calculate the probability that he was confused in the first 5 sessions, that is calculated like this:
(0.7)^{5} = 0.168
As we already calculated the probability that he is confused in the first 5 session, we need to calculate the probability that he will say his lines correctly in the sixth session, that is calculated by multiplying the probability that he has said his lines wrong the first 5 sessions by the probability that he says his line correctly in the sixth session:
(0.168)(0.3) = 0.0504
So the probability that the actor said his lines correctly in the sixth session is 0.0504.
- What is the average number of session it takes for the actor to say correctly his lines?
To determine the average number of session that will be required, you only have to apply the equation of the expectation of the geometric distribution:
E[X] = \cfrac{1}{p}
Where p is the probability that the session succeeds:
E[X] = \cfrac{1}{0.3} = \cfrac{10}{3} = 3.33
Therefore, the average number of session that will be required is 3.33, it would have to be left in integers because complete session are taken, so on average 3 sessions will be required.
Thank you for being in this moment with us : )
